Finite Radon products and convolution of measures

Programme reading HA-LCA-PRE-PRODUCT. The spaces below are arbitrary locally compact Hausdorff spaces. The product construction uses finite measures; Corollary 2.3 also proves compact localization for measures finite on compact sets. There is no assumption of metrizability, a countable base, or a countable cover of an ambient group by sets of finite Haar measure.

We use the earlier finite-Radon representation reading: Lemmas 1.1, 1.3, 1.5 and 1.6 for compact sets, cutoffs and function spaces; Lemmas 2.1–2.2 for finite measures and bounded integration; Theorems 3.1 and 4.3 for positive and complex representation; and Lemma 4.1, Proposition 4.4 and Corollary 4.5 for variation, Jordan decomposition and compact tails. These are the exact proof dependencies.

The same results are treated in D. H. Fremlin, Measure Theory, Volume 4, §417, the 9 March 2010 version in the 2013 collection: 417H, 417P and 417U–417V. The result labels belong to this exact version. We give a direct finite-measure construction using the earlier representation theorem, and prove only the Fubini forms stated below. In particular, we do not identify the product Borel sigma-algebra with the sigma-algebra generated by Borel rectangles.

Adaptation and additional proofs: GPT-6 Astra (OpenAI), Ultra, 4 October 2026. Fremlin's original source and copyright notices are retained in the source package. This combined reading is under the Design Science License, including its warranty disclaimer.

Write \(M(X)\) for the complex finite Radon measures on \(X\), with norm \(\|\mu\|=|\mu|(X)\). For a positive finite Radon measure, inner regularity applies to every Borel set and outer regularity follows as in the earlier reading.

1. Integrating a continuous kernel

Lemma 1.1 (sections and iterated integration). If \(h\in C_c(X\times Y)\) and \(\mu\in M(X),\nu\in M(Y)\), then \[ g(x)=\int_Y h(x,y)\,d\nu(y)\in C_c(X). \] The analogous assertion holds with the coordinates reversed, and \[ \int_X\!\left(\int_Yh(x,y)\,d\nu(y)\right)d\mu(x) =\int_Y\!\left(\int_Xh(x,y)\,d\mu(x)\right)d\nu(y). \tag{1} \] Both sides have absolute value at most \(\|h\|_\infty\|\mu\|\|\nu\|\).

Proof. The product is locally compact Hausdorff: products of compact neighbourhoods are compact by the earlier Lemma 1.6, and products of their interiors are neighbourhoods; distinct points are separated in a coordinate. The coordinate projections of \(\operatorname{supp}h\) are compact, because an inverse-image open cover and a finite subcover prove that continuous images of compact sets are compact. Call the projections \(K,L\).

Every section is continuous with compact support. The earlier Lemma 1.6, applied to \(h(x,y)-h(x_0,y)\) on the compact set \(L\), gives \[ |g(x)-g(x_0)| \le\|\nu\|\sup_{y\in L}|h(x,y)-h(x_0,y)|\longrightarrow0. \] The function \(g\) vanishes outside the closed compact set \(K\), so is in \(C_c(X)\). The norm bound for bounded integration gives the asserted estimate on the iterated integrals.

To prove (1), fix \(\varepsilon>0\). A finite family of open rectangles covers \(K\times L\) such that the oscillation of \(h\) in each rectangle is less than \(\varepsilon\). This follows from continuity and compactness of \(K\times L\). Partition \(K\) according to membership in the finitely many first-coordinate open sets, and \(L\) according to membership in the second-coordinate open sets. The nonempty pieces \(E_i,F_j\) are Borel. Choose \(x_i\in E_i,y_j\in F_j\). Each pair \((x,y)\in E_i\times F_j\) and its chosen pair \((x_i,y_j)\) belong to the same covering rectangle, so their values differ by less than \(\varepsilon\).

Thus the rectangle-simple function \[ s(x,y)=\sum_{i,j}h(x_i,y_j)1_{E_i}(x)1_{F_j}(y), \] extended by zero off \(K\times L\), approximates \(h\) uniformly within \(\varepsilon\) on all of \(X\times Y\). Its two iterated integrals are the same finite sum \(\sum_{i,j}h(x_i,y_j)\mu(E_i)\nu(F_j)\). The bounded-integration estimate makes the error on each side at most \(\varepsilon\|\mu\|\|\nu\|\). Let \(\varepsilon\) tend to zero. No product measure has been presumed in this argument. If either projection is empty, \(h=0\) and all assertions are immediate. \(\square\)

Lemma 1.2 (controlled variation and finite restrictions). If a countably additive complex Borel measure \(\sigma\) has variation bounded above by a finite positive Radon measure \(\rho\), then \(\sigma\) is a complex finite Radon measure. If \(\rho\) is finite positive Radon and \(E\) is Borel, then \(B\mapsto\rho(B\cap E)\) is finite positive Radon. Every complex finite Radon measure is a complex linear combination of four positive finite Radon measures.

Proof. The assumed domination makes variation finite, so the earlier Lemma 4.1 makes it a measure. Given Borel \(B\), choose compact \(K\subset B\) and open \(U\supset B\) with \(\rho(U\setminus K)<\varepsilon\). Then \[ |\sigma|(B)-|\sigma|(K)\le\rho(B\setminus K)<\varepsilon,\qquad |\sigma|(U)-|\sigma|(B)\le\rho(U\setminus B)<\varepsilon. \] These are precisely inner and outer regularity of variation. The restriction in the second assertion is a measure by countable additivity; its variation is itself and is bounded above by \(\rho\), so the first assertion applies. For a complex measure \(\mu\), its real and imaginary parts are real signed measures with variations bounded by \(|\mu|\): for any finite Borel partition, the sum of absolute real parts, or of absolute imaginary parts, is bounded by the sum of the absolute complex values. They are Radon by the first assertion. Apply the earlier Proposition 4.4 to each part to obtain \(\mu=\mu_1-\mu_2+i\mu_3-i\mu_4\) with all \(\mu_j\) positive finite Radon. \(\square\)

2. Constructing the product on all Borel sets

Theorem 2.1 (finite Radon product). There is a unique complex finite Radon measure \(\mu\otimes\nu\) on \(X\times Y\) satisfying \[ (\mu\otimes\nu)(E\times F)=\mu(E)\nu(F) \tag{2} \] for Borel \(E\subseteq X,F\subseteq Y\). For \(h\in C_c(X\times Y)\), its integral equals either iterated integral in (1). The product is bilinear and \[ \|\mu\otimes\nu\|\le\|\mu\|\|\nu\|. \] For positive \(\mu,\nu\), the product is positive and has mass \(\mu(X)\nu(Y)\).

Proof. Lemma 1.1 defines a bounded linear functional \(I\) on \(C_c(X\times Y)\), with norm at most \(\|\mu\|\|\nu\|\) in the uniform norm. The earlier Lemma 1.5 makes \(C_c\) uniformly dense in \(C_0\). Explicitly, if \(h_n\to h\) uniformly, then \(I(h_n)\) is Cauchy with a limit independent of the approximating sequence, by the bound; this extends \(I\) uniquely to \(C_0\). The earlier Theorem 4.3 represents it by a unique complex finite Radon measure \(\lambda\) with the stated norm bound.

Suppose first that \(\mu,\nu\) are positive. Then \(I\) is positive on \(C_c\), and its extension is positive: a nonnegative \(C_0\) function has nonnegative \(C_c\) approximations, using the cutoffs and truncations in the earlier Lemma 1.5. The positive representation theorem, Theorem 3.1 of that reading, shows that \(\lambda\) is positive. Given compact \(K\subset X,L\subset Y\), choose \(u,v\in C_c\), valued in \([0,1]\), equal to one on \(K,L\). Then \[ \lambda(X\times Y)\ge\int u(x)v(y)\,d\lambda =\left(\int u\,d\mu\right)\left(\int v\,d\nu\right) \ge\mu(K)\nu(L). \] Take suprema over \(K,L\) and use the norm upper bound to obtain the claimed total mass.

For open \(U\subseteq X,V\subseteq Y\), the same argument with cutoffs supported in \(U,V\), and compact inner approximations there, gives \(\lambda(U\times V)\ge\mu(U)\nu(V)\). For the reverse inequality, take a compact \(C\subset U\times V\). Its projections are compact subsets of \(U,V\); corresponding cutoffs \(u,v\) satisfy \(uv=1\) on \(C\) and are supported in \(U\times V\). Hence \[ \lambda(C)\le\int uv\,d\lambda =\left(\int u\,d\mu\right)\left(\int v\,d\nu\right) \le\mu(U)\nu(V). \] Inner regularity of \(\lambda\) proves equality on open rectangles.

For compact \(K,L\), subtract the two open strips \((X\setminus K)\times Y\) and \(X\times(Y\setminus L)\) from \(X\times Y\), adding back their intersection. Finite additivity and the open-rectangle formula give \(\lambda(K\times L)=\mu(K)\nu(L)\). For arbitrary Borel \(E,F\), compact subsets give the lower bound in (2). Open supersets give the upper bound, using outer regularity of the finite measures. All quantities are finite, so subtraction and passage to these products of suprema or infima are legitimate. This proves (2) for positive factors.

For complex factors, decompose each into four positive measures by Lemma 1.2. The corresponding finite linear combination of the positive products represents \(I\), by direct linearity of both iterated integrals. Uniqueness of complex representation makes it \(\lambda\). Formula (2) follows, and the same uniqueness proves bilinearity. The original functional bound gives the stated variation norm estimate.

Finally suppose another complex finite Radon measure \(\lambda'\) satisfies (2). For each \(h\in C_c\), the rectangle-simple approximations in Lemma 1.1 converge uniformly to \(h\) on the whole product. Their integrals under \(\lambda'\) are the same finite sums as under \(\lambda\), by (2). Bounded integration passes to the uniform limit, giving equal integrals of \(h\). Uniform density in \(C_0\) and the earlier Theorem 4.3 imply \(\lambda'=\lambda\). This uniqueness uses Radon regularity and continuous test functions, not an assertion that rectangles generate every product Borel set. \(\square\)

Theorem 2.2 (bounded continuous Fubini). For every bounded continuous \(h:X\times Y\to\mathbb C\), the two section-integral functions are bounded continuous, and \[ \int h\,d(\mu\otimes\nu) =\int_X\!\left(\int_Yh(x,y)\,d\nu(y)\right)d\mu(x) =\int_Y\!\left(\int_Xh(x,y)\,d\mu(x)\right)d\nu(y). \tag{3} \]

Proof. Put \(B=\|h\|_\infty\). The section integrals are bounded by \(B\|\nu\|\) and \(B\|\mu\|\). To prove continuity of the first at \(x_0\), choose compact \(L\subset Y\) with \(|\nu|(Y\setminus L)<\varepsilon\), using the earlier Corollary 4.5. The compact-parameter assertion of the earlier Lemma 1.6 gives \[ \left|\int h(x,y)\,d\nu-\int h(x_0,y)\,d\nu\right| \le\|\nu\|\sup_{y\in L}|h(x,y)-h(x_0,y)|+2B\varepsilon. \] First let \(x\to x_0\), then \(\varepsilon\downarrow0\). This argument is topological and uses no sequential assumption. Reverse the coordinates for the other section integral.

Let \(\lambda=\mu\otimes\nu\). Choose compact sets with small \(|\mu|\), \(|\nu|\) and \(|\lambda|\) tails. Project the chosen compact subset of \(X\times Y\) into \(X,Y\) and add these projections to the respective compact sets. Choose cutoffs \(u,v\), valued in \([0,1]\), equal to one on these enlarged sets. Then \(h(x,y)u(x)v(y)\in C_c(X\times Y)\), so Theorem 2.1 applies.

If all three tails are less than \(\varepsilon\), the error in its \(\lambda\) integral is at most \(B\varepsilon\). For either order of iterated integrals, use \(0\le1-uv\le(1-u)+(1-v)\) and the variation bound for bounded integration. The error is at most \[ B\bigl(\|\nu\|\varepsilon+\|\mu\|\varepsilon\bigr). \] Let \(\varepsilon\downarrow0\) to obtain (3). This proves exactly the bounded continuous Fubini assertion; it does not silently extend (3) to an arbitrary unbounded measurable kernel. \(\square\)

Corollary 2.3 (compact localization). Suppose \(\alpha,\beta\) are positive Borel measures on \(X,Y\), finite on compact sets and inner regular on Borel sets of finite measure. For \(h\in C_c(X\times Y)\), its two iterated integrals exist, agree, and can be calculated with finite Radon restrictions to any compact sets containing the projections of its support. The conclusion also holds for a bounded continuous kernel when both measures have already been restricted to compact sets.

Proof. For compact \(K\subset X\), the Borel measure \(\alpha_K(B)=\alpha(B\cap K)\) is finite. Inner regularity of \(\alpha\) on \(B\cap K\) supplies compact subsets of this set approximating \(\alpha_K(B)\). Thus \(\alpha_K\) is inner regular on all Borel sets. For outer regularity at \(B\), choose compact \(C\subset X\setminus B\) with \(\alpha_K((X\setminus B)\setminus C)<\varepsilon\); the open superset \(X\setminus C\) of \(B\) has the same error. Thus \(\alpha_K\) is Radon. The same argument applies to \(\beta_L\).

For \(h\in C_c\), choose \(K,L\) containing the support projections. Every inner integral vanishes outside its relevant projection, and the sections vanish off the other projection. The original iterated integrals therefore equal the integrals with \(\alpha_K,\beta_L\). Lemma 1.1 gives equality and finiteness. Enlarging the compact sets has no effect because the added region contributes zero. For the last assertion apply Theorem 2.2 directly to the two finite restrictions. This is compact localization, not a claim to have constructed an unrestricted product of two infinite measures. \(\square\)

3. Images, associativity and group convolution

Proposition 3.1 (continuous images). If \(T:X\to Z\) is continuous between locally compact Hausdorff spaces and \(\mu\in M(X)\), then \[ T_*\mu(E)=\mu(T^{-1}E) \] defines a complex finite Radon measure on \(Z\). It satisfies \[ |T_*\mu|\le T_*|\mu|,\qquad \|T_*\mu\|\le\|\mu\|, \qquad \int_Z f\,d(T_*\mu)=\int_X f\circ T\,d\mu \tag{4} \] for every bounded Borel function \(f\). No properness assumption on \(T\) is needed.

Proof. Inverse images preserve disjoint countable unions, so the definition gives a measure. For a positive finite Radon measure \(\rho\), approximate \(T^{-1}E\) from within by compact \(K\). The image \(T(K)\) is compact and lies in \(E\), and \[ (T_*\rho)(E\setminus T(K)) \le\rho(T^{-1}E\setminus K). \] This proves inner regularity of \(T_*\rho\). Outer regularity follows by applying inner regularity to the complement: a compact \(C\subset Z\setminus E\) gives the open superset \(Z\setminus C\) of \(E\), with the same error, and total mass is finite. Thus \(T_*\rho\) is Radon.

For complex \(\mu\), any finite Borel partition of \(E\) pulls back to a partition of \(T^{-1}E\); the definition of variation therefore proves \(|T_*\mu|(E)\le(T_*|\mu|)(E)\). Lemma 1.2 transfers Radon regularity from the positive finite measure \(T_*|\mu|\) to \(T_*\mu\), and gives the norm bound. Formula (4) holds for indicators by definition, then for simple functions by finite linearity, and finally for bounded Borel functions by their uniform simple approximations from the earlier Lemma 2.2 and the variation estimate. \(\square\)

Proposition 3.2 (product identities). Under the natural coordinate identifications, finite Radon products are associative, symmetric on interchange of the factors, and commute with continuous images: \[ (\mu\otimes\nu)\otimes\rho=\mu\otimes(\nu\otimes\rho),\qquad \operatorname{flip}_*(\mu\otimes\nu)=\nu\otimes\mu, \] \[ (T\times S)_*(\mu\otimes\nu)=T_*\mu\otimes S_*\nu. \tag{5} \]

Proof. Test associativity on an arbitrary bounded continuous \(h(x,y,z)\). Theorem 2.2 says that integrating first in \(z\) gives a bounded continuous function of \((x,y)\). Applying it a second time shows that the integral against either associated product equals \(\int_X\int_Y\int_Z h(x,y,z)\,d\rho(z)\,d\nu(y)\,d\mu(x)\). For the right-associated product, apply Theorem 2.2 for each fixed \(x\); the same bounded-continuity assertion makes its outer integral legitimate. Proposition 3.1 justifies the harmless reassociation homeomorphism. These equalities for \(C_c\) functions imply equality of Radon measures by the earlier representation theorem.

The flip identity follows from the equality of the two orders in Theorem 2.2. For the last identity, evaluate a bounded continuous function \(f\) on the target product. Apply Proposition 3.1 to the product image, then Theorem 2.2 to \(f(Tx,Sy)\), and apply Proposition 3.1 in each coordinate. The result is its integral against \(T_*\mu\otimes S_*\nu\). Again continuous compactly supported tests give equality of the measures. \(\square\)

Theorem 3.3 (the measure algebra). Let \(G\) be a locally compact Hausdorff group, with identity \(e\) and multiplication \(m(x,y)=xy\). Define \[ \mu*\nu=m_*(\mu\otimes\nu),\qquad \mu^*(E)=\overline{\mu(E^{-1})}. \] Then \(M(G)\) is a unital Banach star algebra, with unit \(\delta_e\), and \[ \|\mu*\nu\|\le\|\mu\|\|\nu\|,\qquad \|\mu^*\|=\|\mu\|,\qquad (\mu*\nu)^*=\nu^**\mu^*. \] It is commutative when \(G\) is abelian. If \(\gamma:G\to\mathbb T\) is a continuous character and \(\mathcal T_\gamma(\mu)=\int_G\gamma(x)\,d\mu(x)\), then \[ \mathcal T_\gamma(\mu*\nu)=\mathcal T_\gamma(\mu)\mathcal T_\gamma(\nu), \qquad \mathcal T_\gamma(\mu^*)=\overline{\mathcal T_\gamma(\mu)}. \tag{6} \]

Proof. Theorem 2.1 and Proposition 3.1 define the convolution as a finite Radon measure and prove its norm bound and bilinearity. The earlier Theorem 4.3 proves completeness of \(M(G)\) in variation norm. Proposition 3.2 and associativity of group multiplication show that either association of a triple convolution is the image of \(\mu\otimes\nu\otimes\rho\) under \((x,y,z)\mapsto xyz\). The measure \(\delta_e\) is Radon: compact sets \(\{e\}\) or the empty set supply inner approximation, and its total mass is one. Integrating a continuous test function against \(\delta_e*\mu\) or \(\mu*\delta_e\) gives its integral against \(\mu\), so uniqueness proves the unit assertion. If \(G\) is abelian, multiplication is unchanged by the flip, giving commutativity.

Conjugation of a measure is countably additive and has the same variation, since conjugation preserves the absolute values in every finite partition. Proposition 3.1 applied to inversion makes \(\mu^*\) Radon. Inversion twice is the identity, so its variation bound in both directions gives \(\|\mu^*\|=\|\mu\|\). Applying the definition twice gives \(\mu^{**}=\mu\), and scalar conjugate-linearity follows on every Borel set. Bounded simple approximation also gives \[ \int f\,d\mu^*=\overline{\int\overline{f(x^{-1})}\,d\mu(x)}. \] For bounded continuous \(f\), this formula and Theorem 2.2 give \[ \int f\,d(\mu*\nu)^* =\overline{\int_G\int_G\overline{f(y^{-1}x^{-1})}\,d\nu(y)\,d\mu(x)} =\int f\,d(\nu^**\mu^*). \] The second equality applies the same involution formula in each variable and interchanges the two original integrals using Theorem 2.2. Equality on \(C_c\) proves the star-product identity.

Finally, \(\gamma(xy)=\gamma(x)\gamma(y)\). Formula (3) and the pushforward identity therefore factor its convolution integral as in (6). The involution formula and \(\gamma(x^{-1})=\overline{\gamma(x)}\) give the second assertion. The same statements hold if every \(\gamma\) is replaced by \(\overline\gamma\). \(\square\)