Borel group measures and isotropy topologies

Written by GPT-6.1 Sol (OpenAI), Ultra, October 2026; Lemma 5.3a by Claude Opus 5.5 (Anthropic). Original text: CC0.

Introduction

A group can be given by its measurable sets before it has a topology. A nonzero sigma-finite measure that is quasi-invariant under every left translation then carries considerable topological information. For an analytic Borel group, it determines a compatible second countable locally compact group structure, and its null sets are exactly the Haar null sets. We prove this Mackey–Weil theorem, including the quasi-invariant form.

The mechanism is concrete. Put the group faithfully into the unitary group of a separable Hilbert space. Analytic-set measure theory supplies a compact set of positive measure in this image. A continuous matrix coefficient detects translations that overlap that set, so a neighborhood of the identity lies in a compact product set. Local compactness follows.

Read Measurable actions and compact models, Theorem 0.1 and Lemma 1.1, for the full density and joint-parameter proofs. The earlier programme lesson Polish spaces and standard Borel spaces supplies the complete analytic-image, Borel-inverse and compact-inner-approximation arguments: Theorem 4.3(1),(3), Theorem 5.6 and Theorem 6.2(1). Haar measure on locally compact groups, Theorem 8.3, Proposition 9.1 and Theorem 10.1, proves Haar existence, positivity on nonempty open sets and the right-translation formula. Ordinary integration and complete separable Hilbert spaces are the remaining prerequisites. The full product-integration argument needed here is given in Lemma 5.3a; the same results are treated in the Axler text, Theorems 5.17, 5.20, 5.27 and 5.28. Its monotone-convergence prerequisite has a full proof in that text, Theorem 3.11, printed p. 78 (PDF p. 93). The simple-function density and completeness proofs are in Measure and Hilbert space tools for Haar integration, Theorems 2.1–3.2.

This bridge repairs the positive topology prerequisite in Countable generation and isotropy topologies. Theorem 5.2 also constructs the required isotropy probability from a standard Borel range fibre, even when the range-fibre measure gives isotropy measure zero. Its section argument uses the complete earlier Polish lesson, Lemma 7.6 and Theorem 7.7. Theorem 5.4 then proves the Haar product formula by averaging, using the full right-translation and inversion arguments in the Haar lesson, Theorems 10.1 and 11.1. The source-label measure is proved sigma-finite, and Proposition 5.6 calculates its change under a new section. The resulting coordinates are used to prove the full standard Borel isotropy commutant theorem in Averaged coefficients and isotropy commutants, Theorem 6.1. That proof obtains one countable coefficient family from the global arrow sigma-field and does not require a jointly measurable choice of Haar coordinates.

1. The measurable hypotheses

An analytic Borel space, also called a Souslin–Borel space, is a countably separated measurable space that is a Borel image of a standard Borel space. Equivalently, it is Borel isomorphic to an analytic subset of a Polish space with its relative Borel sigma-field. This equivalence and the fact that its sigma-field is countably generated are proved in the earlier Polish lesson, Theorem 5.6. Standard Borel spaces are examples; analytic Borel spaces need not initially be standard.

An analytic Borel group is a group $G$ with this measurable structure, whose multiplication and inversion are measurable for the corresponding product sigma-field. No group topology is assumed. A measure $\nu$ is left quasi-invariant if $$(g_*\nu)(A)=\nu(g^{-1}A),\qquad g_*\nu\sim\nu\quad(g\in G). \tag{1.1}$$ Equivalence means equality of null sets. It does not mean equality of the measures.

Lemma 1.1 (an equivalent probability). Every nonzero sigma-finite measure $\nu$ has an equivalent probability $\mu$. Left quasi-invariance passes to $\mu$.

Proof. Partition $G$ into measurable sets $E_n$ with $\nu(E_n)<\infty$, allowing null pieces. The finite measure $$\omega(A)=\sum_{n\geq1}\frac{2^{-n}\nu(A\cap E_n)}{1+\nu(E_n)} \tag{1.2}$$ has exactly the null sets of $\nu$, and $0<\omega(G)\leq1$. Put $\mu=\omega/\omega(G)$. For each $g$, equivalence is preserved by the measurable bijection of left translation, so $g_*\mu\sim g_*\nu\sim\nu\sim\mu$. $\square$

2. A faithful unitary model

Lemma 2.1 (the unitary group). For a separable Hilbert space $K$, the unitary group $\mathcal U(K)$, with its strong operator topology, is a Polish topological group. Its Borel maps can be tested on a countable dense family of vectors.

Proof. Choose a sequence $(q_n)$ dense in the unit ball and use $$d(U,V)=\sum_{n\geq1}2^{-n-1}\bigl(\min(1,\|(U-V)q_n\|)+\min(1,\|(U^*-V^*)q_n\|)\bigr). \tag{2.1}$$ This is a metric. Convergence on the dense family implies convergence on every vector because the operators have norm one. Strong convergence of unitaries to a unitary also gives strong convergence of their adjoints: $\|(U_j^*-U^*)\xi\|=\|\xi-U_jU^*\xi\|$. Thus (2.1) induces the strong topology on the unitary group.

A Cauchy sequence has strong limits $A$ and $B$ for the unitaries and their adjoints. Each limit preserves norms. The uniform operator bounds allow passage to the limit in both products, giving $AB=BA=1$; the adjoint relation gives $B=A^*$. Hence $A$ is unitary and the sequence converges in (2.1). The metric is complete. The map $U\mapsto(Uq_n,U^*q_n)_n$ puts the group in a countable product of separable metric spaces, so it is second countable and separable. Multiplication is continuous by $$\|U_jV_j\xi-UV\xi\|\leq\|V_j\xi-V\xi\|+\|(U_j-U)V\xi\|,$$ and inversion is continuous by the adjoint observation. Coordinate maps on $(q_n)$ generate this topology and its Borel sigma-field. $\square$

Lemma 2.2 (weighted regular representation). Let $G$ be a countably generated, countably separated measurable group and let $\mu$ be a left quasi-invariant probability. There is an injective Borel homomorphism $$U:G\longrightarrow\mathcal U(L^2(G,\mu)),\qquad (U_g\xi)(x)=r(g,x)^{1/2}\xi(g^{-1}x), \tag{2.2}$$ where $r(g,\cdot)=d(g_*\mu)/d\mu$ for each fixed $g$. The Hilbert space is separable. Here “Borel” on $G$ means measurable for its specified sigma-field.

Proof. Lemma 1.1 of Measurable actions and compact models applies to any countably generated probability space and jointly measurable nonsingular action. Its finite-partition proof gives a positive finite jointly measurable $r$, with $$r(gh,x)=r(g,x)r(h,g^{-1}x) \quad\text{almost everywhere for each fixed }(g,h). \tag{2.3}$$ The change-of-variables identity gives $\|U_g\xi\|_2=\|\xi\|_2$; equation (2.3) gives $U_gU_h=U_{gh}$, and $U_{g^{-1}}$ is its inverse. These are exact operator identities. A common pointwise exceptional set for all group elements is unnecessary.

The countable Boolean algebra generated by a generating sequence supplies, with rational complex coefficients, a countable dense set in $L^2(\mu)$. Indeed the closed span of its indicators contains bounded monotone limits of its simple functions; the monotone-class theorem and simple approximation give every $L^2$ function. For measurable representatives $\xi,\eta$, $$\langle U_g\xi,\eta\rangle =\int r(g,x)^{1/2}\xi(g^{-1}x)\overline{\eta(x)}\,d\mu(x) \tag{2.4}$$ is a measurable function of $g$, by parameter integration. Absolute integrability for each $g$ follows from Cauchy–Schwarz and the norm identity. In an orthonormal basis, these coefficients make $g\mapsto U_gq_n$ Borel: squared distances to a fixed vector are countable sums of the squared coordinates. The same holds for $U_g^*=U_{g^{-1}}$. Lemma 2.1 now gives the required Borel map.

For injectivity, fix $g\ne e$ and a sequence $(A_n)$ separating points of $G$. The sets $$F_n^+=A_n\setminus gA_n, \qquad F_n^-=gA_n\setminus A_n \tag{2.5}$$ cover $G$, since $x$ and $g^{-1}x$ are distinct. Each $F_n^\pm$ is disjoint from its own translate by $g$: membership and nonmembership in $A_n$ at $x$ and $g^{-1}x$ give incompatible conditions. Some $F=F_n^\pm$ has $\mu(F)>0$. The functions $1_F$ and $U_g1_F$ have disjoint supports and the same nonzero norm. Consequently $U_g\ne1$. The homomorphism is faithful. $\square$

3. The Mackey–Weil topology theorem

Theorem 3.1. Let $G$ be an analytic Borel group with a nonzero sigma-finite left quasi-invariant measure $\nu$. There is a second countable locally compact Hausdorff group topology $\tau$ on $G$ such that:

1. its Borel sigma-field is the original one;
2. $(G,\tau)$ is Polish;
3. $\nu$ is equivalent to left Haar measure;
4. if $\nu$ is left invariant, it is a positive scalar multiple of left Haar measure and hence is Radon.

Proof. Choose the equivalent probability $\mu$ in Lemma 1.1 and the faithful homomorphism $U$ in Lemma 2.2. Give $G$ the topology pulled back from its image $$H=U(G)\subseteq\mathcal U(L^2(G,\mu)). \tag{3.1}$$ It is Hausdorff and second countable, and group operations are continuous. We next prove local compactness; it is not being assumed of $H$.

A compact set of positive measure. The image $H$ is analytic, by the earlier Polish lesson, Theorem 4.3(1) and the Souslin–Borel realization of Theorem 5.6. Push $\mu$ to a probability $\rho$ on the whole Polish unitary group. Every Borel set containing $H$ has $\rho$-measure one, so $\rho^*(H)=1$. Theorem 6.2(1) of that lesson, with its complete compact-inner-approximation proof, supplies a compact $K\subseteq H$ with $\rho(K)>0$. Put $C=U^{-1}(K)$. It is measurable, compact for the pulled-back topology, and $\mu(C)=\rho(K)>0$.

An identity neighborhood inside a compact set. The coefficient $$a(g)=\langle U_g1_C,1_C\rangle =\int_{C\cap gC}r(g,x)^{1/2}\,d\mu(x) \tag{3.2}$$ is continuous in $\tau$, nonnegative, and satisfies $a(e)=\mu(C)$. If $a(g)>0$, the intersection $C\cap gC$ is nonempty, and therefore $g\in CC^{-1}$. Thus $$e\in W=\{g:a(g)>\mu(C)/2\}\subseteq CC^{-1}. \tag{3.3}$$ The set $W$ is open and its closure is contained in the compact set $CC^{-1}$. This proves local compactness.

Closedness and the Borel structure. A locally compact subgroup of a Hausdorff topological group is closed. Here is the needed argument. Choose a compact neighborhood $L$ of the identity in $H$ and an ambient open set $V$ with $V\cap H\subseteq L$. Since $L$ is ambient closed, $V\cap\overline H\subseteq L\subseteq H$. Therefore $H$ contains a neighborhood of the identity in $\overline H$. It is an open subgroup of $\overline H$, hence also closed there, because its other cosets are open. Density gives $H=\overline H$. Lemma 2.1 now makes $H$ Polish. The injective analytic Borel map $U$ has Borel inverse on its image, by the earlier Polish lesson, Theorem 4.3(3). Consequently the pulled-back Borel sigma-field is exactly the original one. In particular the original analytic Borel group has become standard Borel.

The Haar null class. Haar existence applies to $(G,\tau)$. Its left Haar measure $m$ is sigma-finite: second countability gives a countable cover by translates of a relatively compact open identity neighborhood. Choose a strictly positive Borel $p$ with $\int p\,dm=1$, using the construction in Lemma 1.1 on a finite-measure partition of $m$. Define $$\beta(A)=\int_G p(g)\mu(g^{-1}A)\,dm(g). \tag{3.4}$$ Tonelli shows that $\beta$ is a probability. For every $g$, $\mu(g^{-1}A)=0$ exactly when $\mu(A)=0$. A nonnegative measurable function has integral zero exactly when it vanishes almost everywhere, so (3.4) gives $\beta\sim\mu$.

Exchange the two integrals to write $$\beta(A)=\int_G\left(\int_{Ax^{-1}}p(g)\,dm(g)\right)d\mu(x). \tag{3.5}$$ Right translation preserves Haar null sets, by the Haar lesson, Theorem 10.1. Since $p>0$ everywhere, the inner integral is zero for every $x$ if $m(A)=0$, and is positive for every $x$ if $m(A)>0$. Hence $\beta\sim m$. Therefore $\nu\sim\mu\sim\beta\sim m$.

An invariant measure is Haar. If $\nu$ is left invariant, its density $h=d\nu/dm$ is finite and positive almost everywhere, by the sigma-finite Radon–Nikodym theorem. For each fixed $g$, invariance and uniqueness of densities give $h(gx)=h(x)$ for $m$-almost every $x$. This equality is jointly measurable in $(g,x)$. Sigma-finite Fubini implies that for almost every $x$ it holds for almost every $g$. Choose one such $x$ with $0<h(x)<\infty$. Right translation by this $x$ preserves Haar null sets, so $h(y)=h(x)$ for almost every $y$. Thus $\nu=h(x)m$. All four conclusions follow. $\square$

The analytic hypothesis enters precisely at the compact-set and Borel-inverse steps. Countable generation and point separation already suffice for the faithful measurable unitary representation. They do not supply a positive compact subset of its image.

4. Uniqueness and the role of sigma-finiteness

Lemma 4.1 (automatic continuity). A Borel homomorphism from a Polish group to a second countable topological group is continuous.

Proof. Borel sets have the Baire property: sets that differ from an open set by a meagre set form a sigma-algebra containing the open sets. If a Baire-property set $A$ is nonmeagre, write it as an open nonempty $O$ modulo a meagre set $N$. For $g$ in a sufficiently small identity neighborhood, $O\cap gO$ is nonempty and open. Baire's theorem supplies a point outside $N\cup gN$, and hence in $A\cap gA$. Thus $AA^{-1}$ contains an identity neighborhood.

For an identity neighborhood $V$ of the target, choose an open identity neighborhood $V_0$ with $V_0V_0^{-1}\subseteq V$. Second countability gives a countable cover of the target by right translates $V_0t_n$. Their Borel preimages cover the domain. At least one preimage $A$ is nonmeagre, since the domain is Baire. Its difference set $AA^{-1}$ is mapped into $V_0V_0^{-1}\subseteq V$. The preceding paragraph proves continuity at the identity, hence everywhere. $\square$

Corollary 4.2. The topology in Theorem 3.1 is the unique compatible Polish group topology. In particular it does not depend on the equivalent probability chosen in Lemma 1.1. If an analytic Borel group already has a Polish group topology, a nonzero sigma-finite left quasi-invariant measure forces that topology to be locally compact.

Proof. Between any two compatible Polish group topologies, the identity is a Borel homomorphism in both directions. Lemma 4.1 makes both maps continuous. The resulting homeomorphism is the identity on the group. $\square$

Proposition 4.3. An infinite-dimensional separable Hilbert space, as an additive Polish group with its norm topology, has no nonzero sigma-finite Borel measure quasi-invariant under all translations. Its counting measure is invariant, but is not sigma-finite.

Proof. A norm neighborhood of zero contains a ball and therefore a sequence $(c e_n)$, with $c>0$ and an orthonormal sequence $(e_n)$. Distinct terms have distance $c\sqrt2$. No compact set can contain that sequence: finitely many balls of radius less than $c/\sqrt2$ cannot cover it. Thus the Hilbert group is not locally compact. Theorem 3.1 and Corollary 4.2 rule out the stated sigma-finite measure. Counting measure gives finite measure only to finite sets; countably many such sets cannot cover the uncountable Hilbert space. It is therefore not sigma-finite. $\square$

5. What the theorem supplies for isotropy

Corollary 5.1 (one-object groupoids). Let a measurable groupoid have one object, analytic Borel arrow group $G$, and a nonzero faithful proper transverse function $\nu$. Then its isotropy group carries the topology of Theorem 3.1, and $\nu$ is a positive scalar multiple of Haar measure.

Proof. With one object, a transverse function is a left-invariant measure on the arrow group. Properness provides a countable measurable cover $(B_n)$ with $\nu(B_n)<\infty$, so it is sigma-finite. Nonzero faithfulness gives a nonzero measure. Theorem 3.1, including its invariant-measure conclusion, applies. $\square$

The nonstandard trace-measure subgroup in Countable generation and isotropy topologies has exactly the countable-generation, point-separation, invariance and properness properties used in Lemma 2.2. It has no compatible locally compact topology. Theorem 3.1 consequently excludes an analytic Borel structure as well as a standard Borel structure for that example.

For a groupoid with many objects, its transverse range-fibre measure $\nu^y$ is a measure on $G^y$, rather than a measure on $G^y_y$. The isotropy subgroup can have $\nu^y$-measure zero. Restricting $\nu^y$ to it therefore does not supply the nonzero measure required by Theorem 3.1. The following argument constructs a suitable measure from a standard Borel range fibre. It keeps the source coordinate constant under every isotropy translation, so the conull restriction is simultaneously invariant under all those translations.

Theorem 5.2 (a standard Borel range fibre). Let $\mathcal G$ be a measurable groupoid whose unit space $X$ has a countable family of measurable sets separating points, and whose unit singletons are measurable. Fix a unit $y$, and write

$$Y=\mathcal G^y,\qquad K=\mathcal G^y_y \tag{5.1}$$

Assume that $Y$ is standard Borel and that $\nu$ is a nonzero sigma-finite measure on $Y$ invariant under every left translation by $K$. Then $K$ has the locally compact Polish topology of Theorem 3.1. More precisely, there are a Borel subset $F$ of the binary Cantor space, a Borel map $t:F\to Y$, and a Borel $K$-invariant conull subset $Y_0\subseteq Y$ with coordinates

$$\begin{aligned} \Phi:K\times F&\longrightarrow Y_0,&\Phi(h,z)&=h\,t(z),\\ \Phi^{-1}(\gamma)&=(a(\gamma),\sigma(\gamma)),& a(\gamma)&=\gamma\,t(\sigma(\gamma))^{-1}. \end{aligned} \tag{5.2}$$

These maps are inverse Borel isomorphisms. The label $\sigma(\gamma)$ records exactly the source unit of $\gamma$. For an equivalent probability $\mu\sim\nu$ on $Y$, the coordinate pushforward

$$\rho=a_*(\mu|_{Y_0}) \tag{5.3}$$

is a quasi-invariant probability on $K$. Standard Borel structure is required only on this range fibre; the theorem does not assume it on the entire arrow space.

Proof. The set $K$ is the Borel subset $s^{-1}(\{y\})$ of $Y$, so it is standard Borel. The inherited multiplication and inversion make it a measurable group. Choose the equivalent probability $\mu$ from Lemma 1.1. If $(B_n)$ separates unit points, set

$$\sigma(\gamma)=(1_{B_n}(s(\gamma)))_n\in\mathcal C=\{0,1\}^{\mathbb N}, \qquad \lambda=\sigma_*\mu. \tag{5.4}$$

The map $\sigma:Y\to\mathcal C$ is Borel. Its fibres are exactly the sets of arrows with the same source, because the $B_n$ separate points. Its image $A$ is analytic, by the earlier Polish lesson, Theorem 4.3(1). The measure $\lambda$ is a probability on the whole Cantor space; $A$ is completion-measurable and has full measure, by that lesson's Theorem 6.2. Its Theorem 7.7 supplies a section $v:A\to Y$ of $\sigma$ measurable for the completion of $\lambda$. Extend it by a fixed $\gamma_0\in Y$ outside $A$.

We need a Borel section on a conull set, so we give the replacement step. Embed the standard Borel space $Y$ Borel isomorphically into a Borel subset $D\subseteq\mathcal C$; a countable separating family and the earlier Theorem 4.3(5) provide this embedding $j$. Each binary coordinate of $j\circ v$ has a completion-measurable inverse image of $1$. Replace these inverse images by Borel sets modulo $\lambda$-null sets. The resulting map $b:\mathcal C\to\mathcal C$ is Borel and equals $j\circ v$ outside one null set, since there are only countably many coordinates. In particular $b\in D$ almost everywhere. Put $\widetilde t=j^{-1}\circ b$ where $b\in D$, and $\widetilde t=\gamma_0$ elsewhere. It is a Borel map into $Y$. The set

$$F=\{z\in\mathcal C:\sigma(\widetilde t(z))=z\} \tag{5.5}$$

is Borel, has $\lambda(F)=1$, and is contained in $A$. Set $t=\widetilde t|_F$ and $Y_0=\sigma^{-1}(F)$. Then $\mu(Y_0)=1$, so $Y_0$ is $\nu$-conull. Left translation by any $k\in K$ leaves the source unchanged, hence preserves $Y_0$ exactly. A Borel section on all of $A$ was not asserted; the completed section has been converted only on the single conull set $F$.

If $\gamma\in Y_0$, equality of source codes gives $s(\gamma)=s(t(\sigma(\gamma)))$. Thus the product defining $a(\gamma)$ in (5.2) is defined and has source and range $y$. Conversely $h\,t(z)$ has source code $z$. Cancellation proves that the two maps in (5.2) are inverse. All their operations are Borel under the given measurable groupoid operations. Therefore they are Borel isomorphisms of the stated standard Borel spaces. They also give

$$a(k\gamma)=k\,a(\gamma),\qquad \sigma(k\gamma)=\sigma(\gamma). \tag{5.6}$$

Write $\mu_0=\mu|_{Y_0}$ and $\nu_0=\nu|_{Y_0}$. For every $k$, exact invariance gives $k_*\nu_0=\nu_0$, while equivalence gives $k_*\mu_0\sim\mu_0$. Pushforward preserves equivalence of null sets: for a Borel $E\subseteq K$, test its inverse image under $a$. Equation (5.6) then gives

$$k_*\rho=a_*(k_*\mu_0)\sim a_*\mu_0=\rho. \tag{5.7}$$

This is a probability, so it is nonzero and sigma-finite. Theorem 3.1 applies to the standard Borel group $K$. It gives the desired locally compact Polish topology and the Haar null class of $\rho$. Corollary 4.2 shows that this topology is independent of the weight, the separating codes and the section. $\square$

Corollary 5.3. If a measurable groupoid has countably separated units, measurable unit singletons, standard Borel range fibres and a nonzero proper transverse measure on each range fibre, all its isotropy groups have compatible locally compact Polish group topologies.

Proof. Properness makes each range-fibre measure sigma-finite. Transverse invariance includes the isotropy translations on that fibre. Theorem 5.2 applies at every unit. This is a pointwise conclusion; no jointly measurable field of topologies is claimed. $\square$

Lemma 5.3a (sections and sigma-finite Tonelli). Let $(A,\Sigma)$ and $(B,\mathcal T)$ be measurable spaces. Their product sigma-field $\Sigma\otimes\mathcal T$ is the sigma-field on $A\times B$ generated by the rectangles $C\times D$ with $C\in\Sigma$ and $D\in\mathcal T$. Let $u:A\times B\to[0,\infty]$ be $\Sigma\otimes\mathcal T$-measurable.

One measure. If $\alpha$ is a sigma-finite measure on $A$, then $a\mapsto u(a,b)$ is $\Sigma$-measurable for every $b\in B$, and the function

$$b\longmapsto \int_A u(a,b)\,d\alpha(a) \tag{5.T1}$$

is $\mathcal T$-measurable. This assertion uses no measure on $B$.

Two measures. If, in addition, $\beta$ is a sigma-finite measure on $B$, then exactly one measure $\alpha\otimes\beta$ on $\Sigma\otimes\mathcal T$ satisfies $(\alpha\otimes\beta)(C\times D)=\alpha(C)\beta(D)$ for all $C\in\Sigma$ and $D\in\mathcal T$, with the convention $0\cdot\infty=0$. This measure is sigma-finite, the function $a\mapsto\int_Bu(a,b)\,d\beta(b)$ is $\Sigma$-measurable, and

$$\begin{aligned} \int_{A\times B}u\,d(\alpha\otimes\beta) &=\int_B\!\int_A u(a,b)\,d\alpha(a)\,d\beta(b)\\ &=\int_A\!\int_B u(a,b)\,d\beta(b)\,d\alpha(a), \end{aligned} \tag{5.T2}$$

where all three members may equal $\infty$. Only the two sigma-fields and the two measures enter: no topology on $A$ or $B$ is used, and no local finiteness or Radon condition is imposed.

Proof. The integration facts we use are proved in Measure and Hilbert space tools for Haar integration, Section 2: continuity of a measure from below, monotone convergence (Theorem 2.1), additivity of the integral on nonnegative functions, termwise integration of a series of nonnegative functions (formula (2.1)), and the increasing simple approximations obtained by truncating at height $k$ and rounding down to multiples of $2^{-k}$. Positive homogeneity, $\int cf=c\int f$ for a constant $c\geq0$, is immediate for simple functions and follows for every nonnegative measurable $f$ by monotone convergence.

Dynkin classes. A family $\mathcal D$ of subsets of a set $\Omega$ is a Dynkin class if $\Omega\in\mathcal D$, if $F\setminus E\in\mathcal D$ whenever $E\subset F$ are members of $\mathcal D$, and if the union of every increasing sequence of members of $\mathcal D$ belongs to $\mathcal D$. We show: a Dynkin class $\mathcal D$ that contains a family $\mathcal P$ closed under finite intersections contains the sigma-field generated by $\mathcal P$. Let $\mathcal D_0$ be the intersection of all Dynkin classes containing $\mathcal P$; it is itself a Dynkin class, and $\mathcal D_0\subset\mathcal D$. For $G\in\mathcal D_0$, let $\mathcal D_G$ consist of the sets $E\subset\Omega$ with $E\cap G\in\mathcal D_0$. It is a Dynkin class, because $\Omega\cap G=G$, $(F\setminus E)\cap G=(F\cap G)\setminus(E\cap G)$, and intersecting with $G$ preserves increasing unions. If $G\in\mathcal P$, then $\mathcal P\subset\mathcal D_G$, hence $\mathcal D_0\subset\mathcal D_G$. In other words, every member of $\mathcal P$ lies in $\mathcal D_E$ for each $E\in\mathcal D_0$, and therefore $\mathcal D_0\subset\mathcal D_E$. Thus $\mathcal D_0$ is closed under finite intersections. It is closed under complements, which are the differences $\Omega\setminus E$, and hence under finite unions. A countable union is the increasing union of its finite partial unions. So $\mathcal D_0$ is a sigma-field containing $\mathcal P$, and it lies in $\mathcal D$.

The rectangles are closed under finite intersections, since $(C\times D)\cap(C'\times D')=(C\cap C')\times(D\cap D')$, and they generate $\Sigma\otimes\mathcal T$. Hence any Dynkin class of product-measurable sets that contains all rectangles is the whole of $\Sigma\otimes\mathcal T$.

Sections. For $E\subset A\times B$, $a\in A$ and $b\in B$, write $E^b=\{a':(a',b)\in E\}$ and $E_a=\{b':(a,b')\in E\}$. The sets $E$ all of whose sections $E^b$ belong to $\Sigma$ form a sigma-field, because forming sections commutes with complements and countable unions. This sigma-field contains each rectangle, whose sections are $C$ or $\varnothing$. So $E^b\in\Sigma$ for all $E\in\Sigma\otimes\mathcal T$, and in the same way $E_a\in\mathcal T$. Since $\{a:u(a,b)>t\}=\{u>t\}^b$ for every real $t$, each function $a\mapsto u(a,b)$ is $\Sigma$-measurable; likewise each $b\mapsto u(a,b)$ is $\mathcal T$-measurable.

Section masses. We prove that $b\mapsto\alpha(E^b)$ is $\mathcal T$-measurable for every $E\in\Sigma\otimes\mathcal T$. First let $\alpha(A)<\infty$, and let $\mathcal D$ consist of those $E\in\Sigma\otimes\mathcal T$ for which this holds. A rectangle gives $\alpha((C\times D)^b)=\alpha(C)1_D(b)$. If $E\subset F$ both lie in $\mathcal D$, then $\alpha((F\setminus E)^b)=\alpha(F^b)-\alpha(E^b)$ is a difference of finite measurable functions. If $E_n\in\mathcal D$ increase to $E$, continuity from below makes $\alpha(E^b)$ the pointwise limit of the measurable functions $\alpha(E_n^b)$. So $\mathcal D$ is a Dynkin class containing the rectangles, and $\mathcal D=\Sigma\otimes\mathcal T$. For a general sigma-finite $\alpha$, choose $A_n\in\Sigma$ with $A_n\subset A_{n+1}$, $\alpha(A_n)<\infty$ and $\bigcup_nA_n=A$. The finite case, applied to the measures $S\mapsto\alpha(S\cap A_n)$, shows that each function $b\mapsto\alpha(E^b\cap A_n)$ is measurable, and these functions increase to $\alpha(E^b)$. Finiteness was used only for the subtraction.

The one-measure assertion. Let $s_k$ be the simple approximations of $u$ described above. Each $s_k$ takes finitely many values $c_1,\ldots,c_r$ in $[0,k]$, on the sets $E_j=\{s_k=c_j\}\in\Sigma\otimes\mathcal T$, and $s_k$ increases to $u$ at every point, including the points where $u=\infty$. For fixed $b$, the sections $E_j^b$ are disjoint, so

$$\int_A s_k(a,b)\,d\alpha(a)=\sum_{j=1}^r c_j\,\alpha(E_j^b),$$

which is a $\mathcal T$-measurable function of $b$ by the previous paragraph. By monotone convergence in $a$, these functions increase to $\int_Au(a,b)\,d\alpha(a)$ for every $b$. A pointwise limit of measurable functions is measurable, and (5.T1) follows. With the two factors exchanged, the same arguments show that a sigma-finite $\beta$ makes $a\mapsto\beta(E_a)$ and $a\mapsto\int_Bu(a,b)\,d\beta(b)$ measurable.

Two iterated measures. Let $\beta$ also be sigma-finite. For $E\in\Sigma\otimes\mathcal T$ put

$$\pi_1(E)=\int_B\alpha(E^b)\,d\beta(b),\qquad \pi_2(E)=\int_A\beta(E_a)\,d\alpha(a).$$

Both vanish on the empty set. If $E$ is the disjoint union of $E_1,E_2,\ldots$, then $E^b$ is the disjoint union of the sections $E_n^b$, so $\alpha(E^b)=\sum_n\alpha(E_n^b)$, and termwise integration gives $\pi_1(E)=\sum_n\pi_1(E_n)$. Thus $\pi_1$ is a measure, and so, symmetrically, is $\pi_2$. On a rectangle, $\pi_1(C\times D)$ is the integral of the function $\alpha(C)1_D$ with respect to $\beta$, which equals $\alpha(C)\beta(D)$; when $\alpha(C)=\infty$, monotone convergence applied to the functions $k1_D$ gives this value under the convention $0\cdot\infty=0$. In the same way $\pi_2(C\times D)=\alpha(C)\beta(D)$.

Uniqueness. Let $\rho$ and $\rho'$ be measures on $\Sigma\otimes\mathcal T$ with $\rho(C\times D)=\rho'(C\times D)=\alpha(C)\beta(D)$ for every rectangle. Choose increasing sequences $A_n\in\Sigma$ and $B_n\in\mathcal T$ of finite measure with unions $A$ and $B$, and put $R_n=A_n\times B_n$. Fix $n$. The sets $E\in\Sigma\otimes\mathcal T$ with $\rho(E\cap R_n)=\rho'(E\cap R_n)$ include every rectangle, since a rectangle meets $R_n$ in a rectangle. They form a Dynkin class. Indeed, for $E\subset F$ in this class, $\rho((F\setminus E)\cap R_n)=\rho(F\cap R_n)-\rho(E\cap R_n)$ is a difference of finite numbers bounded by $\rho(R_n)=\alpha(A_n)\beta(B_n)$, and the same identity holds for $\rho'$; increasing unions are covered by continuity from below. Hence the equality holds for every $E\in\Sigma\otimes\mathcal T$. The rectangles $R_n$ increase to $A\times B$, so continuity from below gives $\rho(E)=\lim_n\rho(E\cap R_n)=\lim_n\rho'(E\cap R_n)=\rho'(E)$.

In particular $\pi_1=\pi_2$. This common measure is the required $\alpha\otimes\beta$: it has the stated rectangle values, no other measure has them, and it is sigma-finite because the sets $R_n$ have finite measure and cover $A\times B$.

The iterated integrals. For $u=1_E$, the three members of (5.T2) are $(\alpha\otimes\beta)(E)$, $\pi_1(E)$ and $\pi_2(E)$, which are equal. For the simple function $s_k$, the integral over $A\times B$ is $\sum_jc_j(\alpha\otimes\beta)(E_j)$ by the definition of the integral of a simple function. The inner integrals of $s_k$ were computed above, in both orders, and additivity and positive homogeneity of the outer integral give the iterated integrals $\sum_jc_j\pi_1(E_j)$ and $\sum_jc_j\pi_2(E_j)$. So (5.T2) holds for $s_k$. For general $u$, let $k\to\infty$. Monotone convergence applies on $A\times B$; it also applies to each inner integral, and then, since the inner integrals increase with $k$, to the outer integrals. All three members of (5.T2) therefore converge to the corresponding members for $u$, which are consequently equal. $\square$

In the averaging proof below, the first part of Lemma 5.3a makes the Haar section integral measurable on $F$ before any sigma-finiteness of its label measure is known. Formula (5.T2) is used only with two measures already proved sigma-finite. The ordinary nonnegative integration rules invoked in the lemma are proved in [Axler], Theorems 3.4, 3.7–3.11; they require neither sigma-finiteness nor a topology.

The coordinates (5.2) put the invariant measure on $K\times F$. The next argument proves its product structure directly. It applies to any measurable label space; its measure need not already be the pushforward of the original measure.

Theorem 5.4 (an invariant product measure). Let $K$ be a locally compact Polish group with left Haar measure $m$, let $F$ be a measurable space, and let $M$ be a sigma-finite measure on the product sigma-field of $K\times F$. Suppose that $(h,z)\mapsto(kh,z)$ preserves $M$ for every $k\in K$. Then there is a unique sigma-finite measure $\kappa$ on $F$ such that

$$M=m\otimes\kappa. \tag{5.8}$$

For any strictly positive finite Borel function $q$ on $K$ with $\int q\,dm=1$, it is given by

$$\kappa(B)=\int_{K\times F}q(h)1_B(z)\,dM(h,z). \tag{5.9}$$

It is nonzero if $M$ is nonzero. No regularity or local finiteness of $M$ is assumed.

Proof. Haar measure is nonzero and sigma-finite, as in Theorem 3.1. Lemma 1.1 therefore supplies such a function $q$. We use the Haar lesson's convention

$$m(Eh)=\Delta(h)m(E),\qquad \int f(gh)\,dm(g)=\Delta(h)^{-1}\int f(g)\,dm(g), \tag{5.10}$$

with $\Delta$ a continuous positive homomorphism. Its complete Theorem 11.1 gives

$$\int f(g^{-1})\,dm(g) =\int f(g)\Delta(g)^{-1}\,dm(g). \tag{5.11}$$

These formulas apply to all nonnegative Borel functions. Set $p=q\Delta$. Left invariance and (5.11), with $g=hu$, give, for every $h$,

$$\int_K p(g^{-1}h)\,dm(g) =\int_K p(u^{-1})\,dm(u) =\int_K q(u)\,dm(u)=1. \tag{5.12}$$

Define $\kappa$ by (5.9). It is a measure, by monotone convergence, although sigma-finiteness has yet to be proved. Multiplication and inversion on $K$ are jointly Borel, so the averaging integrands below are product-measurable. For a nonnegative product-measurable $f$, the function $Tf(z)=\int_K f(k,z)\,dm(k)$ is measurable by the one-measure assertion of Lemma 5.3a. Apply (5.T2) to the already sigma-finite measures $m$ and $M$, then apply the assumed invariance for each fixed $g$, and exchange these same two integrals again:

$$\begin{aligned} \int f\,dM &=\int_K\!\int_{K\times F} p(g^{-1}h)f(h,z)\,dM(h,z)\,dm(g)\\ &=\int_K\!\int_{K\times F} p(h)f(gh,z)\,dM(h,z)\,dm(g)\\ &=\int_{K\times F}p(h) \left(\int_K f(gh,z)\,dm(g)\right)dM(h,z)\\ &=\int_{K\times F}q(h)Tf(z)\,dM(h,z) =\int_F Tf(z)\,d\kappa(z). \end{aligned} \tag{5.13}$$

The third-to-fourth equality uses (5.10); the last equality is the pushforward integration formula for the weighted measure in (5.9). All integrands are nonnegative. No finite quantity is subtracted from an infinite one, and no Tonelli exchange against an as-yet unproved sigma-finite $\kappa$ occurs.

Choose a measurable cover $(E_n)$ of $K\times F$ with $M(E_n)<\infty$. Write

$$u_n(z)=m(\{h:(h,z)\in E_n\}),\qquad B_{n,j}=\{z:u_n(z)\geq1/j\},\quad j\geq1. \tag{5.14}$$

The functions $u_n$ are measurable. Equation (5.13) for $f=1_{E_n}$ gives

$$\int_F u_n\,d\kappa=M(E_n),\qquad \kappa(B_{n,j})\leq jM(E_n)<\infty. \tag{5.15}$$

For each $z$, the slices of the $E_n$ cover $K$. They cannot all have Haar measure zero, since $m(K)>0$. Thus the countably many $B_{n,j}$ cover $F$, proving sigma-finiteness. Only now does the two-measure part of Lemma 5.3a supply the sigma-finite product $m\otimes\kappa$; its formula (5.T2) and (5.13) agree on every indicator. This proves (5.8). If $\kappa=0$, the same formula gives $M=0$.

For uniqueness choose a Borel $C\subseteq K$ with $0<m(C)<\infty$, for example a compact identity neighborhood. Any product representation has $M(C\times B)=m(C)\kappa(B)$. Dividing by this fixed positive finite number determines $\kappa(B)$ for every measurable $B$, including those of infinite measure. In particular (5.9) is independent of the normalized function $q$. $\square$

The proof uses exact invariance of the whole product measure for every translation. It never chooses conditional measures with translation-dependent exceptional sets. The only product on which the two central integrations are exchanged is $m\otimes M$, whose factors are known sigma-finite at that point.

Corollary 5.5 (the fixed-range-fibre Haar formula). Under Theorem 5.2, fix a left Haar measure $m_y$ on $K$. There is a unique nonzero sigma-finite measure $\kappa_y$ on its Borel label set $F$ such that, for every nonnegative Borel $f$ on $Y_0$,

$$\int_{Y_0} f(\gamma)\,d\nu(\gamma) =\int_F\!\int_K f(h\,t(z))\,dm_y(h)\,d\kappa_y(z). \tag{5.16}$$

Moreover $\kappa_y\sim\lambda|_F$, for the probability $\lambda$ in (5.4).

Proof. Set $M=(\Phi^{-1})_*(\nu|_{Y_0})$. A Borel isomorphism preserves sigma-finiteness. Equations (5.2) and (5.6) transfer every isotropy translation to left translation in the first coordinate, so Theorem 5.4 applies. Its integral formula is (5.16). For a Borel $B\subseteq F$, $\kappa_y(B)=0$ exactly when $M(K\times B)=0$, because $m_y$ is nonzero. This is equivalent to $\nu(\sigma^{-1}B)=0$, then to $\mu(\sigma^{-1}B)=0$, and finally to $\lambda(B)=0$. Nonzero $\nu|_{Y_0}$ gives nonzero $\kappa_y$. $\square$

Proposition 5.6 (a change of section). On the same label set $F$, replace $t(z)$ by $t'(z)=b(z)t(z)$, where $b:F\to K$ is Borel. Keep the same left Haar measure and convention (5.10). Then the source-label measure in (5.16) changes by

$$d\kappa'_y(z)=\Delta(b(z))\,d\kappa_y(z). \tag{5.17}$$

In particular the null class is unchanged; the measure itself need not be unchanged in a nonunimodular group. Replacing $m_y$ by $c m_y$ instead replaces $\kappa_y$ by $c^{-1}\kappa_y$.

Proof. The old and new coordinates satisfy $h=h'b(z)$, so the map from old coordinates to new ones is $(h,z)\mapsto(hb(z)^{-1},z)$. For each fixed $z$, (5.10) gives $$\int_K f(hb(z)^{-1},z)\,dm_y(h) =\Delta(b(z))\int_K f(h',z)\,dm_y(h').$$ Tonelli proves the asserted product formula for the new coordinates. The density is positive and finite; it preserves the null class and sigma-finiteness, the latter by subdividing a finite-measure cover according to $\Delta(b)\leq n$. Uniqueness in Theorem 5.4 identifies $\kappa'_y$. The scalar change of Haar normalization follows directly from (5.16). $\square$

Connes's author-hosted Proposition 15, PDF 38–39, uses this topology and an isotropy Haar product decomposition in its positive argument. The unrestricted countably generated assertions have complete counterexamples in Countable generation and isotropy topologies and Principal groupoids with extra fibre information. Theorem 5.2 and Corollary 5.5 now provide the topology and full Haar product formula on each fixed standard Borel range fibre. Averaged coefficients and isotropy commutants, Theorems 5.1 and 6.1, supplies the full operator step at the global standard Borel scope, with one countable total family working at every unit. Its completed-probability descent and proper global kernel approximation avoid a jointly measurable choice of these topologies, Haar measures and sections. Such a joint field is therefore neither proved here nor needed for that adopted route. The broader countably generated counterexamples and the separate weak-measurable two-copy factor problem retain their exact scope. Modular spectral disintegration and the flow of weights retain their exact prerequisites in the modular and flow courses.

6. Five concrete models

Example 6.1 (translation overlap). On $\mathbb R$ with Lebesgue measure, the ordinary regular representation is $(\lambda_t f)(x)=f(x-t)$. For $C=[-1,1]$, $$\begin{gathered}\langle\lambda_t1_C,1_C\rangle=(2-|t|)_+,\\\{t:(2-|t|)_+>1\}=(-1,1)\\\subseteq C-C=[-2,2].\end{gathered} \tag{6.1}$$ The integral is the length of $[-1,1]\cap[t-1,t+1]$. Normalizing an equivalent positive-density probability changes the representation by a unitary conjugacy, as Exercise 7.1 proves, so it leaves the constructed topology unchanged.

Open diagram at full size

Figure 6.1. The upper panel shows the proof's exact maps and hypotheses: an analytic Borel group enters its faithful unitary model, a positive compact set is pulled back, and the overlap coefficient places an open identity neighborhood inside $CC^{-1}$. The lower panel is the exact real translation calculation (6.1), with $C=[-1,1]$, $m(C)=2$ for this Haar example, threshold $1$, and $C-C=[-2,2]$. The general probability-space construction uses its own $\mu(C)/2$; the plotted Haar coefficient is a concrete model, not a formula for every measure. Proof locators: Lemma 2.2, Theorem 3.1 and Example 6.1. Classical source context: [Mackey] and [Ramsay].

Example 6.2 (the rationals). Give $\mathbb Q$ its usual relative Borel structure. Every subset is Borel, since it is a countable union of singletons. An enumeration $(q_n)$ and masses $\mu(\{q_n\})=2^{-n}$, for $n\geq1$, give a quasi-invariant probability. The topology of Theorem 3.1 is discrete: a countable locally compact Polish group is Baire, so its decomposition into closed singletons forces an isolated point, and translations make every point isolated. This differs from the inherited real topology, which is not locally compact. That topology is not Polish: in a countable nondiscrete metrizable group every singleton is nowhere dense, contradicting the Baire theorem if a compatible complete metric existed.

Example 6.3 (affine transformations). On $G=(0,\infty)\times\mathbb R$, let $$(a,b)(c,d)=(ac,b+ad),\qquad d\nu(a,b)=a^{-2}\,da\,db. \tag{6.2}$$ The usual topology is Polish and locally compact. Left translation has Jacobian $a_0^2$, which cancels the denominator $(a_0a)^2$, so $\nu$ is left invariant. It is sigma-finite on the rectangles $[1/n,n]\times[-n,n]$. Right translation by $(c,d)$ has Jacobian $c$, and hence $$\nu(E(c,d))=c^{-1}\nu(E). \tag{6.3}$$ This is a nonunimodular example. The theorem produces the usual topology, by Corollary 4.2. The use of right translation in (3.5) requires preservation of null sets, rather than right invariance of the Haar measure.

Example 6.4 (zero measure on isotropy). Let the units be $\mathbb R$, and write arrows as $(x,h,z):z\to x$, with

$$(x,h,z)(z,k,u)=(x,h+k,u),\qquad (x,h,z)^{-1}=(z,-h,x). \tag{6.4}$$

Give each range fibre its measure $d\nu^x(h,z)=dh\,dz$. Left translation changes $h$ by a constant and keeps $z$ fixed, so this is a transverse function. The sets where all three coordinates have absolute value at most $n$ cover the arrow space and have range-fibre measure at most $4n^2$. Thus it is faithful and proper. The isotropy at $y$ is the line $z=y$ and has $\nu^y$-measure zero. Its group is the additive real line.

Choose $t(z)=(y,0,z)$. Then $a(y,h,z)=h$, and the equivalent probability

$$d\mu_y(h,z)=\tfrac14e^{-|h|-|z|}\,dh\,dz, \qquad d\rho(h)=\tfrac12e^{-|h|}\,dh \tag{6.5}$$

has the asserted nonzero pushforward. Every translated density is strictly positive, so $\rho$ is quasi-invariant. This constructs the isotropy measure without restricting $\nu^y$ to its null isotropy line. Pushing the unweighted $\nu^y$ to the same coordinate would give infinite mass to every set of positive Lebesgue measure; Exercise 7.9 explains why that measure is not sigma-finite.

Open diagram at full size

Figure 6.2. The model fixes $y=0$, so $Y=\mathcal G^0$ has coordinates $(h,z)$. Its Haar-isotropy line $z=0$ is null for $dh\,dz$; the section is $t(z)=(0,0,z)$, and $a(0,h,z)=h$. The square is the schematic window $[-2,2]^2$ in the full plane; the right-hand density curve is sampled every $0.025$ on $[-3,3]$. The marked arrow is exactly $(0,1.4,0.9)$. The probability and its pushforward in (6.5) are defined on the full plane and line, respectively. The diagram traces the mechanism proved in Theorem 5.2 and calculated in Example 6.4; its source context is Connes's Proposition 15 and the programme's full Jankov–von Neumann section theorem.

Example 6.5 (the affine section factor). Use the affine group $K$ of Example 6.3 as the isotropy in the transitive groupoid with arrows $(x,h,z):z\to x$, multiplication $(x,h,z)(z,k,u)=(x,hk,u)$, and range measures $dm(h)\,dz$. Fix a range unit $y$. For the section $t(z)=(y,(1,0),z)$, the label measure is $dz$. Our modular convention is $\Delta(a,b)=a^{-1}$, since (6.3) gives $m(E(c,d))=c^{-1}m(E)$.