# Haar measure on locally compact groups *Written by Claude Opus 5.5 (Anthropic), September 2026. Spot-checked by Claude Opus 5.5 in a separate session. Public domain (CC0).* Source-convention repair and new Proposition 13.1A by GPT-6.1 Sol (OpenAI), Ultra, 5 October 2026. The independently written AI exposition remains CC0. Human sources are credited where compared; no source text or translation is reproduced. This lesson constructs Haar measure on an arbitrary locally compact group and develops the measure theory that goes with it. Sections 2–6 work on locally compact spaces. They prove the Riesz representation theorem for Radon measures, treat image measures and densities, build the product of two Radon measures with its Tonelli and Fubini theorems, and identify \(L^2\) of a product with a Hilbert tensor product. Sections 7–11 turn to groups: topological groups, existence and uniqueness of Haar measure, the modular function, and the inversion formula. Sections 12–15 treat products of groups, groups that are not \(\sigma\)-compact, convolution, and approximate identities. Section 16 has exercises with solutions. No countability assumption is made anywhere. The group need not be \(\sigma\)-compact, second countable or unimodular, and its Haar measure need not be \(\sigma\)-finite. This generality needs care: a Borel set of infinite measure can have only null compact subsets, and Fubini's theorem can fail. Section 13 shows what goes wrong, and the rest of the lesson is arranged so that it does no harm. Haar measure makes a locally compact group \(G\) into a measure space on which \(G\) acts by translations. Then \(L^1(G)\) is a Banach \(*\)-algebra under convolution, and \(L^2(G)\) carries the left and right regular representations. These objects are the starting point of abstract harmonic analysis, and of the group von Neumann algebras studied in the course on modular theory and weights. The lesson assumes measure theory (outer measures, the convergence theorems, \(L^p\) spaces, and the Radon–Nikodym theorem for \(\sigma\)-finite measures), point-set topology, and Hilbert spaces. The facts used without proof are stated at the end, under "Background used without proof". One of them, on multiplication operators, is proved in the lesson Decomposable operators and the diagonal algebra. One example uses the Stone–Weierstrass theorem. The free companion [Tornier, §§1–3] uses the same Borel regularity and modular-function conventions as this lesson. It states general Haar existence and uniqueness without proving them; Sections 8 and 9 below give complete proofs. [Fremlin, 441C–441E and 442B] supplies a different construction through invariant compact-set contents and a proof of uniqueness. His completed, locally determined measure convention must be distinguished from the outer-regular Borel convention here; Proposition 13.1A makes the comparison explicitly. These comparisons do not impose a countability hypothesis. ## 1. Conventions *Spaces.* An *LCH space* is a Hausdorff space that is locally compact. Its compact subsets are closed. \(C_c(X)\) is the space of continuous complex functions with compact support, \(C_c^+(X)\) is the set of \(f\in C_c(X)\) with \(f\geq0\) and \(f\neq0\), and \(\|f\|_{\sup}=\sup_x|f(x)|\). For an open set \(U\) we write \(f\prec U\) when \(f\in C_c(X)\), \(0\leq f\leq1\) and \(\operatorname{supp}f\subseteq U\). For a compact set \(K\) we write \(K\prec f\) when \(f\in C_c(X)\), \(0\leq f\leq 1\) and \(f=1\) on \(K\). A function \(h\colon X\to[0,\infty]\) is *lower semicontinuous* (lsc) if every set \(\{h>c\}\) is open; lsc functions are Borel. *Four tools from topology.* The proofs rest on two standard facts, recalled under "Background used without proof": every point of an LCH space has a base of compact neighbourhoods, and Tietze's extension theorem holds for compact subsets of an LCH space. From them we derive four tools. - **(T1)** (*Shrinking*) If \(K\subseteq U\) with \(K\) compact and \(U\) open, there is an open \(V\) with compact closure such that \(K\subseteq V\subseteq\overline V\subseteq U\). *Proof.* Since compact neighbourhoods form a base, each point of \(K\) has a compact neighbourhood inside \(U\). Finitely many of their interiors cover \(K\); let \(V\) be the union of these interiors. Then \(\overline V\) lies in the union of the finitely many compact neighbourhoods, which is compact and contained in \(U\). \(\square\) - **(T2)** (*Urysohn's lemma*) If \(K\subseteq U\) as in (T1), there is \(f\) with \(K\prec f\prec U\). *Proof.* Take \(V\) from (T1). The boundary \(\partial V=\overline V\setminus V\) is compact and misses \(K\). The function that is \(1\) on \(K\) and \(0\) on \(\partial V\) is continuous on the compact set \(K\cup\partial V\), so Tietze's extension theorem, applied with the arc \([0,1]\), extends it to a continuous \(g\colon X\to[0,1]\). Let \(f=g\) on \(\overline V\) and \(f=0\) on \(X\setminus V\). These two closed sets cover \(X\), and the two definitions agree on their intersection \(\partial V\). So \(f\) is continuous, \(f=1\) on \(K\), and \(\operatorname{supp}f\subseteq\overline V\subseteq U\). \(\square\) - **(T3)** (*Partitions of unity*) If \(K\) is compact and \(K\subseteq U_1\cup\dots\cup U_n\) with each \(U_i\) open, there are \(h_i\prec U_i\) with \(\sum_ih_i=1\) on \(K\) and \(\sum_ih_i\leq1\) everywhere. *Proof.* For each \(x\in K\) pick an index \(i\) with \(x\in U_i\), and by (T1) an open \(V_x\ni x\) whose compact closure lies in \(U_i\). Finitely many \(V_x\) cover \(K\). Let \(F_i\) be the union of the closures of the chosen \(V_x\) that were assigned to \(i\). Then \(F_i\) is compact, \(F_i\subseteq U_i\), and \(K\subseteq\bigcup_iF_i\). Choose \(F_i\prec g_i\prec U_i\) by (T2) and set \(h_1=g_1\) and \(h_k=(1-g_1)\cdots(1-g_{k-1})g_k\). By induction, \(\sum_{k\leq m}h_k=1-\prod_{k\leq m}(1-g_k)\). This lies in \([0,1]\), and it equals \(1\) on each \(F_i\). \(\square\) - **(T4)** (*Tube lemma*) If \(K\subseteq X\) and \(L\subseteq Y\) are compact and \(K\times L\subseteq W\) with \(W\) open in \(X\times Y\), there are open \(U\supseteq K\) and \(V\supseteq L\) with \(U\times V\subseteq W\). *Proof.* Fix \(x\in K\). Each \((x,y)\), \(y\in L\), lies in an open box inside \(W\); finitely many boxes \(P_j\times Q_j\) of this kind have \(Q_j\) covering \(L\). Put \(P^x=\bigcap_jP_j\) and \(Q^x=\bigcup_jQ_j\); then \(P^x\times Q^x\subseteq W\). Cover \(K\) by finitely many \(P^{x_i}\), and let \(U=\bigcup_iP^{x_i}\) and \(V=\bigcap_iQ^{x_i}\). \(\square\) *Measures and functions.* A measure is positive and countably additive. \(\mathcal B(X)\) denotes the Borel sets of \(X\), the \(\sigma\)-algebra generated by the open sets; functions on \(X\) are Borel unless said otherwise. A Borel set is *\(\sigma\)-finite* for a measure when countably many Borel sets of finite measure cover it. For \(1\leq p<\infty\), \(L^p(\mu)\) consists of the Borel functions with \(\int|f|^p\,d\mu<\infty\), modulo functions that vanish almost everywhere. Hilbert spaces are complex, and the inner product \(\langle\xi,\eta\rangle=\int\xi\bar\eta\,d\mu\) is linear in \(\xi\). For a Borel \(f\colon X\to[0,\infty]\) we use the *layer functions* \[ s_n=2^{-n}\sum_{i=1}^{n2^n}1_{\{f>i2^{-n}\}} . \tag{1.1} \] They are simple Borel functions, and they vanish where \(f\) does. They increase, \(s_n\leq s_{n+1}\), because the \(i\)-th term of \(s_n\) is at most the sum of the terms of \(s_{n+1}\) with indices \(2i-1\) and \(2i\). And \(s_n(x)\to f(x)\) at every point: if \(f(x)0\), choose \(f\prec V\cap U\) with \(I(f)>\mu(V\cap U)-\varepsilon\). The set \(V\setminus\operatorname{supp}f\) is open; choose \(g\prec V\setminus\operatorname{supp}f\) with \(I(g)>\mu(V\setminus\operatorname{supp}f)-\varepsilon\). The supports of \(f\) and \(g\) are disjoint, so \(f+g\prec V\) and \[ \mu(V)\geq I(f+g)>\mu(V\cap U)+\mu(V\setminus\operatorname{supp}f)-2\varepsilon\geq\mu^*(V\cap U)+\mu^*(V\setminus U)-2\varepsilon , \] since \(V\setminus U\subseteq V\setminus\operatorname{supp}f\). For general \(A\), take an open \(V\supseteq A\) with \(\mu(V)\leq\mu^*(A)+\varepsilon\), apply the open case, and use monotonicity. By Carathéodory's theorem the \(\mu^*\)-measurable sets form a \(\sigma\)-algebra on which \(\mu^*\) is countably additive. It contains the open sets, hence \(\mathcal B(X)\). Let \(\mu\) be the restriction of \(\mu^*\) to \(\mathcal B(X)\). It satisfies (R2) by construction. *Step 4: compact sets.* Let \(K\) be compact and \(1_K\leq f\in C_c(X)\); in particular \(f\geq0\). For \(0c\}\) contains \(K\), and every \(g\prec U_c\) satisfies \(g\leq f/c\). So \(\mu(K)\leq\mu(U_c)\leq I(f)/c\), and letting \(c\to1\) gives \(\mu(K)\leq I(f)\). Some \(f\) with \(K\prec f\) exists by (T2) with \(U=X\), so \(\mu(K)<\infty\), which is (R1). Given \(\varepsilon>0\), choose an open \(U\supseteq K\) with \(\mu(U)\leq\mu(K)+\varepsilon\) and then \(f\) with \(K\prec f\prec U\): \(I(f)\leq\mu(U)\leq\mu(K)+\varepsilon\). This proves the second formula of (2.1). *Step 5: (R3).* If \(f\prec U\), then \(f\prec V\) for every open \(V\supseteq\operatorname{supp}f\), so \(I(f)\leq\mu(V)\), and \(I(f)\leq\mu(\operatorname{supp}f)\) by (R2). Taking the supremum over \(f\prec U\) gives \(\mu(U)\leq\sup\{\mu(K):K\subseteq U\ \text{compact}\}\). The reverse inequality is monotonicity. *Step 6: \(\mu\) represents \(I\).* By linearity it suffices to treat real \(f\in C_c(X)\) with \(0\leq f\leq1\). Fix \(N\geq1\). Put \(K_0=\operatorname{supp}f\), \(K_j=\{f\geq j/N\}\) for \(1\leq j\leq N\) (compact sets), and \(f_j=\min(\max(f-\frac{j-1}N,0),\frac1N)\). Then \(f=\sum_{j=1}^Nf_j\), each \(f_j\) is in \(C_c(X)\) with \(\operatorname{supp}f_j\subseteq K_{j-1}\), and \(\frac1N1_{K_j}\leq f_j\leq\frac1N1_{K_{j-1}}\). Integrating, \(\frac1N\mu(K_j)\leq\int f_j\,d\mu\leq\frac1N\mu(K_{j-1})\). On the other side, \(\mu(K_j)\leq I(Nf_j)\) by Step 4. Also \(Nf_j\prec U\) for every open \(U\supseteq K_{j-1}\), so \(I(Nf_j)\leq\mu(U)\), and \(I(Nf_j)\leq\mu(K_{j-1})\) by (R2). Summing over \(j\), both \(I(f)\) and \(\int f\,d\mu\) lie between \(\frac1N\sum_{j=1}^N\mu(K_j)\) and \(\frac1N\sum_{j=0}^{N-1}\mu(K_j)\). These bounds differ by at most \(\mu(\operatorname{supp}f)/N\). Let \(N\to\infty\). \(\square\) **Proposition 2.3** (Regularity on \(\sigma\)-finite sets). Every Radon measure \(\mu\) on \(X\) has the following properties. 1. If \(E\) is Borel and \(\mu(E)<\infty\), then \(\mu(E)=\sup\{\mu(K):K\subseteq E\ \text{compact}\}\). 2. The same holds for every Borel set that is \(\sigma\)-finite for \(\mu\). 3. Every \(\sigma\)-finite Borel set \(E\) is the union of a \(\sigma\)-compact set and a Borel null set. **Proof.** (1) Let \(\varepsilon>0\). By (R2) choose an open \(U\supseteq E\) with \(\mu(U)<\mu(E)+\varepsilon\), so \(\mu(U\setminus E)<\varepsilon\). Again by (R2) choose an open \(W\supseteq U\setminus E\) with \(\mu(W)<\varepsilon\). By (R3) choose a compact \(F\subseteq U\) with \(\mu(F)>\mu(U)-\varepsilon\). The compact set \(K=F\setminus W\) lies in \(E\): a point of \(F\) outside \(E\) is in \(U\setminus E\subseteq W\). And \(\mu(K)\geq\mu(F)-\mu(W)>\mu(E)-2\varepsilon\). (2) Write \(E\) as an increasing union of Borel sets \(E_n\) of finite measure; then \(\mu(E)=\lim\mu(E_n)\), and (1) applies to each \(E_n\). (3) With \(E_n\) as in (2), choose compact \(K_{n,m}\subseteq E_n\) with \(\mu(E_n\setminus K_{n,m})<1/m\). The \(\sigma\)-compact set \(S=\bigcup_{n,m}K_{n,m}\) lies in \(E\), and \(\mu(E_n\setminus S)=0\) for every \(n\), so \(\mu(E\setminus S)=0\). \(\square\) *Where the hypotheses enter.* Local compactness and the Hausdorff property are used only through (T1)–(T3): they supply enough functions \(f\prec U\). No countability is used, and the measure \(\mu\) need not be \(\sigma\)-finite. A measure that is inner and outer regular on every Borel set would be more convenient, but Example 13.4 shows that on the group \(\mathbb R\times\mathbb R_d\) no measure of that kind represents Haar integration. So the lesson never uses inner regularity on arbitrary Borel sets. ## 3. Image measures, densities, and approximation by continuous functions This section collects four operations on a Radon measure that the rest of the lesson uses constantly: moving it by a homeomorphism, multiplying it by a positive continuous density, integrating lower semicontinuous functions, and approximating \(L^p\) functions by continuous ones. **Proposition 3.1.** Fix a Radon measure \(\mu\) on an LCH space \(X\). 1. (*Homeomorphisms*) Let \(\theta\colon X\to X'\) be a homeomorphism onto an LCH space. Then \(\theta_*\mu(E)=\mu(\theta^{-1}(E))\) defines a Radon measure on \(X'\), and \(\int f\,d\theta_*\mu=\int f\circ\theta\,d\mu\) for every Borel \(f\geq0\) and every \(f\in L^1(\theta_*\mu)\). 2. (*Positive continuous densities*) Let \(\varphi\colon X\to(0,\infty)\) be continuous. Then \(\varphi\mu(E)=\int_E\varphi\,d\mu\) defines a Radon measure; \(\int f\,d(\varphi\mu)=\int f\varphi\,d\mu\) for every Borel \(f\geq0\); and \(\varphi\mu\) has the same null sets and the same \(\sigma\)-finite sets as \(\mu\). Consequently, if a Radon measure \(\nu\) satisfies \(\int f\,d\nu=\int f\varphi\,d\mu\) for all \(f\in C_c(X)\), then \(\nu=\varphi\mu\). 3. (*Lower semicontinuous functions*) For every lsc \(h\colon X\to[0,\infty]\), \[ \int h\,d\mu=\sup\Big\{\int g\,d\mu:g\in C_c(X),\ 0\leq g\leq h\Big\}. \tag{3.1} \] 4. (*Density*) Let \(1\leq p<\infty\). Every \(f\in L^p(\mu)\) vanishes outside a \(\sigma\)-finite Borel set, and hence, after a change on a null set, outside a \(\sigma\)-compact set. The space \(C_c(X)\) is dense in \(L^p(\mu)\). **Proof.** (1) Since \(\theta\) is a homeomorphism, \(\theta^{-1}\) maps compact, open and Borel sets of \(X'\) onto sets of the same kind in \(X\), and it preserves inclusions. So (R1)–(R3) transfer from \(\mu\) to \(\theta_*\mu\). The integral formula holds for indicators by definition, and by linearity for simple functions. The layer functions (1.1) and monotone convergence extend it to Borel \(f\geq0\), and splitting \(f\) into four nonnegative parts extends it to integrable \(f\). (2) The formula for integrals holds for indicators by definition, hence for Borel \(f\geq0\) by the layer functions (1.1) and monotone convergence; in particular \(\varphi\mu\) is countably additive, so it is a measure. Since \(\varphi>0\), \(\int_E\varphi\,d\mu=0\) exactly when \(\mu(E)=0\). The sets \(\{1/n\leq\varphi\leq n\}\) cover \(X\), and on each of them the two measures are comparable, so the \(\sigma\)-finite sets agree. It remains to check (R1)–(R3) for \(\varphi\mu\). (R1) holds because \(\varphi\) is bounded on compact sets. (R2): Let \(E\) be Borel with \(\varphi\mu(E)<\infty\) (otherwise take \(U=X\)), and let \(\varepsilon>0\). For \(j\in\mathbb Z\), the open sets \(V_j=\{2^{j-1}<\varphi<2^{j+1}\}\) cover \(X\). Put \(E_j=E\cap V_j\). Then \(\mu(E_j)\leq2^{1-j}\varphi\mu(E_j)<\infty\), so by (R2) for \(\mu\) there is an open \(U_j\) with \(E_j\subseteq U_j\subseteq V_j\) and \(\mu(U_j\setminus E_j)<\varepsilon2^{-|j|-j-3}\). Then \(\varphi\mu(U_j\setminus E_j)\leq2^{j+1}\mu(U_j\setminus E_j)<\varepsilon2^{-|j|-2}\). The set \(U=\bigcup_jU_j\) is open, contains \(E\), and \(U\setminus E\subseteq\bigcup_j(U_j\setminus E_j)\). So \(\varphi\mu(U\setminus E)<\varepsilon\sum_j2^{-|j|-2}<\varepsilon\). (R3): Let \(U\) be open and \(c<\varphi\mu(U)\). If \(\mu(U\cap\{\varphi>s\})=\infty\) for some \(s>0\), then (R3) for \(\mu\) gives a compact \(K\subseteq U\cap\{\varphi>s\}\) with \(\mu(K)>c/s\), and \(\varphi\mu(K)\geq s\mu(K)>c\). Otherwise all these sets have finite \(\mu\)-measure. Put \(D_j=U\cap\{2^j<\varphi\leq2^{j+1}\}\) for \(j\in\mathbb Z\). These Borel sets are disjoint, cover \(U\), and have finite \(\mu\)-measure. Choose a finite set \(J\subseteq\mathbb Z\) with \(\sum_{j\in J}\varphi\mu(D_j)>c+\eta\) for some \(\eta>0\). By inner regularity on sets of finite measure (Proposition 2.3(1)), choose compact \(K_j\subseteq D_j\) with \(2^{j+1}\mu(D_j\setminus K_j)<\eta/|J|\). The compact set \(K=\bigcup_{j\in J}K_j\subseteq U\) satisfies \(\varphi\mu(K)\geq\sum_{j\in J}\big(\varphi\mu(D_j)-2^{j+1}\mu(D_j\setminus K_j)\big)>c\). The last sentence of (2) follows from the uniqueness in the Riesz theorem 2.2. (3) The inequality \(\geq\) is clear. For \(\leq\), let \(c<\int h\,d\mu\). The layer functions (1.1) of \(h\) have the form \(\sum_i(t_i-t_{i-1})1_{\{h>t_i\}}\) with \(t_i=i2^{-n}\), and their integrals tend to \(\int h\,d\mu\) by monotone convergence. So there is a partition \(0=t_0t_i\})>c\). Each set \(\{h>t_i\}\) is open, so (R3) gives compact \(K_i\subseteq\{h>t_i\}\) with \(\sum_i(t_i-t_{i-1})\mu(K_i)>c\). Choose \(K_i\prec\varphi_i\prec\{h>t_i\}\) by (T2) and put \(g=\sum_i(t_i-t_{i-1})\varphi_i\in C_c(X)\). At a point \(x\), \(\varphi_i(x)\neq0\) forces \(h(x)>t_i\), so \(g(x)\) is at most the sum of \(t_i-t_{i-1}\) over the indices \(i\) with \(t_ic\). (4) The set \(\{f\neq0\}\) is the union of the sets \(\{|f|>1/n\}\), and by Chebyshev's inequality each has measure at most \(n^p\|f\|_p^p\). So \(\{f\neq0\}\) is \(\sigma\)-finite, and Proposition 2.3(3) gives the \(\sigma\)-compact set. Simple functions are dense in \(L^p(\mu)\), and a simple function in \(L^p\) is a combination of indicators of Borel sets of finite measure. So it suffices to approximate \(1_E\) with \(\mu(E)<\infty\). Given \(\varepsilon>0\), inner regularity (Proposition 2.3(1)) and outer regularity (R2) give a compact \(K\) and an open \(U\) with \(K\subseteq E\subseteq U\) and \(\mu(U\setminus K)<\varepsilon\). Take \(K\prec f\prec U\). Then \(|f-1_E|\leq1_{U\setminus K}\), so \(\|f-1_E\|_p\leq\varepsilon^{1/p}\). \(\square\) Part (4) needs no \(\sigma\)-finiteness of \(\mu\): an \(L^p\) function with \(p<\infty\) automatically lives on a \(\sigma\)-finite set, and there inner regularity holds. For a Haar measure (Section 8) and \(p=2\), it says that \(C_c(G)\) is dense in \(L^2(G)\). In (2) the density must be strictly positive; Exercise 16.1 shows what goes wrong with a density that vanishes on a closed set. ## 4. Products of Radon measures Let \(X,Y\) be LCH spaces with Radon measures \(\mu,\nu\). For \(F\in C_c(X\times Y)\) write \(F_x=F(x,\cdot)\), and let \(K_F\) and \(L_F\) be the projections of \(\operatorname{supp}F\) to \(X\) and \(Y\); they are compact. The product of \(\mu\) and \(\nu\) is built from iterated integrals of such \(F\), and the first step is to approximate \(F\) by sums of products. **Lemma 4.1** (Tensor approximation). Let \(F\in C_c(X\times Y)\). 1. The map \(x\mapsto F_x\) is continuous from \(X\) to \(C_c(Y)\) with the norm \(\|\cdot\|_{\sup}\), and \(F_x\) vanishes outside \(L_F\). 2. Let \(U\supseteq K_F\) be open with compact closure. For every \(\varepsilon>0\) there are \(\varphi_1,\dots,\varphi_n\prec U\) and \(g_1,\dots,g_n\in C_c(Y)\) vanishing outside \(L_F\) such that \(|F(x,y)-\sum_i\varphi_i(x)g_i(y)|\leq\varepsilon\) for all \((x,y)\). **Proof.** (1) Fix \(x_0\) and \(\varepsilon>0\). For each \(y\in L_F\), continuity of \(F\) at \((x_0,y)\) gives open \(P_y\ni x_0\) and \(Q_y\ni y\) with \(|F(x,y')-F(x_0,y)|<\varepsilon/4\) on \(P_y\times Q_y\). Finitely many \(Q_{y_k}\) cover \(L_F\); let \(P=\bigcap_kP_{y_k}\). For \(x\in P\) and \(y'\in L_F\), pick \(k\) with \(y'\in Q_{y_k}\); comparing both \(F(x,y')\) and \(F(x_0,y')\) with \(F(x_0,y_k)\) gives \(|F(x,y')-F(x_0,y')|<\varepsilon/2\). Off \(L_F\) both values are \(0\). (2) By (1), each \(x\in K_F\) has an open neighbourhood \(P_x\subseteq U\) with \(\|F_{x'}-F_x\|_{\sup}<\varepsilon\) for \(x'\in P_x\). Cover \(K_F\) by \(P_{x_1},\dots,P_{x_n}\), take a partition of unity \(\varphi_i\prec P_{x_i}\) from (T3), and let \(g_i=F_{x_i}\). Then \[ F(x,y)-\sum_i\varphi_i(x)F(x_i,y)=\Big(1-\sum_i\varphi_i(x)\Big)F(x,y)+\sum_i\varphi_i(x)\big(F(x,y)-F(x_i,y)\big). \] The first term vanishes: \(\sum_i\varphi_i=1\) on \(K_F\), and \(F_x=0\) off \(K_F\). In the second, \(\varphi_i(x)\neq0\) forces \(x\in P_{x_i}\), so the term is at most \(\varepsilon\sum_i\varphi_i(x)\leq\varepsilon\) in absolute value. \(\square\) **Proposition 4.2** (Iterated integrals). Let \(F\in C_c(X\times Y)\). 1. The function \(x\mapsto\int F(x,y)\,d\nu(y)\) is in \(C_c(X)\) and vanishes off \(K_F\); similarly with the roles of the factors exchanged. 2. The two iterated integrals agree: \[ \int\Big(\int F(x,y)\,d\nu(y)\Big)d\mu(x)=\int\Big(\int F(x,y)\,d\mu(x)\Big)d\nu(y). \tag{4.1} \] 3. \(F\) is measurable for the product \(\sigma\)-algebra \(\mathcal B(X)\otimes\mathcal B(Y)\). **Proof.** (1) \(\big|\int F(x,y)\,d\nu(y)-\int F(x_0,y)\,d\nu(y)\big|\leq\|F_x-F_{x_0}\|_{\sup}\,\nu(L_F)\), which tends to \(0\) as \(x\to x_0\) by Lemma 4.1(1). (2) Choose an open \(U\supseteq K_F\) with compact closure by (T1). For a sum \(F'=\sum_i\varphi_i\otimes g_i\) as in Lemma 4.1(2), both sides equal \(\sum_i\int\varphi_i\,d\mu\int g_i\,d\nu\). All these functions vanish outside the compact set \(\overline U\times L_F\), and \(|F-F'|\leq\varepsilon\), so each side of (4.1) for \(F\) differs from the same side for \(F'\) by at most \(\varepsilon\,\mu(\overline U)\,\nu(L_F)\). Let \(\varepsilon\to0\). (3) The approximants are \(\mathcal B(X)\otimes\mathcal B(Y)\)-measurable, and \(F\) is their pointwise limit as \(\varepsilon=1/m\to0\). \(\square\) **Definition 4.3** (Radon product). The *Radon product* \(\mu\hat\times\nu\) is the Radon measure on \(X\times Y\) that the Riesz theorem 2.2 gives for the positive linear functional \(F\mapsto\int\big(\int F(x,y)\,d\nu(y)\big)d\mu(x)\) on \(C_c(X\times Y)\). By (4.1) it also represents the other iterated integral. **Proposition 4.4** (Rectangles). For open \(U\subseteq X\), \(V\subseteq Y\) and compact \(K\subseteq X\), \(L\subseteq Y\), with the convention \(0\cdot\infty=0\), \[ (\mu\hat\times\nu)(U\times V)=\mu(U)\,\nu(V),\qquad(\mu\hat\times\nu)(K\times L)=\mu(K)\,\nu(L). \tag{4.2} \] **Proof.** If \(F\prec U\times V\), then \(x\mapsto\int F(x,y)\,d\nu(y)\) vanishes off \(U\) and is at most \(\nu(V)\), so \(\int F\,d(\mu\hat\times\nu)\leq\mu(U)\nu(V)\); when \(\mu(U)=0\) the integral is \(0\). If \(f\prec U\) and \(g\prec V\), then \(f\otimes g\prec U\times V\) and its integral is \(\int f\,d\mu\int g\,d\nu\). Taking suprema and using (2.1) for \(\mu\), \(\nu\) and \(\mu\hat\times\nu\) gives the first formula. For the second, if \(F\in C_c(X\times Y)\) and \(1_{K\times L}\leq F\), then \(\int F(x,y)\,d\nu(y)\geq\nu(L)1_K(x)\), so \(\int F\,d(\mu\hat\times\nu)\geq\mu(K)\nu(L)\). If \(1_K\leq f\) and \(1_L\leq g\), then \(1_{K\times L}\leq f\otimes g\), whose integral is \(\int f\,d\mu\int g\,d\nu\). Taking infima and using (2.1) gives the second formula. \(\square\) ## 5. Tonelli's and Fubini's theorems for Radon products Let \(X,Y\) be LCH spaces with Radon measures \(\mu,\nu\), and let \(\pi=\mu\hat\times\nu\) be their Radon product (Definition 4.3). For \(E\subseteq X\times Y\) put \(E_x=\{y:(x,y)\in E\}\) and \(E^y=\{x:(x,y)\in E\}\); if \(E\) is Borel, so are all sections, because \(y\mapsto(x,y)\) is continuous. A function on \(X\) is *\(\mu\)-a.e. Borel* if it agrees with a Borel function outside a Borel \(\mu\)-null set; its integral is that of the Borel function. **Theorem 5.1.** Everything below also holds with the roles of \(X\) and \(Y\) exchanged, since \(\pi\) represents both iterated integrals (4.1). 1. For every open \(W\subseteq X\times Y\), the function \(x\mapsto\nu(W_x)\) is lsc and \(\pi(W)=\int\nu(W_x)\,d\mu(x)\). 2. For every compact \(C\subseteq X\times Y\), the function \(x\mapsto\nu(C_x)\) is Borel and \(\pi(C)=\int\nu(C_x)\,d\mu(x)\). 3. For every Borel \(E\) with \(\pi(E)<\infty\), the function \(x\mapsto\nu(E_x)\) is \(\mu\)-a.e. Borel and \(\pi(E)=\int\nu(E_x)\,d\mu(x)\). If \(\pi(E)=0\), then \(\nu(E_x)=0\) for \(\mu\)-almost every \(x\). 4. (*Tonelli*) Let \(F\colon X\times Y\to[0,\infty]\) be Borel and vanish outside a set that is \(\sigma\)-finite for \(\pi\). Then \(x\mapsto\int F(x,y)\,d\nu(y)\) is \(\mu\)-a.e. Borel, \(y\mapsto\int F(x,y)\,d\mu(x)\) is \(\nu\)-a.e. Borel, and \[ \int F\,d\pi=\int\Big(\int F(x,y)\,d\nu(y)\Big)d\mu(x)=\int\Big(\int F(x,y)\,d\mu(x)\Big)d\nu(y). \tag{5.1} \] 5. (*Fubini*) Let \(F\in L^1(\pi)\). Then \(F(x,\cdot)\in L^1(\nu)\) for \(\mu\)-almost every \(x\), the \(\mu\)-a.e. defined function \(x\mapsto\int F(x,y)\,d\nu(y)\) is \(\mu\)-integrable, and (5.1) holds. 6. (*Rectangles*) Let \(A\subseteq X\) and \(B\subseteq Y\) be Borel and \(\sigma\)-finite for \(\mu\) and \(\nu\). Then \(A\times B\) is \(\sigma\)-finite for \(\pi\), and \(\pi(A\times B)=\mu(A)\nu(B)\) with \(0\cdot\infty=0\). In particular \(A\times B\) is \(\pi\)-null when \(A\) is \(\mu\)-null. 7. If \(X\) and \(Y\) are \(\sigma\)-compact, \(\pi(E)=\int\nu(E_x)\,d\mu(x)\) for every \(E\in\mathcal B(X)\otimes\mathcal B(Y)\), so \(\pi\) extends the usual product measure. If \(X\) and \(Y\) are second countable, then \(\mathcal B(X\times Y)=\mathcal B(X)\otimes\mathcal B(Y)\). *Support and regularity:* A function supported on a \(\sigma\)-compact set satisfies the support requirement in (4), since compact sets have finite product measure; (4) and (5) allow the more general \(\sigma\)-finite support. [Tornier, Theorem 1.11] states the usual theorem for two globally \(\sigma\)-finite factors. The compact approximation and finite-measure reductions below prove precisely the larger scope needed here. Example 13.5 proves that removing the support condition, or demanding a product measure both inner and outer regular on every Borel set, fails on \(\mathbb R_d\times\mathbb R\). **Proof.** (1) *Lower semicontinuity.* Let \(c<\nu(W_{x_0})\). The section \(W_{x_0}\) is open, so (R3) gives a compact \(L\subseteq W_{x_0}\) with \(\nu(L)>c\). By the tube lemma (T4) applied to \(\{x_0\}\times L\subseteq W\), there is an open \(P\ni x_0\) with \(P\times L\subseteq W\). For \(x\in P\), \(L\subseteq W_x\), so \(\nu(W_x)>c\). *The inequality \(\leq\).* If \(F\prec W\), then \(\int F(x,y)\,d\nu(y)\leq\nu(W_x)\), and integrating gives \(\int F\,d\pi\leq\int\nu(W_x)\,d\mu(x)\). Take the supremum over \(F\) and use (2.1). *The inequality \(\geq\).* Put \(h(x)=\nu(W_x)\) and let \(c<\int h\,d\mu\). By (3.1) choose \(g\in C_c(X)\) with \(0\leq g\leq h\) and \(\int g\,d\mu>c\). Let \(K=\operatorname{supp}g\), and choose \(\eta>0\) with \(\int g\,d\mu-\eta\,\mu(K)>c\). For each \(x\in K\), (R3) gives a compact \(L_x\subseteq W_x\) with \(\nu(L_x)\geq g(x)-\eta/2\) (empty if \(g(x)\leq\eta/2\)). By (T4) there are open \(P_x\ni x\) and \(Q_x\supseteq L_x\) with \(P_x\times Q_x\subseteq W\); shrinking \(P_x\), we may also assume \(gg(x)-\eta\). Hence \(\pi(W)\geq\int F\,d\pi\geq\int_K(g-\eta)\,d\mu>c\). (2) Let \(K\) and \(L\) be the projections of \(C\). By (T1) choose open \(U\supseteq K\) and \(V\supseteq L\) with compact closures, and put \(W=U\times V\). Then \(\nu(W_x)=1_U(x)\nu(V)<\infty\), \(\pi(W)=\mu(U)\nu(V)<\infty\) by (4.2), and \(W\setminus C\) is open. So \(\nu(C_x)=\nu(W_x)-\nu((W\setminus C)_x)\) is a difference of finite lsc functions, hence Borel, and by (1), \(\pi(C)=\pi(W)-\pi(W\setminus C)=\int\nu(C_x)\,d\mu(x)\). (3) By outer regularity (R2) and inner regularity on sets of finite measure (Proposition 2.3(1)), choose open \(W_n\supseteq E\) and compact \(C_n\subseteq E\) with \(\pi(W_n)<\pi(E)+1/n\) and \(\pi(C_n)>\pi(E)-1/n\). Replacing \(W_n\) by \(W_1\cap\dots\cap W_n\) and \(C_n\) by \(C_1\cup\dots\cup C_n\), we may assume \(W_n\) decreases and \(C_n\) increases. Let \(a(x)=\lim_n\nu((W_n)_x)\) and \(b(x)=\lim_n\nu((C_n)_x)\); both are Borel, and \(b(x)\leq\nu(E_x)\leq a(x)\). By (1), \(\int\nu((W_1)_x)\,d\mu=\pi(W_1)<\infty\), so dominated convergence gives \(\int a\,d\mu=\lim\pi(W_n)=\pi(E)\). By (2) and monotone convergence, \(\int b\,d\mu=\lim\pi(C_n)=\pi(E)\). Since \(0\leq a-b\) and \(\int(a-b)\,d\mu=0\), we get \(a=b\) outside a Borel \(\mu\)-null set, and there \(\nu(E_x)=b(x)\). If \(\pi(E)=0\), then \(\int b\,d\mu=0\), so \(b=0\) almost everywhere. (4) By (3), (5.1) holds for \(F=1_E\) when \(\pi(E)<\infty\). If \(E\) is \(\sigma\)-finite, write it as an increasing union of Borel sets of finite measure and use monotone convergence on all three terms; countably many null sets have a null union. By linearity (5.1) holds for nonnegative simple Borel functions vanishing outside a \(\sigma\)-finite set. For general \(F\), the layer functions (1.1) of \(F\) are such simple functions and increase to \(F\); apply monotone convergence again. (5) Split \(F\) into the four nonnegative functions \((\operatorname{Re}F)^\pm\) and \((\operatorname{Im}F)^\pm\), and apply (4) to each. They vanish outside \(\{F\neq0\}\), which is \(\sigma\)-finite by Proposition 3.1(4), and their iterated integrals are finite, so their inner integrals are finite almost everywhere. Subtract. (6) By Proposition 2.3(3), write \(A=S_A\cup N_A\) and \(B=S_B\cup N_B\) with \(S_A,S_B\) \(\sigma\)-compact and \(N_A,N_B\) null. For compact \(L\subseteq Y\) and \(\varepsilon>0\), choose an open \(U\supseteq N_A\) with \(\mu(U)<\varepsilon\) and an open \(V\supseteq L\) with compact closure; by (4.2), \(\pi(N_A\times L)\leq\mu(U)\nu(V)\leq\varepsilon\,\nu(\overline V)\). So \(N_A\times L\) is null, and so is \(N_A\times S_B\). Choosing also an open \(V'\supseteq N_B\) with \(\nu(V')<\varepsilon\), we get \(\pi(N_A\times N_B)\leq\mu(U)\nu(V')<\varepsilon^2\). In the same way \(S_A\times N_B\) is null. So \(A\times B\) differs from the \(\sigma\)-compact set \(S_A\times S_B\) by a null set, and it is \(\sigma\)-finite by (R1). Now (4) with \(F=1_{A\times B}\) gives \(\pi(A\times B)=\int1_A(x)\nu(B)\,d\mu(x)=\mu(A)\nu(B)\). (7) If \(X\) and \(Y\) are \(\sigma\)-compact, so is \(X\times Y\), and every Borel set is \(\sigma\)-finite by (R1); apply (4) to \(1_E\), noting \(\mathcal B(X)\otimes\mathcal B(Y)\subseteq\mathcal B(X\times Y)\) because the projections are continuous. For \(E\) in the product \(\sigma\)-algebra, \(\int\nu(E_x)\,d\mu(x)\) is the usual product measure of \(\sigma\)-finite factors. If \(X\) and \(Y\) have countable bases, every open \(W\subseteq X\times Y\) is the union of the countably many products of basic sets contained in \(W\), so \(W\in\mathcal B(X)\otimes\mathcal B(Y)\). \(\square\) *Where the hypotheses enter, and why they are needed.* Parts (1)–(3) use only the Radon properties. \(\sigma\)-finiteness enters in (4)–(6), and it cannot be dropped. Example 13.5 gives a closed set in \(\mathbb R_d\times\mathbb R\) whose two iterated integrals are \(0\) and \(\infty\), and Example 13.6 gives a set \(\{e\}\times G\) with \(\pi(\{e\}\times G)=\infty\) although \(\mu(\{e\})=0\). In particular, (4) applies to every Borel \(F\geq0\) that vanishes outside a \(\sigma\)-compact set, since \(\sigma\)-compact sets are \(\sigma\)-finite by (R1). **Example 5.2** (Borel sets of products). If \(X\) and \(Y\) are second countable, \(\mathcal B(X\times Y)=\mathcal B(X)\otimes\mathcal B(Y)\) (Theorem 5.1(7)). Without second countability this fails even for discrete groups. Let \(G\) be a discrete group of cardinality greater than \(2^{\aleph_0}\), for instance a free abelian group on such a set. The diagonal \(\Delta_G\subseteq G\times G\) is open, but it is not in \(\mathcal B(G)\otimes\mathcal B(G)\), which here is the \(\sigma\)-algebra generated by all rectangles. Proof: a set in the \(\sigma\)-algebra generated by a family \(\mathcal C\) lies in the \(\sigma\)-algebra generated by some countable subfamily, because the sets with this property form a \(\sigma\)-algebra containing \(\mathcal C\). So \(\Delta_G\) lies in the \(\sigma\)-algebra generated by countably many rectangles \(A_n\times B_n\). Put \(\varphi(x)=(1_{A_n}(x))_n\) and \(\psi(y)=(1_{B_n}(y))_n\) in \(\{0,1\}^{\mathbb N}\). The sets \((\varphi\times\psi)^{-1}(S)\), \(S\subseteq\{0,1\}^{\mathbb N}\times\{0,1\}^{\mathbb N}\), form a \(\sigma\)-algebra containing each \(A_n\times B_n\), so \(\Delta_G=(\varphi\times\psi)^{-1}(S)\) for some \(S\). If \(\varphi(x)=\varphi(x')\), then \((x,x')\in\Delta_G\) because \((\varphi(x),\psi(x'))=(\varphi(x'),\psi(x'))\in S\); so \(x=x'\). Hence \(\varphi\) is injective and \(|G|\leq2^{\aleph_0}\), a contradiction. So the Radon product (here counting measure on \(G\times G\)) lives on a \(\sigma\)-algebra strictly larger than the product \(\sigma\)-algebra. This is why Theorem 5.1 and Theorem 6.2 are proved for Borel functions on \(X\times Y\), not only for product-measurable ones. ## 6. Hilbert tensor products and L² of a product *Hilbert tensor products.* For Hilbert spaces \(H,K\) the Hilbert tensor product \(H\otimes K\) is a Hilbert space with a bilinear map \((\xi,\eta)\mapsto\xi\otimes\eta\) such that \(\langle\xi\otimes\eta,\xi'\otimes\eta'\rangle=\langle\xi,\xi'\rangle\langle\eta,\eta'\rangle\) and the elementary tensors span a dense subspace. One construction takes \(H\otimes K\) to be the space of antilinear Hilbert–Schmidt maps from \(K\) to \(H\), with the Hilbert–Schmidt norm \(|||\cdot|||\) (see "Background used without proof"). The one step of that construction that needs an argument is completeness. Let \((A_n)\) be Cauchy in \(|||\cdot|||\). Since the operator norm is at most \(|||\cdot|||\), the sequence is Cauchy in operator norm, so \(A_n\to A\) in operator norm for some bounded antilinear \(A\). For any finite set \(F\) of vectors of an orthonormal basis \((v_\beta)\) of \(K\), \(\sum_{\beta\in F}\|(A-A_n)v_\beta\|^2=\lim_m\sum_{\beta\in F}\|(A_m-A_n)v_\beta\|^2\leq\sup_{m\geq n}|||A_m-A_n|||^2\). Taking the supremum over \(F\) shows \(|||A-A_n|||\to0\), and in particular \(A\) is Hilbert–Schmidt. **Lemma 6.1** (Isometric extension). Suppose \(B\colon H\times K\to\mathcal K\) is bilinear, where \(\mathcal K\) is a third Hilbert space, and \(\langle B(\xi,\eta),B(\xi',\eta')\rangle=\langle\xi,\xi'\rangle\langle\eta,\eta'\rangle\). There is a unique isometry \(U\colon H\otimes K\to\mathcal K\) with \(U(\xi\otimes\eta)=B(\xi,\eta)\). **Proof.** For finite sums, \(\|\sum_iB(\xi_i,\eta_i)\|^2=\sum_{i,j}\langle\xi_i,\xi_j\rangle\langle\eta_i,\eta_j\rangle=\|\sum_i\xi_i\otimes\eta_i\|^2\). So \(\sum_i\xi_i\otimes\eta_i\mapsto\sum_iB(\xi_i,\eta_i)\) is well defined (a sum of tensors that is \(0\) goes to a vector of norm \(0\)) and isometric on the span of the elementary tensors. It extends by continuity to the closure, which is \(H\otimes K\). Uniqueness holds because the elementary tensors are total. \(\square\) **Theorem 6.2** (\(L^2\) of the Radon product). Let \(\mu,\nu\) be Radon measures on LCH spaces \(X,Y\), and \(\pi=\mu\hat\times\nu\). For functions \(\xi\) on \(X\) and \(\eta\) on \(Y\) write \((\xi\boxtimes\eta)(x,y)=\xi(x)\eta(y)\). 1. For \(\xi\in L^2(\mu)\) and \(\eta\in L^2(\nu)\), \(\xi\boxtimes\eta\in L^2(\pi)\); its class depends only on the classes of \(\xi\) and \(\eta\); and \(\langle\xi\boxtimes\eta,\xi'\boxtimes\eta'\rangle=\langle\xi,\xi'\rangle\langle\eta,\eta'\rangle\). 2. There is a unique unitary \(U\colon L^2(\mu)\otimes L^2(\nu)\to L^2(\pi)\) with \(U(\xi\otimes\eta)=\xi\boxtimes\eta\). 3. Every \(\zeta\in L^2(\pi)\) satisfies \(\zeta(x,\cdot)\in L^2(\nu)\) for \(\mu\)-almost every \(x\), and \(\|\zeta\|^2=\int\|\zeta(x,\cdot)\|_{L^2(\nu)}^2\,d\mu(x)\); likewise in the other variable. 4. For a bounded Borel function \(a\) on \(X\), the operator \(U^*M_{a\circ\mathrm{pr}_1}U\), where \(M\) denotes multiplication on \(L^2(\pi)\), is the unique bounded operator on \(L^2(\mu)\otimes L^2(\nu)\) that sends every \(\xi\otimes\eta\) to \((a\xi)\otimes\eta\). It is the operator \(m_a\otimes1\). The same holds for the second factor. **Proof.** (1) Take Borel representatives. The sets \(S_\xi=\{\xi\neq0\}\) and \(S_\eta=\{\eta\neq0\}\) are \(\sigma\)-finite by Proposition 3.1(4). The function \(\xi\boxtimes\eta\) is Borel on \(X\times Y\) and vanishes outside \(S_\xi\times S_\eta\), which is \(\sigma\)-finite by Theorem 5.1(6). Tonelli's theorem 5.1(4) gives \(\int|\xi\boxtimes\eta|^2\,d\pi=\|\xi\|^2\|\eta\|^2\). If \(\xi=\xi'\) outside a \(\mu\)-null Borel set \(N\), then \(\xi\boxtimes\eta=\xi'\boxtimes\eta\) outside \(N\times S_\eta\), which is \(\pi\)-null by Theorem 5.1(6); similarly in the second variable. The function \((\xi\bar\xi')\boxtimes(\eta\bar\eta')\) is in \(L^1(\pi)\) by the Cauchy–Schwarz inequality, and Fubini's theorem 5.1(5) gives the inner product formula. (2) Lemma 6.1, with \(B(\xi,\eta)=\xi\boxtimes\eta\), gives the isometry \(U\). Its range is closed, because \(U\) is an isometry on a complete space, and it contains \(\varphi\boxtimes g\) for \(\varphi\in C_c(X)\) and \(g\in C_c(Y)\). These span a dense subspace of \(L^2(\pi)\). Indeed, let \(\zeta\in L^2(\pi)\) and \(\varepsilon>0\). Since \(C_c\) is dense in \(L^2\) (Proposition 3.1(4)), there is \(F\in C_c(X\times Y)\) with \(\|\zeta-F\|_2<\varepsilon\). Choose an open \(O\supseteq K_F\) with compact closure (T1). The tensor approximation (Lemma 4.1(2)), applied with \(O\), gives \(F'=\sum_i\varphi_i\boxtimes g_i\) with \(|F-F'|\leq\delta\), where \(F\) and \(F'\) vanish outside the compact set \(\overline O\times L_F\). Then \(\|F-F'\|_2\leq\delta\,\pi(\overline O\times L_F)^{1/2}=\delta\,(\mu(\overline O)\nu(L_F))^{1/2}\) by (4.2), which is less than \(\varepsilon\) for small \(\delta\). So \(U\) is onto. Uniqueness holds because elementary tensors are total. (3) Apply Tonelli's theorem 5.1(4) to \(|\zeta|^2\). It vanishes outside the set \(\{\zeta\neq0\}\), which is \(\sigma\)-finite by Proposition 3.1(4) applied to \(\pi\). (4) \(M_{a\circ\mathrm{pr}_1}(\xi\boxtimes\eta)=(a\xi)\boxtimes\eta\), so \(U^*M_{a\circ\mathrm{pr}_1}U(\xi\otimes\eta)=(a\xi)\otimes\eta\). Two bounded operators that agree on a total set are equal. \(\square\) In words: **\(L^2(\mu)\otimes L^2(\nu)\cong L^2(X\times Y,\mu\hat\times\nu)\) through \(\xi\otimes\eta\mapsto\xi\boxtimes\eta\), for arbitrary Radon measures, with the Radon product and not the ordinary product measure.** Every element of \(L^2(\pi)\) has \(\sigma\)-finite support, so the pathological sets of Examples 13.5 and 13.6 never carry \(L^2\) mass. ## 7. Topological groups **Proposition 7.1.** Let \(G\) be a group, topologized so that multiplication \(G\times G\to G\) and inversion are continuous. The Hausdorff property is not assumed in this proposition. 1. Left and right translations and inversion are homeomorphisms of \(G\). If \(U\) is open and \(A\subseteq G\) is any set, then \(AU\) and \(UA\) are open. 2. Every neighbourhood \(U\) of \(e\) contains a neighbourhood \(V\) of \(e\) with \(V=V^{-1}\) and \(VV\subseteq U\). 3. The closure of a subgroup is a subgroup. An open subgroup is closed. 4. If \(A\) and \(B\) are compact, so is \(AB\). 5. Let \(H\) be a subgroup, \(G/H\) the set of left cosets with the quotient topology, and \(q\colon G\to G/H\) the quotient map. Then \(q\) is open. If \(H\) is closed, \(G/H\) is Hausdorff. If \(G\) is locally compact, every point of \(G/H\) has a compact neighbourhood. If \(H\) is normal, \(G/H\) is a group with continuous multiplication and inversion. 6. If \(\{e\}\) is closed, \(G\) is Hausdorff. In general \(N=\overline{\{e\}}\) is a normal subgroup as well as closed, and \(G/N\) is Hausdorff. **Proof.** (1) Translation by \(x\) has the continuous inverse translation by \(x^{-1}\), and inversion is its own inverse. Also \(AU=\bigcup_{a\in A}aU\), and each \(aU\) is open. (2) Continuity of multiplication at \((e,e)\) gives neighbourhoods \(W_1,W_2\) of \(e\) with \(W_1W_2\subseteq U\). Put \(W=W_1\cap W_2\) and \(V=W\cap W^{-1}\). (3) The continuous map \((x,y)\mapsto xy^{-1}\) sends \(H\times H\) into \(H\), hence sends \(\overline H\times\overline H=\overline{H\times H}\) into \(\overline H\). If \(H\) is open, its complement is a union of cosets \(xH\), which are open by (1). (4) Multiplication is continuous and maps the compact set \(A\times B\) onto \(AB\). (5) For open \(V\subseteq G\), \(q^{-1}(q(V))=VH\) is open by (1), so \(q(V)\) is open. Let \(H\) be closed and \(y\notin xH\). The coset \(xH\) is closed, so some neighbourhood \(U\) of \(e\) has \(Uy\cap xH=\varnothing\); by (2) take \(V\) symmetric with \(VV\subseteq U\). If \(q(Vx)\) and \(q(Vy)\) met, we would have \(v_1xh_1=v_2yh_2\) with \(v_i\in V\) and \(h_i\in H\), so \(v_1^{-1}v_2\,y=xh_1h_2^{-1}\in xH\) with \(v_1^{-1}v_2\in VV\subseteq U\), which is impossible. So \(q(Vx)\) and \(q(Vy)\) are disjoint neighbourhoods of \(xH\) and \(yH\) (\(q\) is open). If \(C\) is a compact neighbourhood of \(e\) in \(G\), then \(q(xC)\) is compact and contains the open set \(q(x\operatorname{int}C)\ni xH\). If \(H\) is normal and \(W\) is an open neighbourhood of \(q(xy)\), continuity of multiplication in \(G\) gives open \(V_1\ni x\) and \(V_2\ni y\) with \(V_1V_2\subseteq q^{-1}(W)\), so \(q(V_1)q(V_2)\subseteq W\) with \(q(V_i)\) open. Inversion is handled the same way. (6) If \(\{e\}\) is closed, apply (5) with \(H=\{e\}\). In general \(N\) is a subgroup by (3), and it lies in every closed subgroup. For \(z\in G\), \(zNz^{-1}\) is a closed subgroup (conjugation is a homeomorphism and an automorphism), so it contains \(N\). Applying this to \(z^{-1}\) gives \(zNz^{-1}=N\). Now apply (5) with \(H=N\). \(\square\) By (6), dividing by \(\overline{\{e\}}\) removes any failure of the Hausdorff property, which is why we may assume it throughout. From now on \(G\) is a locally compact group. Two more facts are needed: \(G\) splits into \(\sigma\)-compact pieces, and compactly supported continuous functions are uniformly continuous. **Proposition 7.2** (An open \(\sigma\)-compact subgroup). Some subgroup \(G_0\) of \(G\) is at once open, closed and \(\sigma\)-compact. Its left cosets \(xG_0\) partition \(G\) into open, closed, \(\sigma\)-compact pieces. **Proof.** Let \(C\) be a compact neighbourhood of \(e\) and \(U=C\cap C^{-1}\), a compact symmetric neighbourhood of \(e\). The sets \(U^n=U\cdots U\) (\(n\) factors) are compact by Proposition 7.1(4), and \(G_0=\bigcup_nU^n\) is the subgroup generated by \(U\). It contains the neighbourhood \(xU\) of each of its points \(x\), since \(xU\subseteq U^{n+1}\) when \(x\in U^n\). So it is open, hence closed by Proposition 7.1(3), and it is \(\sigma\)-compact. If \(G\) is connected, the open and closed subgroup \(G_0\) is all of \(G\), so every connected locally compact group is \(\sigma\)-compact. \(\square\) **Proposition 7.3** (Uniform continuity). For every \(f\in C_c(G)\), \(\|R_yf-f\|_{\sup}\to0\) and \(\|L_yf-f\|_{\sup}\to0\) as \(y\to e\). **Proof.** Let \(K=\operatorname{supp}f\) and \(\varepsilon>0\). For \(x\in K\) choose a neighbourhood \(U_x\) of \(e\) with \(|f(xz)-f(x)|<\varepsilon/2\) for \(z\in U_x\), and a symmetric neighbourhood \(V_x\) with \(V_xV_x\subseteq U_x\); note \(V_x\subseteq U_x\). Finitely many sets \(x_jV_{x_j}\) cover \(K\); let \(V=\bigcap_jV_{x_j}\), a symmetric neighbourhood of \(e\). Let \(y\in V\). If \(x\in K\), write \(x=x_ju\) with \(u\in V_{x_j}\). Then \(xy=x_j(uy)\) with \(uy\in V_{x_j}V_{x_j}\subseteq U_{x_j}\), so \(f(xy)\) and \(f(x)\) are both within \(\varepsilon/2\) of \(f(x_j)\), and \(|f(xy)-f(x)|<\varepsilon\). If \(xy\in K\), the same argument applied to the point \(xy\) and to \(y^{-1}\in V\) gives \(|f(x)-f(xy)|<\varepsilon\). Otherwise both values are \(0\). So \(\|R_yf-f\|_{\sup}\leq\varepsilon\) for \(y\in V\). For left translations, \(L_yf(x)=\check f(x^{-1}y)=R_y\check f(x^{-1})\), so \(\|L_yf-f\|_{\sup}=\|R_y\check f-\check f\|_{\sup}\) with \(\check f\in C_c(G)\). \(\square\) Proposition 7.3 underlies the continuity of translation in \(L^p\) (Theorem 14.2(6)), and with it the strong continuity of the regular representations (Exercise 16.3). A bounded function \(f\) with \(\|L_yf-f\|_{\sup}\to0\) as \(y\to e\) is called *left uniformly continuous*, and one with \(\|R_yf-f\|_{\sup}\to0\) *right uniformly continuous*. [Tornier, Proposition 1.7] writes uniform continuity directly as small left and right multiplication increments. To avoid ambiguity between translation conventions, this lesson always writes the condition out. ## 8. Existence of Haar measure **Definition 8.1.** Fix a locally compact group \(G\). We call a Radon measure \(\mu\neq0\) on \(G\) a *left* (*right*) *Haar measure* when \(\mu(xE)=\mu(E)\) (\(\mu(Ex)=\mu(E)\)) for every Borel set \(E\) and every \(x\in G\). A *left Haar integral* is a left-invariant positive linear functional \(I\neq0\) on \(C_c(G)\), that is, \(I(L_yf)=I(f)\) for all \(f\) and \(y\). **Proposition 8.2** (Invariance through integrals). Fix a Radon measure \(\mu\) on \(G\). 1. \(\mu\) is a left Haar measure exactly when \(\tilde\mu(E)=\mu(E^{-1})\) is a right Haar measure. 2. \(\mu(yE)=\mu(E)\) for all Borel \(E\) and all \(y\) exactly when \(\int L_yf\,d\mu=\int f\,d\mu\) for all \(f\in C_c(G)\) and all \(y\). In that case \(\int f(yx)\,d\mu(x)=\int f\,d\mu\) for all \(y\), all Borel \(f\geq0\), and all \(f\in L^1(\mu)\). So, through the Riesz theorem 2.2, left Haar measures and left Haar integrals are the same objects. **Proof.** (1) Inversion \(\iota\) is a homeomorphism, so \(\tilde\mu=\iota_*\mu\) is Radon by Proposition 3.1(1). Also \(\tilde\mu(Ex)=\mu(x^{-1}E^{-1})\), and this equals \(\tilde\mu(E)=\mu(E^{-1})\) for all \(E\) and \(x\) exactly when \(\mu\) is left invariant. (2) Let \(\ell_z(x)=zx\) and \(\mu_y=(\ell_{y^{-1}})_*\mu\), so \(\mu_y(E)=\mu(yE)\). By Proposition 3.1(1), \(\mu_y\) is Radon and \(\int f\,d\mu_y=\int f(y^{-1}x)\,d\mu(x)=\int L_yf\,d\mu\). If \(\int L_yf\,d\mu=\int f\,d\mu\) on \(C_c(G)\), the Radon measures \(\mu_y\) and \(\mu\) agree by the uniqueness in Theorem 2.2. The converse and the last sentence follow from the same integral formula, applied to \(y^{-1}\). \(\square\) **Theorem 8.3** (Existence). Every locally compact group has a left Haar measure. *Source comparison:* [Tornier, Theorem 2.2] states existence at exactly this generality. [Fremlin, 441C and 441E] proves it through invariant compact-set contents. The argument below instead constructs a positive invariant functional on \(C_c(G)\), so Theorem 2.2 produces the Borel measure with our specified regularity. The proof builds a left Haar integral as a limit of normalized "covering numbers", which compare a function \(f\) with translates of a function \(\varphi\) of small support. *Covering numbers.* For \(f,\varphi\in C_c^+(G)\), let \((f:\varphi)\) be the infimum of \(\sum_jc_j\) over all finite families \(c_1,\dots,c_n\geq0\) and \(x_1,\dots,x_n\in G\) with \(f\leq\sum_jc_jL_{x_j}\varphi\). This number is finite. Indeed, the open set \(W=\{\varphi>\|\varphi\|_{\sup}/2\}\) is nonempty; fix \(w\in W\). The open sets \(xw^{-1}W\), \(x\in\operatorname{supp}f\), cover \(\operatorname{supp}f\), so finitely many of them, \(x_jw^{-1}W\) for \(j\leq n\), do. On \(x_jw^{-1}W\) we have \(L_{x_jw^{-1}}\varphi>\|\varphi\|_{\sup}/2\). Hence \(f\leq\sum_j\frac{2\|f\|_{\sup}}{\|\varphi\|_{\sup}}L_{x_jw^{-1}}\varphi\), and \((f:\varphi)\leq2n\|f\|_{\sup}/\|\varphi\|_{\sup}\). For \(f,f_1,f_2,\varphi,\psi\in C_c^+(G)\), \(y\in G\) and \(c>0\): \[ \begin{gathered} (L_yf:\varphi)=(f:\varphi),\quad (f_1+f_2:\varphi)\leq(f_1:\varphi)+(f_2:\varphi),\quad (cf:\varphi)=c(f:\varphi),\\ f_1\leq f_2\Rightarrow(f_1:\varphi)\leq(f_2:\varphi),\quad (f:\varphi)\geq\frac{\|f\|_{\sup}}{\|\varphi\|_{\sup}},\quad (f:\psi)\leq(f:\varphi)(\varphi:\psi). \end{gathered} \tag{8.1} \] For the first rule, \(f\leq\sum_jc_jL_{x_j}\varphi\) holds exactly when \(L_yf\leq\sum_jc_jL_{yx_j}\varphi\). The next three follow by adding, scaling or comparing dominations. For the fifth, evaluate a domination at a point \(t_0\) where \(f(t_0)=\|f\|_{\sup}\): \(\|f\|_{\sup}\leq\sum_jc_j\varphi(x_j^{-1}t_0)\leq\|\varphi\|_{\sup}\sum_jc_j\). For the last, if \(f\leq\sum_ic_iL_{x_i}\varphi\) and \(\varphi\leq\sum_kd_kL_{z_k}\psi\), then \(f\leq\sum_{i,k}c_id_kL_{x_iz_k}\psi\), and \(\sum_{i,k}c_id_k=(\sum_ic_i)(\sum_kd_k)\). Fix \(f_0\in C_c^+(G)\) and put \(I_\varphi(f)=(f:\varphi)/(f_0:\varphi)\). The last rule of (8.1), used twice, gives \[ (f_0:f)^{-1}\leq I_\varphi(f)\leq(f:f_0). \tag{8.2} \] Each \(I_\varphi\) is left invariant, subadditive, positively homogeneous and monotone. It is additive only approximately: **Lemma 8.4** (Almost additivity). Let \(f_1,f_2\in C_c^+(G)\) and \(\varepsilon>0\). There is a neighbourhood \(V\) of \(e\) such that \(I_\varphi(f_1)+I_\varphi(f_2)\leq I_\varphi(f_1+f_2)+\varepsilon\) whenever \(\operatorname{supp}\varphi\subseteq V\). **Proof.** By Urysohn's lemma (T2) choose \(g\in C_c^+(G)\) with \(g=1\) on \(\operatorname{supp}(f_1+f_2)\). Let \(\delta>0\), to be fixed at the end, and put \(h=f_1+f_2+\delta g\). Let \(h_i=f_i/h\) where \(h>0\), and \(h_i=0\) elsewhere (\(i=1,2\)). Since \(h\geq\delta\) on \(\operatorname{supp}f_i\), \(h_i\) is continuous on the open set \(\{h>0\}\) and vanishes on the open set \(G\setminus\operatorname{supp}f_i\); these two open sets cover \(G\). So \(h_i\in C_c(G)\), and \(h_1+h_2\leq1\). By uniform continuity (Proposition 7.3) there is a neighbourhood \(V\) of \(e\) with \(|h_i(xv)-h_i(x)|<\delta\) for all \(x\in G\), \(v\in V\) and \(i=1,2\). Let \(\operatorname{supp}\varphi\subseteq V\) and \(h\leq\sum_jc_jL_{x_j}\varphi\). If \(\varphi(x_j^{-1}x)\neq0\), then \(x_j^{-1}x\in V\), so \(|h_i(x)-h_i(x_j)|<\delta\). Hence \[ f_i(x)=h(x)h_i(x)\leq\sum_jc_j\varphi(x_j^{-1}x)h_i(x)\leq\sum_jc_j\varphi(x_j^{-1}x)\big(h_i(x_j)+\delta\big). \] So \((f_i:\varphi)\leq\sum_jc_j(h_i(x_j)+\delta)\), and \((f_1:\varphi)+(f_2:\varphi)\leq(1+2\delta)\sum_jc_j\). Taking the infimum over dominations of \(h\) and using (8.1), \[ (f_1:\varphi)+(f_2:\varphi)\leq(1+2\delta)(h:\varphi)\leq(1+2\delta)\big((f_1+f_2:\varphi)+\delta(g:\varphi)\big). \] Divide by \((f_0:\varphi)\) and use (8.2): \(I_\varphi(f_1)+I_\varphi(f_2)\leq I_\varphi(f_1+f_2)+2\delta(f_1+f_2:f_0)+\delta(1+2\delta)(g:f_0)\). Choose \(\delta\) so that the last two terms add up to less than \(\varepsilon\). \(\square\) **Proof of Theorem 8.3.** Let \(\Omega=\prod_{f\in C_c^+(G)}\big[(f_0:f)^{-1},(f:f_0)\big]\). It is compact Hausdorff by Tychonoff's theorem, and by (8.2) each \(I_\varphi\) is a point of \(\Omega\). For a neighbourhood \(V\) of \(e\), let \(\Phi_V\) be the closure in \(\Omega\) of \(\{I_\varphi:\varphi\in C_c^+(G),\ \operatorname{supp}\varphi\subseteq V\}\). It is nonempty by (T2), and \(\Phi_{V_1}\cap\dots\cap\Phi_{V_n}\supseteq\Phi_{V_1\cap\dots\cap V_n}\neq\varnothing\). By compactness some point \(I\) lies in every \(\Phi_V\). By the definition of the product topology this means: for every neighbourhood \(V\) of \(e\), all \(f_1,\dots,f_m\in C_c^+(G)\) and all \(\varepsilon>0\), there is \(\varphi\) with \(\operatorname{supp}\varphi\subseteq V\) and \(|I(f_k)-I_\varphi(f_k)|<\varepsilon\) for \(k\leq m\). Let \(f_1,f_2\in C_c^+(G)\) and \(\varepsilon>0\). Take \(V\) from Lemma 8.4 and such a \(\varphi\) for \(f_1\), \(f_2\) and \(f_1+f_2\). Then \[ I(f_1)+I(f_2)0\), because each \(I_\varphi\) has these properties exactly. Also \(I(f)\geq(f_0:f)^{-1}>0\) and \(I(f_0)=1\). Put \(I(0)=0\). A real \(f\in C_c(G)\) is \(f^+-f^-\) with \(f^\pm\in C_c^+(G)\cup\{0\}\); set \(I(f)=I(f^+)-I(f^-)\). If also \(f=g-h\) with \(g,h\in C_c^+(G)\cup\{0\}\), then \(g+f^-=h+f^+\), and additivity gives \(I(g)-I(h)=I(f^+)-I(f^-)\). So \(I\) is well defined and linear on real functions; for complex \(f\) put \(I(f)=I(\operatorname{Re}f)+iI(\operatorname{Im}f)\). Then \(I\) is a left Haar integral with \(I(f_0)=1\). By the Riesz theorem 2.2 and Proposition 8.2, its Radon measure is a left Haar measure. \(\square\) *Where the hypotheses enter.* Compact supports make every \((f:\varphi)\) finite. Local compactness and the Hausdorff property supply the small functions \(\varphi\), through Urysohn's lemma (T2), and they make the Riesz theorem available. Continuity of the group operations gives the uniform continuity used in Lemma 8.4. The compactness step uses Tychonoff's theorem, so this particular argument uses the corresponding choice principle. The compact-set-content construction in [Fremlin, 441C] provides a second route, also using an ultrafilter compactness step. The full functional argument given here does not assert a choice-free construction. Right Haar measures exist too: by Proposition 8.2(1), \(\tilde\mu\) is a right Haar measure. ## 9. Positivity and uniqueness of Haar measure **Proposition 9.1** (Positivity). A left Haar measure \(\mu\) satisfies \(\mu(U)>0\) for every nonempty open set \(U\), and \(\int f\,d\mu>0\) for every \(f\in C_c^+(G)\). **Proof.** Suppose \(U\neq\varnothing\) is open and \(\mu(U)=0\). Fix \(u\in U\). For compact \(K\), the null open sets \(xu^{-1}U\), \(x\in K\), cover \(K\), and finitely many suffice; so \(\mu(K)=0\). Then \(\mu(G)=0\) by (R3), which contradicts \(\mu\neq0\). For \(f\in C_c^+(G)\), the open set \(\{f>\|f\|_{\sup}/2\}\) is nonempty, so \(\int f\,d\mu\geq\frac12\|f\|_{\sup}\,\mu(\{f>\|f\|_{\sup}/2\})>0\). \(\square\) **Theorem 9.2** (Uniqueness). Any two left Haar measures \(\mu,\nu\) on \(G\) are proportional: \(\mu=c\nu\) for some \(c\in(0,\infty)\). Equivalently, every left Haar integral is a positive multiple of \(f\mapsto\int f\,d\mu\). *Source comparison:* [Fremlin, 442B] proves general uniqueness by comparing ratios on small symmetric neighbourhoods. The proof below compares ratios of compactly supported integrals instead, using only the locally finite compact-support Fubini identity (4.1). [Tornier, Theorem 2.2] states the same uniqueness conclusion. **Proof.** Fix \(f,g\in C_c^+(G)\). It suffices to show \(\int f\,d\mu/\int f\,d\nu=\int g\,d\mu/\int g\,d\nu\); the denominators are positive by Proposition 9.1. Fix a compact symmetric neighbourhood \(V_0\) of \(e\), and let \(A=(\operatorname{supp}f)V_0\cup V_0(\operatorname{supp}f)\) and \(B=(\operatorname{supp}g)V_0\cup V_0(\operatorname{supp}g)\), which are compact by Proposition 7.1(4). For \(y\in V_0\), the function \(x\mapsto f(xy)-f(yx)\) vanishes off \(A\): \(f(xy)\neq0\) forces \(x\in(\operatorname{supp}f)y^{-1}\), and \(f(yx)\neq0\) forces \(x\in y^{-1}\operatorname{supp}f\). Let \(\varepsilon>0\). Since \(|f(xy)-f(yx)|\leq|R_yf(x)-f(x)|+|f(x)-L_{y^{-1}}f(x)|\), uniform continuity (Proposition 7.3) gives a symmetric neighbourhood \(V\subseteq V_0\) of \(e\) with \(|f(xy)-f(yx)|<\varepsilon\) and \(|g(xy)-g(yx)|<\varepsilon\) for all \(x\in G\) and \(y\in V\). Choose \(h_1\in C_c^+(G)\) with \(\operatorname{supp}h_1\subseteq V\) and put \(h=h_1+\check h_1\). Then \(h\in C_c^+(G)\), \(\operatorname{supp}h\subseteq V\), and \(h(x^{-1})=h(x)\). Every integrand below is a continuous function with compact support on \(G\times G\), so (4.1) allows the order of integration to be exchanged. By left invariance of \(\mu\) in \(x\), \[ \int h\,d\nu\int f\,d\mu=\iint h(y)f(yx)\,d\mu(x)\,d\nu(y). \] By left invariance of \(\mu\) in \(x\), then an exchange and the symmetry of \(h\), then left invariance of \(\nu\) in \(y\), then another exchange, \[ \int h\,d\mu\int f\,d\nu=\iint h(y^{-1}x)f(y)\,d\mu(x)\,d\nu(y)=\iint h(x^{-1}y)f(y)\,d\nu(y)\,d\mu(x)=\iint h(y)f(xy)\,d\nu(y)\,d\mu(x)=\iint h(y)f(xy)\,d\mu(x)\,d\nu(y). \] Subtract. For \(y\in\operatorname{supp}h\subseteq V\) the inner integrand \(h(y)(f(yx)-f(xy))\) is at most \(\varepsilon h(y)\) in absolute value and vanishes for \(x\notin A\). Hence \[ \Big|\int h\,d\nu\int f\,d\mu-\int h\,d\mu\int f\,d\nu\Big|\leq\varepsilon\,\mu(A)\int h\,d\nu . \] Dividing by \(\int h\,d\nu\int f\,d\nu\) gives \(\big|\frac{\int f\,d\mu}{\int f\,d\nu}-\frac{\int h\,d\mu}{\int h\,d\nu}\big|\leq\frac{\varepsilon\mu(A)}{\int f\,d\nu}\). The same holds for \(g\) with \(B\). So the two ratios for \(f\) and \(g\) differ by at most \(\varepsilon\big(\mu(A)/\int f\,d\nu+\mu(B)/\int g\,d\nu\big)\), where \(A\) and \(B\) do not depend on \(\varepsilon\). Hence the ratio is a constant \(c>0\) on \(C_c^+(G)\). By linearity \(\int f\,d\mu=c\int f\,d\nu\) on \(C_c(G)\), and \(\mu=c\nu\) by the uniqueness in Theorem 2.2. \(\square\) Only compactly supported continuous integrands on \(G\times G\) occur, so this proof needs nothing beyond (4.1); in particular it needs no \(\sigma\)-finiteness. From now on, \(\mu\) denotes a fixed left Haar measure on \(G\), and we write \(\int f(t)\,dt=\int f\,d\mu\). ## 10. The modular function A left Haar measure need not be right invariant. The modular function measures how far it is from right invariance. **Theorem 10.1.** Let \(\mu\) be a left Haar measure on \(G\). 1. For each \(x\in G\), \(E\mapsto\mu(Ex)\) is a left Haar measure. So there is a unique \(\Delta(x)\in(0,\infty)\) with \(\mu(Ex)=\Delta(x)\mu(E)\) for every Borel \(E\). Another left Haar measure in place of \(\mu\) gives the same \(\Delta(x)\). 2. \(\Delta\) is a continuous homomorphism from \(G\) to the multiplicative group \((0,\infty)\). 3. For every Borel \(f\geq0\), and for every \(f\in L^1(\mu)\), \[ \int f(tx)\,dt=\Delta(x)^{-1}\int f(t)\,dt . \tag{10.1} \] In particular \(\|R_xf\|_p=\Delta(x)^{-1/p}\|f\|_p\) for \(f\in L^p(\mu)\), \(1\leq p<\infty\). 4. \(\Delta=1\) on every compact subgroup. Every compact, abelian or discrete group is *unimodular* (\(\Delta\equiv1\)). \(G\) is unimodular exactly when \(\mu\) is right invariant as well. If \(G/[G,G]\) is compact, where \([G,G]\) is the closure of the subgroup generated by the commutators \(xyx^{-1}y^{-1}\), then \(G\) is unimodular. **Proof.** (1) Right translation by \(x\) is a homeomorphism, so \(\mu_x(E)=\mu(Ex)\) is Radon by Proposition 3.1(1). It is nonzero and left invariant, since \(y(Ex)=(yE)x\). By uniqueness (Theorem 9.2), \(\mu_x=\Delta(x)\mu\) with \(\Delta(x)>0\). Replacing \(\mu\) by \(c\mu\) does not change \(\Delta\). (2) Let \(E\) be a compact neighbourhood of \(e\), so \(0<\mu(E)<\infty\) by Proposition 9.1 and (R1). Then \(\Delta(xy)\mu(E)=\mu(Exy)=\Delta(y)\mu(Ex)=\Delta(y)\Delta(x)\mu(E)\). So \(\Delta\) is a homomorphism, and \(\Delta(x^{-1})=\Delta(x)^{-1}\). Continuity is proved after (3). (3) For \(f=1_E\), \(\int1_E(tx)\,dt=\mu(Ex^{-1})=\Delta(x)^{-1}\mu(E)\). Linearity, then the layer functions (1.1) with monotone convergence, then splitting into four nonnegative parts extend this to simple functions, to Borel \(f\geq0\), and to \(L^1\). Apply it to \(|f|^p\) for the norm formula. *Continuity of \(\Delta\).* Fix \(f\in C_c^+(G)\); by (3), \(\Delta(x)^{-1}=\int R_xf\,dt/\int f\,dt\), where the denominator is positive by Proposition 9.1. Let \(W\) be a compact neighbourhood of \(e\) and \(x\in x_0W\). Then \(R_xf\) vanishes off the compact set \((\operatorname{supp}f)W^{-1}x_0^{-1}\), and \(\|R_xf-R_{x_0}f\|_{\sup}=\|R_{x_0}(R_{x_0^{-1}x}f-f)\|_{\sup}=\|R_{x_0^{-1}x}f-f\|_{\sup}\), which tends to \(0\) as \(x\to x_0\) by uniform continuity (Proposition 7.3). So \(\int R_xf\,dt\to\int R_{x_0}f\,dt\). (4) \(\Delta(K)\) is a compact subgroup of \((0,\infty)\) for a compact subgroup \(K\). If it contained some \(r\neq1\), it would contain all powers \(r^n\), \(n\in\mathbb Z\), which form an unbounded set. So \(\Delta(K)=\{1\}\); take \(K=G\) for compact \(G\). For abelian \(G\), \(\mu(Ex)=\mu(xE)=\mu(E)\). On a discrete group, counting measure is invariant on both sides, and it is Radon: compact sets are finite and every set is open. The equivalence with right invariance is the definition of \(\Delta\). Finally, \(\Delta\) takes values in an abelian group, so it is \(1\) on every commutator, hence on the subgroup they generate, and by continuity on its closure \(N=[G,G]\). This \(N\) is closed, and it is normal because conjugates of commutators are commutators and conjugation is a homeomorphism. So \(\Delta=\bar\Delta\circ q\) for a homomorphism \(\bar\Delta\) on the locally compact group \(G/N\) (Proposition 7.1(5)), and \(\bar\Delta\) is continuous because \(q\) is open: \(\bar\Delta^{-1}(W)=q(\Delta^{-1}(W))\). If \(G/N\) is compact, \(\Delta(G)=\bar\Delta(G/N)\) is a compact subgroup of \((0,\infty)\), hence \(\{1\}\). \(\square\) In shorthand, (10.1) reads \(d\mu(tx)=\Delta(x)\,d\mu(t)\). ## 11. Inversion and right Haar measure **Theorem 11.1** (Inversion). For every Borel \(f\colon G\to[0,\infty]\), \[ \int f(t^{-1})\,dt=\int f(t)\,\Delta(t)^{-1}\,dt , \tag{11.1} \] and the same holds for every Borel \(f\) with \(\int|f|\Delta^{-1}\,d\mu<\infty\). Equivalently, the right Haar measure \(\tilde\mu(E)=\mu(E^{-1})\) equals \(\Delta^{-1}\mu\), that is, \(\tilde\mu(E)=\int_E\Delta(t)^{-1}\,dt\). Moreover \(\int f(t^{-1})\Delta(t)^{-1}\,dt=\int f(t)\,dt\) for every Borel \(f\geq0\). **Proof.** For \(f\in C_c(G)\) put \(f^\natural(t)=f(t^{-1})\Delta(t)^{-1}\), again in \(C_c(G)\), and \(J(f)=\int f^\natural\,d\mu\). The functional \(J\) is linear, and \(J(f)>0\) for \(f\in C_c^+(G)\) by Proposition 9.1. It is left invariant: with \(k=f^\natural\) and \(y\in G\), \[ (L_yf)^\natural(t)=f\big((ty)^{-1}\big)\Delta(t)^{-1}=f\big((ty)^{-1}\big)\Delta(ty)^{-1}\Delta(y)=\Delta(y)\,k(ty), \] so \(J(L_yf)=\Delta(y)\Delta(y)^{-1}\int k\,d\mu=J(f)\) by (10.1). So \(J\) is a left Haar integral, and by Proposition 8.2 and uniqueness (Theorem 9.2), \(J=c\int\cdot\,d\mu\) on \(C_c(G)\) for some \(c>0\). Now \((f^\natural)^\natural(t)=f^\natural(t^{-1})\Delta(t)^{-1}=f(t)\Delta(t)\Delta(t)^{-1}=f(t)\). So for \(f\in C_c^+(G)\), \(\int f\,d\mu=J(f^\natural)=c\int f^\natural\,d\mu=cJ(f)=c^2\int f\,d\mu\), and \(c=1\). Apply \(J=\int\cdot\,d\mu\) to \(g=f\Delta^{-1}\in C_c(G)\): since \(g(t^{-1})\Delta(t)^{-1}=f(t^{-1})\), this gives (11.1) for \(f\in C_c(G)\). The measure \(\tilde\mu=\iota_*\mu\) is Radon (Proposition 3.1(1)) with \(\int f\,d\tilde\mu=\int f(t^{-1})\,dt\). The measure \(\Delta^{-1}\mu\) is Radon by Proposition 3.1(2), because \(\Delta^{-1}\) is continuous and positive. By the previous paragraph they integrate every \(f\in C_c(G)\) alike, so they are equal by the uniqueness in Theorem 2.2. Integrating a Borel \(f\geq0\) against both (Proposition 3.1(1) and (2)) gives (11.1). For \(f\) with \(\int|f|\Delta^{-1}d\mu<\infty\), apply this to the four nonnegative parts of \(f\). For the last formula apply (11.1) to the Borel function \(t\mapsto f(t)\Delta(t)\), using \(\Delta(t^{-1})=\Delta(t)^{-1}\). \(\square\) Formula (11.1) holds for all nonnegative Borel functions, whatever their support. **Corollary 11.2.** 1. \(\mu\) and \(\tilde\mu\) have the same null sets and the same \(\sigma\)-finite sets. 2. For \(1\leq p<\infty\), \((S_pf)(t)=\Delta(t)^{-1/p}f(t^{-1})\) defines a linear isometry of \(L^p(\mu)\) onto itself with \(S_p^2=1\). For \(p=2\), \((J\xi)(t)=\Delta(t)^{-1/2}\overline{\xi(t^{-1})}\) is a conjugate-linear isometric involution of \(L^2(G)\). 3. \(f\mapsto\check f\) and \(f\mapsto\Delta^{1/p}f\) are isometries of \(L^p(\mu)\) onto \(L^p(\tilde\mu)\). 4. If \(G\) is not unimodular, \(\Delta\) is unbounded above and below. 5. The measure \((1+\Delta)\mu\) is a Radon measure, and \(C_c(G)\) is dense in \(L^2((1+\Delta)\mu)\). Consequently \(C_c(G)\) is a core for multiplication by \(\Delta^{1/2}\) on its maximal domain \(\{\xi\in L^2(G):\Delta^{1/2}\xi\in L^2(G)\}\), whose graph norm is \(\big(\int|\xi|^2(1+\Delta)\,d\mu\big)^{1/2}\). **Proof.** (1) Since \(\tilde\mu=\Delta^{-1}\mu\) with a continuous positive density, this is Proposition 3.1(2). (2) By the last formula of the theorem applied to \(|f|^p\), \(\int\Delta(t)^{-1}|f(t^{-1})|^p\,dt=\int|f|^p\,dt\). Also \(S_p(S_pf)(t)=\Delta(t)^{-1/p}\Delta(t^{-1})^{-1/p}f(t)=f(t)\), so \(S_p\) is onto. Complex conjugation commutes with \(S_2\) and is a conjugate-linear isometry. (3) \(\int|\check f|^p\,d\tilde\mu=\int|f(t)|^p\,dt\) by the definition of \(\tilde\mu\), and \(\int\Delta|f|^p\,d\tilde\mu=\int|f|^p\,d\mu\) by the theorem; the inverse maps have the same form. (4) \(\Delta(G)\) is a subgroup of \((0,\infty)\) containing some \(r\neq1\), hence all \(r^n\). (5) Apply Proposition 3.1(2) with the positive continuous density \(1+\Delta\), and then the density of \(C_c\) in \(L^2\) (Proposition 3.1(4)). The graph norm identity is immediate, and a function \(\xi\) lies in the maximal domain exactly when it lies in \(L^2((1+\Delta)\mu)\). \(\square\) **Example 11.3** (Explicit Haar measures on \(\mathbb R^\times\) and on the \(ax+b\) group). Integrals of continuous compactly supported functions of one real variable below are Riemann integrals, and substitutions use the change-of-variables rule for Riemann integrals (see "Background used without proof"). (a) On the multiplicative group \(\mathbb R^\times=\mathbb R\setminus\{0\}\), \(I(f)=\int f(x)\,dx/|x|\) is a Haar integral: the substitution \(u=cx\) gives \(\int f(cx)\,dx/|x|=\int f(u)\,du/|u|\) for \(c\neq0\). The group is abelian, so \(\Delta\equiv1\). (b) Let \(G=\{(a,b):a>0,\ b\in\mathbb R\}\) with \((a,b)(a',b')=(aa',ab'+b)\), identity \((1,0)\), inverse \((a,b)^{-1}=(1/a,-b/a)\), and the topology of the open half-plane. Put \(I(f)=\int_0^\infty\big(\int_{\mathbb R}f(a,b)\,db\big)a^{-2}\,da\). For \(x_0=(a_0,b_0)\), the substitutions \(u=a_0b+b_0\) (inner) and \(v=a_0a\) (outer) give \[ \int_0^\infty\!\!\int_{\mathbb R}f(a_0a,a_0b+b_0)\,db\,\frac{da}{a^2}=\int_0^\infty\frac1{a_0}\int_{\mathbb R}f(a_0a,u)\,du\,\frac{da}{a^2}=\int_0^\infty\!\!\int_{\mathbb R}f(v,u)\,du\,\frac{dv}{v^2}, \] so \(I\) is a left Haar integral, \(d\mu=a^{-2}\,da\,db\). The substitutions \(u=ab_0+b\) and \(v=aa_0\) give \(\int f(tx_0)\,d\mu(t)=a_0\int f\,d\mu\), so by (10.1) \[ \Delta(a,b)=1/a . \] The inversion formula (11.1) can be checked directly: the substitutions \(u=-b/a\) and \(v=1/a\) give \(\int f(t^{-1})\,d\mu(t)=\int_0^\infty\int_{\mathbb R}f(v,u)\,du\,v^{-1}\,dv\), which is \(\int f\Delta^{-1}\,d\mu\). So the right Haar measure is \(a^{-1}\,da\,db\). The commutators \((a,b)(a',b')(a,b)^{-1}(a',b')^{-1}\) have first coordinate \(1\), and they fill the normal subgroup \(\{1\}\times\mathbb R\); the quotient \(G/(\{1\}\times\mathbb R)\cong(0,\infty)\) is not compact, in line with Theorem 10.1(4). This group is a good test case for the signs in (10.1) and (11.1). ## 12. Products of groups The Radon product of Haar measures is a Haar measure. For two copies of \(G\) this gives the measure on \(G\times G\) behind convolution (Section 14); for infinitely many compact groups it gives the Haar measure of an infinite product. **Proposition 12.1** (The group \(G\times G\)). Keep \(G\) and its left Haar measure \(\mu\), and let \(\pi=\mu\hat\times\mu\) on \(G\times G\). 1. On the locally compact group \(G\times G\), \(\pi\) is invariant under left translations, so it is a left Haar measure; its modular function is \((s,t)\mapsto\Delta(s)\Delta(t)\). In the same way, if each \(G_j\) (\(j=1,\dots,n\)) carries a left Haar measure \(\mu_j\), the iterated Radon product \((\cdots(\mu_1\hat\times\mu_2)\cdots)\hat\times\mu_n\) is one on \(G_1\times\dots\times G_n\), and it integrates functions in \(C_c\) by iterated integrals. 2. For every continuous \(a\colon G\to G\), the map \(\Phi_a(s,t)=(s,a(s)t)\) is a homeomorphism of \(G\times G\) that preserves \(\pi\). So \((W_a\zeta)(s,t)=\zeta(s,a(s)t)\) is a unitary on \(L^2(G\times G,\pi)\), with inverse \(\zeta\mapsto\zeta(s,a(s)^{-1}t)\). 3. Under the unitary \(U\) of Theorem 6.2, the operator \(1\otimes\lambda(g)\) is \(W_a\) for the constant \(a\equiv g^{-1}\): \(\zeta(s,t)\mapsto\zeta(s,g^{-1}t)\). For \(a(s)=s\), \(W_a\) is the operator \((W\zeta)(s,t)=\zeta(s,st)\), with inverse \(\zeta(s,s^{-1}t)\). **Proof.** (1) For \(F\in C_c(G\times G)\) and \((g,h)\in G\times G\), \(\int\big(\int F(gs,ht)\,dt\big)ds=\int\big(\int F(gs,t)\,dt\big)ds=\int\big(\int F(s,t)\,dt\big)ds\), by left invariance applied first to the inner function \(t\mapsto F(gs,t)\) and then to the outer function \(s\mapsto\int F(s,t)\,dt\), which lies in \(C_c(G)\) by Proposition 4.2(1). So \(\pi\) is a left Haar measure by Proposition 8.2 (it is nonzero by (4.2)). In the same way, (10.1) gives \(\iint F(sg,th)\,dt\,ds=\Delta(g)^{-1}\Delta(h)^{-1}\iint F\,dt\,ds\), which identifies the modular function by Theorem 10.1(3). The statement for \(n\) factors follows by induction, using Proposition 4.2 for the iterated integral. (2) \(\Phi_a\) is continuous, with continuous inverse \((s,t)\mapsto(s,a(s)^{-1}t)\). For \(F\in C_c(G\times G)\), also \(F\circ\Phi_a\in C_c(G\times G)\), and \(\int F\circ\Phi_a\,d\pi=\int\big(\int F(s,a(s)t)\,dt\big)ds=\int\big(\int F(s,t)\,dt\big)ds\) by left invariance in \(t\) for each fixed \(s\). So the Radon measures \((\Phi_a)_*\pi\) (Proposition 3.1(1)) and \(\pi\) agree by the uniqueness in Theorem 2.2. Hence \(\zeta\mapsto\zeta\circ\Phi_a\) preserves null sets and the \(L^2(\pi)\) norm, and its inverse is composition with \(\Phi_a^{-1}\). (3) \(U(\xi\otimes L_g\eta)(s,t)=\xi(s)\eta(g^{-1}t)=U(\xi\otimes\eta)(s,g^{-1}t)\). Both operators are bounded and agree on elementary tensors. \(\square\) **Example 12.2** (Infinite products of compact groups, and \((\mathbb Z_2)^{\mathbb N}\) versus \([0,1]\)). Let \((G_\alpha)_{\alpha\in A}\) be compact groups with Haar measures \(\mu_\alpha(G_\alpha)=1\) (compact groups are unimodular by Theorem 10.1(4)). The product \(G=\prod_\alpha G_\alpha\), with the product topology and the group law taken coordinate by coordinate, is a compact group by Tychonoff's theorem. Write \(C_F(G)\) for the continuous functions that depend on only finitely many coordinates, \(f(x)=f_0(x_{\alpha_1},\dots,x_{\alpha_n})\) with \(f_0\) continuous. It is a self-adjoint algebra, it contains the constant functions, and it separates the points of \(G\): if \(x_\alpha\neq x'_\alpha\), (T2) on \(G_\alpha\) gives a continuous function of the coordinate \(x_\alpha\) that separates them. By the Stone–Weierstrass theorem, \(C_F(G)\) is dense in \(C(G)\). Put \(I(f)=\int f_0\,d(\mu_{\alpha_1}\hat\times\cdots\hat\times\mu_{\alpha_n})\). The value is independent of how \(f\) is written: by (4.1) the order of the factors is irrelevant, and a coordinate on which \(f_0\) does not depend contributes a factor \(\mu_\beta(G_\beta)=1\). So \(I\) is linear and positive on \(C_F(G)\), \(|I(f)|\leq\|f\|_{\sup}\), \(I(1)=1\), and \(I\) is left invariant because each finite product measure is a left Haar measure (Proposition 12.1(1)). It extends uniquely to a positive, left-invariant, norm-one functional on \(C(G)\) (positivity passes to the limit because \(\operatorname{Re}f_n+\|f-f_n\|_{\sup}\geq0\) when \(f\geq0\)). Its Radon measure is the Haar measure of \(G\) with total mass \(1\). For \(G=(\mathbb Z_2)^{\mathbb N}\), each factor with mass \(\frac12\) on each point, let \(\Phi(a)=\sum_ja_j2^{-j}\in[0,1]\). This map is continuous and onto, and it is one-to-one except over the dyadic rationals \(j2^{-k}\) in \((0,1)\), each of which has two preimages. It carries Haar measure to Lebesgue measure \(\lambda_1\) on \([0,1]\): \(\lambda_1(E)=\mu(\Phi^{-1}(E))\) for every Borel \(E\). Proof: \(\Phi_*\mu\) is a finite Borel measure on \([0,1]\). It is outer regular: by (R2) for \(\mu\) there is an open \(O\supseteq\Phi^{-1}(E)\) with \(\mu(O\setminus\Phi^{-1}(E))<\varepsilon\); then \(U=[0,1]\setminus\Phi(G\setminus O)\) is open (\(\Phi\) is a closed map, since \(G\) is compact), contains \(E\), and satisfies \(\Phi^{-1}(U)\subseteq O\), so \(\Phi_*\mu(U\setminus E)<\varepsilon\). A finite outer regular measure on a compact space is also inner regular on open sets (take complements), so \(\Phi_*\mu\) is Radon. For continuous \(f\) on \([0,1]\), the cylinder sets fixing \(a_1,\dots,a_k\) have measure \(2^{-k}\) and are mapped onto the intervals \([j2^{-k},(j+1)2^{-k}]\), so \(\int f\circ\Phi\,d\mu\) differs from the Riemann sum \(\sum_j2^{-k}f(j2^{-k})\) by at most the oscillation of \(f\) on intervals of length \(2^{-k}\). Letting \(k\to\infty\), \(\int f\,d\Phi_*\mu=\int_0^1f\,dx\). By the uniqueness in Theorem 2.2, \(\Phi_*\mu=\lambda_1\). ## 13. Groups that are not σ-compact When \(G\) is not \(\sigma\)-compact, its Haar measure is not \(\sigma\)-finite, and several familiar facts need care: inner regularity, the duality between \(L^1\) and \(L^\infty\), and Fubini's theorem. This section describes the measure through the cosets of an open \(\sigma\)-compact subgroup, sets up the right notion of \(L^\infty\), and ends with the examples that show what fails. Fix an open, closed, \(\sigma\)-compact subgroup \(G_0\) (Proposition 7.2) and a set \(Y\subseteq G\) that contains exactly one point of each left coset of \(G_0\). So the open \(\sigma\)-compact sets \(yG_0\), \(y\in Y\), partition \(G\), and \(\mu\) restricted to any one of them is \(\sigma\)-finite by (R1). The restriction of \(\mu\) to the Borel subsets of \(G_0\) is a left Haar measure of the locally compact group \(G_0\): the restriction of a Radon measure to an open set is Radon there, it is nonzero by Proposition 9.1, and it is invariant under left translation by elements of \(G_0\). **Proposition 13.1** (Cosets). Let \(E\subseteq G\) be Borel. 1. If \(E\) lies in countably many cosets \(y_jG_0\), then \(\mu(E)=\sum_j\mu(E\cap y_jG_0)\). 2. If \(E\) meets uncountably many cosets, then \(\mu(E)=\infty\). 3. Hence a Borel set is \(\sigma\)-finite exactly when it lies in countably many cosets. In particular every Borel set of finite measure does. Moreover \(G\) is \(\sigma\)-compact \(\iff\) \(\mu\) is \(\sigma\)-finite \(\iff\) \(Y\) is countable. **Proof.** (1) is countable additivity. (2) By (R2) it suffices to show \(\mu(U)=\infty\) for every open \(U\supseteq E\). Such \(U\) meets uncountably many cosets, and each nonempty open set \(U\cap yG_0\) has positive measure by Proposition 9.1. For some \(n\), uncountably many of these measures exceed \(1/n\); adding \(k\) of them gives \(\mu(U)\geq k/n\) for every \(k\). (3) A \(\sigma\)-finite set is covered by countably many sets of finite measure, each lying in countably many cosets by (2); conversely each coset is \(\sigma\)-finite. If \(Y\) is countable, countably many \(\sigma\)-compact cosets cover \(G\); if \(G\) is \(\sigma\)-compact, \(\mu\) is \(\sigma\)-finite by (R1); if \(\mu\) is \(\sigma\)-finite, \(Y\) is countable by the first claim. \(\square\) **Proposition 13.1A** (Comparing the two measure conventions). Let \(\mu\), \(G_0\) and \(Y\) be as in Proposition 13.1. On Borel subsets of \(G\) define \[ \mu_{\mathrm{in}}(E)=\sum_{y\in Y}\mu(E\cap yG_0), \] where an arbitrary nonnegative sum means the supremum of its finite partial sums. Then \(\mu_{\mathrm{in}}\) is a Borel measure, finite on compact sets and inner regular on every Borel set. It agrees with \(\mu\) on every \(\mu\)-\(\sigma\)-finite Borel set. It need not be outer regular on all Borel sets: for \(Y=\{0\}\times\mathbb R_d\) in Example 13.4, \(\mu_{\mathrm{in}}(Y)=0\) but every open neighbourhood of \(Y\) has infinite \(\mu_{\mathrm{in}}\)-measure. The two measures nonetheless give canonically isometric \(L^p\) spaces for \(1\leq p<\infty\), by taking representatives supported on countably many cosets. **Proof.** Countable additivity follows by interchanging two nonnegative sums: both sides are the supremum of the same finite sums over pairs \((y,n)\). A compact set meets only finitely many of the disjoint open cosets, so the sum is finite there and equals \(\mu\). On each coset, \(\mu\) is \(\sigma\)-finite, and Proposition 2.3 gives inner regularity for every Borel subset of that coset. Given a number \(c<\mu_{\mathrm{in}}(E)\), finitely many cosets already contribute more than \(c\); choosing one compact approximation inside each of them produces a compact subset of \(E\) with measure greater than \(c\). This proves inner regularity. Agreement on \(\mu\)-\(\sigma\)-finite sets follows from Proposition 13.1(1) and (3). In Example 13.4 every line contributes zero to \(Y\), while an open neighbourhood meets every line in a nonempty open interval. Its positive line measures have unbounded finite partial sums, proving the outer-regularity failure. If \(f\) is \(\mu_{\mathrm{in}}\)-\(p\)-integrable, the sum of its \(p\)-integrals on the cosets is finite. Only countably many of those integrals are positive: for each integer \(n\), only finitely many can exceed \(1/n\). Erasing \(f\) outside their countable union changes it on a \(\mu_{\mathrm{in}}\)-null set and gives a representative supported on countably many cosets. On that union the two measures agree, so its \(\mu\)-\(p\)-norm is the same. Conversely a \(\mu\)-\(p\)-integrable function is supported on countably many cosets by Proposition 3.1(4) and Proposition 13.1(3), and its two norms agree. If two such representatives have the same \(\mu_{\mathrm{in}}\)-class, their difference is supported on a countable union of cosets, where equality almost everywhere for one measure is equality for the other. The identification is therefore well defined and isometric in both directions. \(\square\) This explains how the inner-regular convention of [Fremlin, §416 and 441E] can be used for \(L^p\) comparison without replacing (R2), the measure of a locally null infinite set, or the product conventions in this lesson. The local \(L^\infty\) convention in Definition 13.2 retains the information on every finite-measure patch. In particular, the general Haar theorem in either convention does not justify unrestricted Borel Tonelli for the product used here. **Definition 13.2.** A set \(E\subseteq G\) is *locally Borel* when its intersection with each Borel set of finite measure is Borel, and *locally null* when, in addition, all these intersections are null. A property holds *locally almost everywhere* when the points where it fails form a locally null set. Call \(f\colon G\to\mathbb C\) *locally measurable* when the preimage of each Borel subset of \(\mathbb C\) is locally Borel. \(L^\infty(G)\) consists of the locally measurable \(f\) with \(|f|\leq c\) locally almost everywhere for some constant \(c\); two such functions are identified when they agree locally almost everywhere; and \(\|f\|_\infty\) is the infimum of those constants \(c\). Why the local notions? With the ordinary definitions, \(L^\infty\) need not be the dual of \(L^1\) when the measure is not \(\sigma\)-finite. For Haar measure on \(\mathbb R\times\mathbb R_d\), the indicator of the locally null set \(Y\) of Example 13.4 is a nonzero element of the ordinary \(L^\infty\), yet it integrates every \(L^1\) function to \(0\). The local versions repair this. **Theorem 13.3.** 1. \(E\) is locally Borel exactly when every \(E\cap yG_0\) is Borel, and locally null exactly when every \(E\cap yG_0\) is null. \(f\) is locally measurable exactly when every restriction \(f|_{yG_0}\) is Borel. A Borel set is locally null exactly when all its compact subsets are null. A locally null Borel set is null exactly when it is \(\sigma\)-finite. Countable unions of locally null sets are locally null, and \(|f|\leq\|f\|_\infty\) locally almost everywhere. 2. For \(f\in L^\infty(G)\) and \(g\in L^p(G)\), \(1\leq p<\infty\), the product \(fg\) is Borel, its class depends only on the classes of \(f\) and \(g\), and \(\|fg\|_p\leq\|f\|_\infty\|g\|_p\). 3. (*\(L^1{}^*=L^\infty\)*) For every bounded linear functional \(\Phi\) on \(L^1(G)\) there is a unique \(f\in L^\infty(G)\) with \(\Phi(g)=\int fg\,d\mu\) for all \(g\in L^1(G)\), and \(\|\Phi\|=\|f\|_\infty\). 4. The map \(f\mapsto m_f\), \(m_f\xi=f\xi\), is an isometric \(*\)-isomorphism of \(L^\infty(G)\) onto a von Neumann algebra on \(L^2(G)\) that is maximal abelian. 5. (*Radon–Nikodym, restricted form*) Let \(\nu\) and \(\mu'\) be Radon measures on an LCH space \(X\), with \(\nu\) \(\sigma\)-finite and \(\nu(E)=0\) whenever \(\mu'(E)=0\). Then \(\nu(E)=\int_Eh\,d\mu'\) for all Borel \(E\), for some Borel \(h\geq0\). **Proof.** (1) Each coset \(yG_0\) is the union of countably many compact sets \(yK_n\), each of finite measure, so \(E\cap yG_0=\bigcup_nE\cap yK_n\) is Borel (null) if \(E\) is locally Borel (locally null). Conversely, a Borel \(F\) of finite measure lies in countably many cosets \(y_jG_0\), so \(E\cap F=\bigcup_j(E\cap y_jG_0)\cap F\) is Borel (null). The statement about functions follows. If every compact subset of a Borel set \(E\) is null and \(\mu(F)<\infty\), then \(\mu(E\cap F)=0\) by inner regularity on sets of finite measure (Proposition 2.3(1)); conversely, compact sets have finite measure. A \(\sigma\)-finite locally null Borel set is the union of its intersections with countably many sets of finite measure, each null. If \(E_n\) are locally null and \(\mu(F)<\infty\), then \(\mu(\bigcup_nE_n\cap F)\leq\sum_n\mu(E_n\cap F)=0\). Finally \(\{|f|>\|f\|_\infty\}=\bigcup_n\{|f|>\|f\|_\infty+1/n\}\) is a countable union of locally null sets. (2) By Proposition 3.1(4), \(\{g\neq0\}\) is a disjoint union of Borel sets \(F_n\) of finite measure. Each \(f1_{F_n}\) is Borel, so \(fg=\sum_n(f1_{F_n})g\) is Borel (at each point at most one term is nonzero). If \(f=f'\) locally a.e., the set \(\{f\neq f'\}\cap\{g\neq0\}\) is Borel, \(\sigma\)-finite and locally null, hence null by (1); so \(fg=f'g\) a.e. In the same way \(|fg|\leq\|f\|_\infty|g|\) a.e. (3) For \(y\in Y\), the functions in \(L^1(G)\) that vanish off \(yG_0\) form a copy of \(L^1(yG_0,\mu)\), and \(\mu\) is \(\sigma\)-finite there. The duality \(L^1{}^*=L^\infty\) for \(\sigma\)-finite measures provides a Borel function \(f_y\) on \(yG_0\) with \(|f_y|\leq\|\Phi\|\) and \(\Phi(g)=\int_{yG_0}f_yg\,d\mu\) for those \(g\). Let \(f=f_y\) on \(yG_0\). By (1), \(f\) is locally measurable, and \(\|f\|_\infty\leq\|\Phi\|\). Let \(g\in L^1(G)\). The set \(\{g\neq0\}\) is \(\sigma\)-finite, so it lies in countably many cosets \(y_jG_0\), and \(g=\sum_jg1_{y_jG_0}\) in \(L^1\) by dominated convergence. Hence \(\Phi(g)=\sum_j\int_{y_jG_0}fg\,d\mu=\int fg\,d\mu\), again by dominated convergence, since \(|fg|\leq\|\Phi\||g|\). By (2) with \(p=1\), \(\|\Phi\|\leq\|f\|_\infty\). For uniqueness, suppose \(\int hg\,d\mu=0\) for all \(g\in L^1(G)\) but \(h\neq0\) on a set that is not locally null. Then for some \(c>0\) and some Borel \(F\) of finite measure, \(D=\{|h|>c\}\cap F\) has positive measure; with \(g=1_D\bar h/|h|\) we get \(\int hg\,d\mu=\int_D|h|\,d\mu>0\), a contradiction. (4) By (2), \(m_f\) is bounded with \(\|m_f\|\leq\|f\|_\infty\). If \(c<\|f\|_\infty\), the set \(\{|f|>c\}\) is not locally null, so for some Borel \(F\) of finite measure the set \(D=\{|f|>c\}\cap F\) has \(\mu(D)>0\); then \(\|m_f1_D\|_2^2\geq c^2\mu(D)=c^2\|1_D\|_2^2\). So \(m\) is isometric. Clearly \(m_{fg}=m_fm_g\) and \(m_{\bar f}=m_f^*\). Let \(T\) commute with every \(m_f\). Then \(T\) commutes with the projections \(P_y=m_{1_{yG_0}}\), so it maps each \(H_y=P_yL^2(G)\cong L^2(yG_0,\mu)\) into itself. On \(H_y\) it commutes with multiplication by every bounded Borel function on \(yG_0\) (extended by \(0\), such a function lies in \(L^\infty(G)\)). On a \(\sigma\)-finite measure space the multiplication operators form a maximal abelian algebra (Decomposable operators and the diagonal algebra). Applied to the \(\sigma\)-finite measure space \(yG_0\), this shows that \(T|_{H_y}\) is multiplication by a bounded Borel function \(f_y\) on \(yG_0\), which we may take with \(|f_y|\leq\|T\|\) by the norm formula just proved. Let \(f=f_y\) on \(yG_0\); then \(f\in L^\infty(G)\). A vector \(\xi\in L^2(G)\) lives on countably many cosets (Proposition 3.1(4) and Proposition 13.1(3)), so \(\xi=\sum_jP_{y_j}\xi\) and \(T\xi=\sum_jf_{y_j}P_{y_j}\xi=f\xi\). So \(T=m_f\). Thus \(m(L^\infty)'\subseteq m(L^\infty)\), and the reverse inclusion holds because \(m(L^\infty)\) is abelian. Hence \(m(L^\infty)'=m(L^\infty)\), which is maximal abelian, and \(m(L^\infty)''=m(L^\infty)\). (5) By Proposition 2.3(3), \(X=S\cup N\) with \(S\) \(\sigma\)-compact and \(\nu(N)=0\). On the Borel subsets of \(S\), \(\mu'\) is \(\sigma\)-finite by (R1), and the Radon–Nikodym theorem for \(\sigma\)-finite measures gives a Borel \(h_0\geq0\) on \(S\) with \(\nu(E)=\int_Eh_0\,d\mu'\) for Borel \(E\subseteq S\). Put \(h=h_0\) on \(S\) and \(h=0\) on \(N\). \(\square\) Part (4) carries the maximal abelian property of \(L^\infty\) from \(\sigma\)-finite measure spaces to Haar measure on every locally compact group, and no lifting theorem is needed. The following examples show what fails without \(\sigma\)-finiteness. In each of them the bad set is locally null, so it carries no \(L^p\) mass for \(p<\infty\). **Example 13.4** (\(\mathbb R\times\mathbb R_d\): a Haar measure that is not \(\sigma\)-finite). Let \(\mathbb R_d\) be the additive group of reals with the discrete topology, and \(G=\mathbb R\times\mathbb R_d\). The subgroup \(G_0=\mathbb R\times\{0\}\) is open and \(\sigma\)-compact, and its cosets are the lines \(\mathbb R\times\{d\}\). A compact subset of \(G\) meets only finitely many lines, so \(I(f)=\sum_d\int_{\mathbb R}f(x,d)\,dx\) is a finite sum for \(f\in C_c(G)\); it is a left Haar integral, by translation invariance of the Riemann integral. On each line, the Haar measure \(\mu\) is Lebesgue measure: the restriction of a Radon measure to an open set is Radon on that set, and both measures integrate the compactly supported continuous functions on the open set \(\mathbb R\times\{d\}\) alike. By Proposition 13.1, \(\mu(E)=\sum_d\lambda_1(\{x:(x,d)\in E\})\) if \(E\) meets countably many lines, and \(\mu(E)=\infty\) otherwise. The closed set \(Y=\{0\}\times\mathbb R_d\) meets every line, so \(\mu(Y)=\infty\). Its compact subsets are finite and null. So \(\mu\) is not inner regular on \(Y\), and \(Y\) is locally null (Theorem 13.3(1)) but not null. No measure that is inner and outer regular on every Borel set can represent the Haar integral of \(G\): it would be Radon, hence equal to \(\mu\) by the uniqueness in Theorem 2.2, and \(\mu\) fails inner regularity on \(Y\). Nevertheless every \(f\in L^p(G)\), \(p<\infty\), lives on countably many lines; \(C_c(G)\) is dense in \(L^p(G)\) (Proposition 3.1(4)); and \(L^\infty(G)\) consists of the functions whose restrictions to the lines are Borel and essentially bounded, uniformly in \(d\); it is the dual of \(L^1(G)\) (Theorem 13.3(3)). **Example 13.5** (The diagonal of \(\mathbb R_d\times\mathbb R\): Tonelli's theorem needs \(\sigma\)-finiteness). Let \(X=\mathbb R_d\) with counting measure \(c\) and \(Y=\mathbb R\) with Lebesgue measure \(m\). Their Radon product is the Haar measure of the group \(\mathbb R_d\times\mathbb R\) (Proposition 12.1(1)). The diagonal \(D=\{(t,t)\}\) is a closed subgroup. Its sections are single points: \(m(D_s)=0\) for every \(s\), while \(c(D^t)=1\) for every \(t\). So \[ \int m(D_s)\,dc(s)=0,\qquad\int c(D^t)\,dm(t)=\infty,\qquad(c\hat\times m)(D)=\infty, \] the last because \(D\) meets every coset \(\{s\}\times\mathbb R\) of the open subgroup \(\{0\}\times\mathbb R\) (Proposition 13.1(2)). Thus Tonelli's theorem 5.1(4) fails for \(1_D\), whose support is not \(\sigma\)-finite. The compact subsets of \(D\) are finite, and each point has measure \(c(\{s\})m(\{s\})=0\) by (4.2); so \(D\) is locally null (Theorem 13.3(1)). The example also shows that the rectangle function \(A\times B\mapsto c(A)m(B)\) has no extension to a measure \(m'\) on \(\mathcal B(X\times Y)\) that is inner and outer regular on every Borel set. Such an \(m'\) would be Radon. It would integrate \(\varphi\boxtimes g\) to \(\int\varphi\,dc\int g\,dm\) for \(\varphi\in C_c(X)\) and \(g\in C_c(Y)\), by uniform approximation with simple functions on the compact rectangle \(\operatorname{supp}\varphi\times\operatorname{supp}g\), which has finite \(m'\)-measure. By the tensor approximation (Lemma 4.1) it would then agree with \(c\hat\times m\) on \(C_c(X\times Y)\), hence equal it by the uniqueness in Theorem 2.2. But inner regularity would force \(m'(D)=0\). **Example 13.6** (\(\{e\}\times G\): the rectangle rule needs \(\sigma\)-finite factors). Let \(G=\mathbb R\times\mathbb R_d\) and \(\pi=\mu\hat\times\mu\). The closed set \(\{e\}\times G\) has \(\mu(\{e\})=0\) but \(\pi(\{e\}\times G)=\infty\). Indeed \(\pi\) is a Haar measure on \(G\times G\) (Proposition 12.1(1)), \(G_0\times G_0\) is an open \(\sigma\)-compact subgroup of \(G\times G\), and \(\{e\}\times G\) meets each of the uncountably many cosets \(G_0\times(\mathbb R\times\{d\})\); apply Proposition 13.1(2). So Theorem 5.1(6) fails when \(B\) is not \(\sigma\)-finite, and "\(\pi(A\times B)=\mu(A)\mu(B)\) with \(0\cdot\infty=0\)" is false for the Radon product. The set is locally null by Theorem 13.3(1): its intersection with a compact \(K\subseteq G\times G\) lies in \(\{e\}\times\mathrm{pr}_2(K)\), which is null by (4.2). So, like \(Y\) and \(D\), it carries no \(L^2\) mass. ## 14. Convolution and continuity of translation In this section and the next, \(\pi=\mu\hat\times\mu\) on \(G\times G\), and integrals over \(G\) are taken with respect to \(\mu\). Convolution integrals are iterated integrals over \(G\times G\) in disguise, so we first record how \(\pi\) behaves under the changes of variables that occur. **Lemma 14.1** (Changes of variables on \(G\times G\)). The maps \[ \Theta_1(x,y)=(y,y^{-1}x),\qquad\Theta_2(x,y)=(y,xy^{-1}),\qquad\Theta_3(x,y)=(y,xy),\qquad\sigma(x,y)=(y,x) \] are homeomorphisms of \(G\times G\) with \((\Theta_1)_*\pi=\sigma_*\pi=\pi\), \((\Theta_2)_*\pi=(\Delta\circ\mathrm{pr}_1)\,\pi\) and \((\Theta_3)_*\pi=(\Delta^{-1}\circ\mathrm{pr}_1)\,\pi\). Consequently each of them pulls \(\pi\)-null sets back to \(\pi\)-null sets and \(\sigma\)-finite sets back to \(\sigma\)-finite sets, and for every Borel \(H\geq0\) on \(G\times G\) \[ \int H\circ\Theta_1\,d\pi=\int H\,d\pi,\qquad\int H\circ\Theta_2\,d\pi=\int H(y,z)\Delta(y)\,d\pi(y,z),\qquad\int H\circ\Theta_3\,d\pi=\int H(y,z)\Delta(y)^{-1}\,d\pi(y,z). \] **Proof.** The inverses are \((y,z)\mapsto(yz,y)\), \((zy,y)\) and \((zy^{-1},y)\), all continuous. For \(F\in C_c(G\times G)\), (4.1) lets us integrate first in \(x\); left invariance and (10.1) in the inner integral give \[ \int F\circ\Theta_1\,d\pi=\int\!\!\int F(y,y^{-1}x)\,dx\,dy=\int F\,d\pi,\qquad \int F\circ\Theta_2\,d\pi=\int\!\!\int F(y,xy^{-1})\,dx\,dy=\int\Delta(y)\!\int F(y,x)\,dx\,dy, \] and similarly \(\int F\circ\Theta_3\,d\pi=\int\Delta(y)^{-1}\int F(y,x)\,dx\,dy\) and \(\int F\circ\sigma\,d\pi=\int F\,d\pi\). The image measures are Radon by Proposition 3.1(1), and so are \((\Delta^{\pm1}\circ\mathrm{pr}_1)\pi\) by Proposition 3.1(2). The uniqueness in Theorem 2.2 gives the four identities of measures. Densities that are continuous and positive do not change null sets or \(\sigma\)-finite sets (Proposition 3.1(2)). The integral formulas follow from Proposition 3.1(1) and (2). \(\square\) For Borel functions \(f,g\) on \(G\), put \[ f*g(x)=\int f(y)\,g(y^{-1}x)\,dy \tag{14.1} \] at every \(x\) where the integral converges absolutely. For \(f,g\geq0\) the integral is defined at every \(x\), with values in \([0,\infty]\). **Theorem 14.2.** 1. (*Tonelli for convolutions*) If \(f,g\geq0\) are Borel and vanish outside \(\sigma\)-finite sets, then \(f*g\) is \(\mu\)-a.e. Borel and \(\int f*g\,dx=\int f\,dx\int g\,dx\). 2. For \(f,g\in L^1(G)\), (14.1) converges absolutely for almost every \(x\), \(\|f*g\|_1\leq\|f\|_1\|g\|_1\), and the class of \(f*g\) depends only on the classes of \(f\) and \(g\). If one of the following integrals converges absolutely, so do the others, and \[ f*g(x)=\int f(xy)g(y^{-1})\,dy=\int f(y^{-1})g(yx)\Delta(y^{-1})\,dy=\int f(xy^{-1})g(y)\Delta(y^{-1})\,dy. \tag{14.2} \] 3. With the involution \(f^*(x)=\Delta(x)^{-1}\overline{f(x^{-1})}\), \(L^1(G)\) is a Banach \(*\)-algebra: convolution is associative, \(\|f^*\|_1=\|f\|_1\), \(f^{**}=f\) and \((f*g)^*=g^**f^*\). Moreover \(L_z(f*g)=(L_zf)*g\) and \(R_z(f*g)=f*(R_zg)\). 4. Let \(1\leq p<\infty\), \(f\in L^1(G)\) and \(g\in L^p(G)\). Then \(f*g(x)\) converges absolutely for almost every \(x\), and \(\|f*g\|_p\leq\|f\|_1\|g\|_p\). If also \(\int|f|\Delta^{-1}\,dx<\infty\) (for instance if \(f\) has compact support, or if \(G\) is unimodular), then \(g*f(x)=\int g(xy^{-1})f(y)\Delta(y^{-1})\,dy\) converges absolutely for almost every \(x\), and \[ \|g*f\|_p\leq\|\Delta^{-1}f\|_1^{1-1/p}\,\|f\|_1^{1/p}\,\|g\|_p . \tag{14.3} \] For \(p=1\) the condition on \(f\) is not needed, and \(\|g*f\|_1\leq\|g\|_1\|f\|_1\). 5. If \(G\) is unimodular, \(11\), Hölder's inequality for the measure \(w(y)\,dy\) of mass \(\|\Delta^{-1}f\|_1\) gives \(\big(\int|g(xy^{-1})|w(y)\,dy\big)^p\leq\|\Delta^{-1}f\|_1^{p-1}\int|g(xy^{-1})|^pw(y)\,dy\), and integrating in \(x\) gives (14.3). If \(f\) vanishes off a compact set \(K\), then \(\|\Delta^{-1}f\|_1\leq(\sup_K\Delta^{-1})\|f\|_1\), and (14.3) gives \(\|g*f\|_p\leq(\sup_K\Delta^{(1/p)-1})\|f\|_1\|g\|_p\). (5) For every \(x\), Hölder's inequality gives \(|f*g(x)|\leq\|f\|_p\big(\int|g(y^{-1}x)|^q\,dy\big)^{1/q}\). By (11.1) and unimodularity, \(\int|g(y^{-1}x)|^q\,dy=\int|g(yx)|^q\Delta(y)^{-1}\,dy=\|g\|_q^q\). For \(f,g\in C_c(G)\), \(f*g\in C_c(G)\) by (3). For general \(f,g\), take \(f_n,g_n\in C_c(G)\) with \(f_n\to f\) in \(L^p\) and \(g_n\to g\) in \(L^q\) (Proposition 3.1(4)). The bound gives \(f_n*g_n\to f*g\) uniformly, and a uniform limit of functions in \(C_c(G)\) lies in \(C_0(G)\). The endpoint cases fail on every noncompact \(G\): for \(f\geq0\) with \(\int f=1\) and \(g\equiv1\), \(f*g\equiv1\notin C_0(G)\). (6) Fix a compact neighbourhood \(V\) of \(e\). For \(g\in C_c(G)\) and \(y\in V\), \(L_yg\) and \(R_yg\) vanish outside the compact set \(K=V(\operatorname{supp}g)\cup(\operatorname{supp}g)V^{-1}\), so \(\|L_yg-g\|_p\leq\mu(K)^{1/p}\|L_yg-g\|_{\sup}\to0\) by uniform continuity (Proposition 7.3), and the same for \(R_y\). For \(f\in L^p\) and \(\varepsilon>0\), choose \(g\in C_c(G)\) with \(\|f-g\|_p<\varepsilon\). Since \(\|L_y\|=1\), and \(\|R_y\|=\Delta(y)^{-1/p}\leq C\) on \(V\) by Theorem 10.1(3) and the continuity of \(\Delta\), we get \(\|R_yf-f\|_p\leq(C+1)\varepsilon+\|R_yg-g\|_p\), and similarly for \(L_y\). (7) Left and right translations and inversion map Borel sets to Borel sets and preserve null sets and \(\sigma\)-finite sets (Theorem 10.1(1) and Corollary 11.2(1)). Hence they map locally Borel and locally null sets to sets of the same kind: for such a map \(\theta\) and \(\mu(F)<\infty\), the set \(\theta^{-1}(F)\) is covered by countably many sets of finite measure, so \(\theta(E)\cap F=\theta\big(E\cap\theta^{-1}(F)\big)\) is Borel (null) when \(E\) is locally Borel (locally null). The map \(y\mapsto y^{-1}x\) and its inverse \(z\mapsto xz^{-1}\) are composites of an inversion and a translation, so \(y\mapsto g(y^{-1}x)\) is locally measurable and bounded by \(\|g\|_\infty\) locally almost everywhere. As in the proof of Theorem 13.3(2), multiplying by \(f\) gives an integrable function with \(|f*g(x)|\leq\|f\|_1\|g\|_\infty\). By (3), \(L_z(f*g)-f*g=(L_zf-f)*g\), and \(\|L_zf-f\|_1\to0\) by (6). For \(g*f\), the substitutions of (2) give \(\int|g(y)||f(y^{-1}x)|\,dy\leq\|g\|_\infty\int|f(y^{-1}x)|\,dy=\|g\|_\infty\int|f(y)|\Delta(y)^{-1}\,dy\). Also \(R_z(g*f)-g*f=g*(R_zf-f)\), and \(\Delta^{-1}R_zf=\Delta(z)R_z(\Delta^{-1}f)\), so \[ \|\Delta^{-1}(R_zf-f)\|_1\leq|\Delta(z)-1|\,\|R_z(\Delta^{-1}f)\|_1+\|R_z(\Delta^{-1}f)-\Delta^{-1}f\|_1\to0 \] by (6) applied to \(\Delta^{-1}f\in L^1\). \(\square\) **Remark 14.3** (The factor \(\Delta^{-1}\) in (7) is needed). In the second half of (7) the bound is \(|R_z(g*f)-g*f|\leq\|g\|_\infty\|\Delta^{-1}(R_zf-f)\|_1\), with \(\Delta^{-1}(R_zf-f)\) where one might expect \(R_zf-f\). This cannot be improved. For \(h\in L^1\) and fixed \(x\), the third form of (14.2) gives \(g*h(x)=\int g(xy^{-1})h(y)\Delta(y)^{-1}\,dy\). The choice \(g(w)=\overline{\operatorname{sgn}h(w^{-1}x)}\) shows that the supremum of \(|g*h(x)|\) over \(\|g\|_\infty\leq1\) is \(\|\Delta^{-1}h\|_1\), which exceeds \(\|h\|_1\) whenever \(h\neq0\) lives where \(\Delta<1\). Such \(h=R_zf-f\) occur on every group that is not unimodular: take \(f\in C_c^+(G)\) supported in the nonempty open set \(\{\Delta<1\}\), and \(z\) with \(\Delta(z)>1\). Then \(R_zf\) is supported in \((\operatorname{supp}f)z^{-1}\), where \(\Delta\) is smaller still, and \(R_zf\neq f\) because \(\int R_zf=\Delta(z)^{-1}\int f\). So the bound \(\|g\|_\infty\|R_zf-f\|_1\) fails for this \(f\), this \(z\) and a suitable \(g\). The extra hypothesis \(\Delta^{-1}f\in L^1\) cannot be dropped either: Exercise 16.4 gives \(f\in L^1\) on the \(ax+b\) group with \(1*f\equiv\infty\). For unimodular groups \(\Delta\equiv1\), and nothing changes. The first half of (7), and (4) for \(f*g\), hold for all groups. ## 15. Approximate identities **Theorem 15.1.** Let the neighbourhoods \(U\) of \(e\) be directed by reverse inclusion. 1. For every neighbourhood \(U\) of \(e\) there is \(\psi_U\in C_c(G)\) with \(\psi_U\geq0\), \(\operatorname{supp}\psi_U\subseteq U\), \(\int\psi_U=1\) and \(\psi_U(x^{-1})=\psi_U(x)\). 2. Let \((\psi_U)\) be any family with \(\psi_U\geq0\), \(\int\psi_U=1\), and \(\psi_U\) vanishing outside a compact subset of \(U\) (continuity is not required). For \(1\leq p<\infty\) and \(f\in L^p(G)\), \[ \|\psi_U*f-f\|_p\leq\sup_{y\in U}\|L_yf-f\|_p\longrightarrow0 . \tag{15.1} \] If \(f\) is bounded with \(\|L_yf-f\|_{\sup}\to0\) as \(y\to e\), then \(\|\psi_U*f-f\|_{\sup}\to0\). 3. If moreover \(\psi_U(x^{-1})=\psi_U(x)\), then \(\|f*\psi_U-f\|_p\leq\sup_{y\in U}\|R_yf-f\|_p\to0\), and \(\|f*\psi_U-f\|_{\sup}\to0\) for bounded \(f\) with \(\|R_yf-f\|_{\sup}\to0\). 4. (*Weak-integral form*) For \(\psi\in L^1(G)\) and \(\xi,\eta\in L^2(G)\), \[ \langle\psi*\xi,\eta\rangle=\int\psi(s)\,\langle L_s\xi,\eta\rangle\,ds , \tag{15.2} \] where \(s\mapsto\langle L_s\xi,\eta\rangle\) is continuous and bounded by \(\|\xi\|\|\eta\|\). So the operator \(\xi\mapsto\psi*\xi\) is the weak integral \(\int\psi(s)\lambda(s)\,ds\) of the left regular representation, its norm is at most \(\|\psi\|_1\), and \(\psi_U*\xi\to\xi\) in \(L^2\) for every \(\xi\). **Proof.** (1) The set \(U'=\operatorname{int}U\cap(\operatorname{int}U)^{-1}\) is an open neighbourhood of \(e\). By (T2) choose \(\varphi\) with \(\{e\}\prec\varphi\prec U'\), put \(\varphi'=\varphi+\check\varphi\), which vanishes off \(\operatorname{supp}\varphi\cup(\operatorname{supp}\varphi)^{-1}\subseteq U'\), and \(\psi_U=\varphi'/\int\varphi'\); the integral is positive by Proposition 9.1. (2) Let \(C\) be a compact set outside of which \(\psi=\psi_U\) vanishes. For almost every \(x\), (14.1) converges absolutely by Theorem 14.2(4), and since \(\int\psi=1\), \(\psi*f(x)-f(x)=\int\psi(y)\big(f(y^{-1}x)-f(x)\big)dy\). Hölder's inequality for the probability measure \(\psi(y)\,dy\) gives \(|\psi*f(x)-f(x)|^p\leq\int\psi(y)|f(y^{-1}x)-f(x)|^p\,dy\). The function \((x,y)\mapsto\psi(y)|f(y^{-1}x)-f(x)|^p\) is Borel and vanishes outside \((\{f\neq0\}\times C)\cup\Theta_1^{-1}(C\times\{f\neq0\})\), which is \(\sigma\)-finite by Theorem 5.1(6) and Lemma 14.1. By Tonelli's theorem, \[ \|\psi*f-f\|_p^p\leq\int\psi(y)\Big(\int|f(y^{-1}x)-f(x)|^p\,dx\Big)dy=\int\psi(y)\,\|L_yf-f\|_p^p\,dy\leq\sup_{y\in U}\|L_yf-f\|_p^p , \] which tends to \(0\) by the continuity of translation (Theorem 14.2(6)). For the uniform statement, \(|\psi*f(x)-f(x)|\leq\int\psi(y)|L_yf(x)-f(x)|\,dy\leq\sup_{y\in U}\|L_yf-f\|_{\sup}\) at every \(x\). (3) By the first form of (14.2) and the symmetry of \(\psi\), \(f*\psi(x)=\int f(xy)\psi(y)\,dy\), so \(f*\psi(x)-f(x)=\int\psi(y)(R_yf(x)-f(x))\,dy\); the integral converges absolutely for almost every \(x\) by Theorem 14.2(4), since \(\psi\) has compact support. Repeat the argument of (2), with \(\Theta_3\) in place of \(\Theta_1\) and \(\int|f(xy)-f(x)|^p\,dx=\|R_yf-f\|_p^p\). (4) The function \((x,s)\mapsto\psi(s)\xi(s^{-1}x)\overline{\eta(x)}\) is Borel, vanishes outside a \(\sigma\)-finite set (as in (2)), and by Tonelli's theorem and the Cauchy–Schwarz inequality in \(x\) it is \(\pi\)-integrable with integral of its absolute value at most \(\int|\psi(s)|\,\|L_s\xi\|\,\|\eta\|\,ds=\|\psi\|_1\|\xi\|\|\eta\|\). Fubini's theorem 5.1(5) gives (15.2). Continuity: \(|\langle L_s\xi-L_{s_0}\xi,\eta\rangle|\leq\|L_{s_0^{-1}s}\xi-\xi\|\,\|\eta\|\to0\) by Theorem 14.2(6). The norm bound is Theorem 14.2(4) with \(p=2\), and the convergence is (2). \(\square\) *Units.* If \(G\) is discrete (with counting measure), \(\delta=1_{\{e\}}\) satisfies \(\delta*f=f*\delta=f\). If \(G\) is not discrete, \(L^1(G)\) has no unit, which is why approximate identities are needed. Indeed, suppose \(u*f=f\) for all \(f\in L^1(G)\). Taking \(f=\psi_U\) symmetric, \(\psi_U=u*\psi_U\to u\) in \(L^1\) by (3). For a compact \(K\) with \(e\notin K\), \(\int_K|\psi_U|=0\) as soon as \(U\cap K=\varnothing\), so \(\int_K|u|=0\). By inner regularity on the \(\sigma\)-finite set \(\{u\neq0\}\) (Proposition 2.3(2)), \(u=0\) almost everywhere off \(\{e\}\). And \(\mu(\{e\})=0\): otherwise every point would have the same positive measure (by left invariance), a compact neighbourhood of \(e\) would be finite by (R1), and \(\{e\}\) would be open. So \(u=0\) almost everywhere, and \(u*f=0\neq f\) for \(f\neq0\), a contradiction. A family as in (1) is a *compactly supported approximate identity*. Its translates behave well: for \(g\in G\), \((L_g\psi_U)*\xi=L_g(\psi_U*\xi)\) by Theorem 14.2(3), so convolution by \(L_g\psi_U\) is \(\lambda(g)\) composed with convolution by \(\psi_U\). ## 16. Exercises **Exercise 16.1** (A density that vanishes on a closed set). On \(G=\mathbb R\times\mathbb R_d\) with Haar measure \(\mu\) (Example 13.4), let \(\varphi(x,d)=|x|\) and \(\nu(E)=\int_E\varphi\,d\mu\). Show that \(\nu\) is a Borel measure, finite on compact sets, with \(\nu(Y)=0\) for \(Y=\{0\}\times\mathbb R_d\) but \(\nu(U)=\infty\) for every open \(U\supseteq Y\). Conclude that the positivity of the density in Proposition 3.1(2) cannot be dropped. *Solution.* \(\nu\) is a measure because \(\varphi\) is a nonnegative Borel function. A compact set \(K\) meets finitely many lines and \(\varphi\) is bounded on it, so \(\nu(K)\leq\max_K\varphi\cdot\mu(K)<\infty\). Since \(\varphi=0\) on \(Y\), \(\nu(Y)=0\). If \(U\supseteq Y\) is open, then for each \(d\) there is \(\delta_d>0\) with \((-\delta_d,\delta_d)\times\{d\}\subseteq U\), and \(\nu\big((-\delta_d,\delta_d)\times\{d\}\big)=\int_{-\delta_d}^{\delta_d}|x|\,dx=\delta_d^2>0\), because \(\mu\) is Lebesgue measure on each line. These sets are disjoint and uncountably many, so, as in the proof of Proposition 13.1(2), \(\nu(U)=\infty\). So \(\nu\) fails (R2) at \(Y\). **Exercise 16.2** (Closed ideals of \(L^1(G)\)). Let \(\mathcal J\subseteq L^1(G)\) be a closed subspace. Show that \(\mathcal J\) is a left ideal (\(g*f\in\mathcal J\) for \(g\in L^1\), \(f\in\mathcal J\)) exactly when \(L_x\mathcal J\subseteq\mathcal J\) for all \(x\), and a right ideal exactly when \(R_x\mathcal J\subseteq\mathcal J\) for all \(x\). *Solution.* *Left, only if.* For \(f\in\mathcal J\) and \(x\in G\), \(L_x(\psi_U*f)=(L_x\psi_U)*f\in\mathcal J\) by Theorem 14.2(3), and \(L_x(\psi_U*f)\to L_xf\) in \(L^1\) by Theorem 15.1(2), since \(L_x\) is isometric. As \(\mathcal J\) is closed, \(L_xf\in\mathcal J\). *Left, if.* Since \(C_c(G)\) is dense in \(L^1\) and \(\|g*f\|_1\leq\|g\|_1\|f\|_1\), it suffices to take \(g\in C_c(G)\). Let \(K=\operatorname{supp}g\) and \(\varepsilon>0\). By Theorem 14.2(6), \(y\mapsto L_yf\) is continuous into \(L^1\), so each \(y\in K\) has an open neighbourhood \(O_y\) with \(\|L_{y'}f-L_yf\|_1<\varepsilon\) for \(y'\in O_y\). Cover \(K\) by \(O_{y_1},\dots,O_{y_n}\), and let \(E_j=(K\cap O_{y_j})\setminus\bigcup_{i1,\ 0