Borel carriers and uniform wandering neighborhoods

Written by GPT-6.1 Sol (OpenAI), Ultra, October 2026. New original text is public domain (CC0).

Introduction

A continuous free action separates a point from all its translates by a compact family avoiding the identity. To obtain a positive measurable wandering set, one needs a point where the original positive set has positive measure in every neighborhood. That point can be found without assuming that the measure is Radon.

This distinction sharpens Takesaki III, XIII.1, Exercise 5. Part (a) asks for a positive compact subset of a given Borel set. Part (b) separates a support point from a compact family of translates. Part (c) asks for measurable compact wandering freeness. We prove part (c) at the printed finite, quasi-invariant, ergodic, full-support topological scope, retaining the chapter's separable locally compact group convention. We do not assume a countable base for the space or Radon regularity in that proof.

For part (a), we give the complete regularity theorem and application when every open set is sigma-compact, including the second countable locally compact case. The exercise block states a general countability convention, “such as separability,” without explicitly choosing a countable base for its topological space. A countable dense set alone is not substituted for that stronger topological hypothesis. We give a complete disposition of the distinct scopes: positive compact extraction holds under the stronger regularity hypothesis and in the atomic standard branch, while Theorem 3.29 gives an atomless counterexample under CH for ordinary separability and an actual conull standard Borel measure. The printed convention does not specify a unique intended topology. The complete part (c) proof holds throughout.

Compact fixed sets and free Polish models, Lemmas 2.1–2.2, proves all three intended assertions for finite Radon measures on arbitrary locally compact Hausdorff spaces. We preserve that proof and now prove the regularity input explicitly, then supply a second argument which bypasses compact extraction for the measurable conclusion. Free actions and the crossed-product diagonal, Definition in Section 4, fixes the compact projection condition. The underlying group action here is an actual jointly continuous point action; every nonidentity element fixes no point.

We use Hausdorff topology, compactness, countable additivity, continuity of measure along monotone sequences, and basic countable ordinal arithmetic. General standard-form, modular and weight theory is not used. Measured covariant-system interpretations retain the chapter's standard sigma-finite background. The atomic branch below uses only countable point separation on a conull standard Borel subspace and proves all three source clauses in that branch. The atomless standard-measure case is treated by the full CH construction below. CH is explicit; existence of such a counterexample without CH is not claimed.

1. Exactly where finite Borel regularity follows

Throughout this section, $X$ is locally compact Hausdorff and every open subset of $X$ is sigma-compact. Let $\mu$ be any finite Borel measure. Here “Radon” means local finiteness and inner approximation of every Borel set by compact subsets. We also prove outer approximation by open supersets.

Lemma 1.1 (the countable cover). A second countable locally compact Hausdorff space satisfies the open-set sigma-compactness hypothesis above.

Proof. First, given $x\in U$ with $U$ open, choose an open neighborhood $V$ of $x$ whose compact closure lies in $U$. Here is the local shrinking argument. Begin with a compact neighborhood $C$ of $x$. In the compact Hausdorff space $C$, separate $x$ from the compact set $C\setminus(U\cap\operatorname{int}C)$: choose disjoint neighborhoods for $x$ and each point of that set, take a finite cover of the latter, and intersect the corresponding neighborhoods of $x$. An ambient open neighborhood inside this intersection and $\operatorname{int}C$ has closure in $C$ disjoint from that compact set. This is the required $V$.

The $V$'s cover $U$. A countable base makes $U$ Lindelöf: for each basic open set contained in a member of this cover choose one such member. These countably many choices still cover $U$. Write them as $V_j$. Then $$C_n=\bigcup_{j\le n}\overline{V_j}\subset U,\qquad C_n\text{ compact},\qquad U=\bigcup_n C_n. \tag{1.1}$$ Thus $U$ is sigma-compact. $\square$

In particular, a separable metrizable locally compact space has this property: balls with centers in a countable dense set and positive rational radii form a countable base. No implication from separability alone for a general Hausdorff space is used.

Theorem 1.2 (finite Borel regularity, with full proof). For every Borel $E\subset X$ and every $\varepsilon>0$, there are compact $K$ and open $U$ such that $$K\subset E\subset U,\qquad \mu(U\setminus K)<\varepsilon. \tag{1.2}$$ The same inner approximation holds for sets measurable in the completion.

Proof. Every open set $O$ is an increasing union of compact sets: take finite unions of any compact sigma-compact cover of $O$. Continuity from below gives $$\mu(O)=\sup\{\mu(K):K\subset O,\ K\text{ compact}\}. \tag{1.3}$$ In particular choose compact $D_n\uparrow X$. Since $\mu(X)<\infty$, $\mu(X\setminus D_n)\to0$.

Every closed $F$ has open outer approximants. Indeed, choose compact $K\subset X\setminus F$ with $\mu((X\setminus F)\setminus K)<\varepsilon$, using (1.3). The open set $U=X\setminus K$ contains $F$, and $\mu(U\setminus F)<\varepsilon$. Every open $O$ already has closed inner approximants by (1.3).

Let $\mathcal R$ be the class of Borel $E$ such that for every positive $\varepsilon$ there are a closed $F$ and an open $U$ with $$F\subset E\subset U,\qquad \mu(U\setminus F)<\varepsilon. \tag{1.4}$$ It contains the open and closed sets. Complementation interchanges $F,U$ with $X\setminus U,X\setminus F$ and preserves the gap, so $\mathcal R$ is closed under complements.

Suppose $E_j\in\mathcal R$ and $E=\bigcup_j E_j$. Choose closed $F_j\subset E_j\subset U_j$ open with $\mu(U_j\setminus F_j)<\varepsilon 2^{-j-3}$. For $U=\bigcup_j U_j$, $$\mu(U\setminus E)\le\sum_j\mu(U_j\setminus E_j)<\varepsilon/4. \tag{1.5}$$ Finiteness and continuity from below give $N$ with $\mu(E\setminus\bigcup_{j\le N}E_j)<\varepsilon/4$. The finite union $F=\bigcup_{j\le N}F_j$ is closed and $$\mu(E\setminus F) \le\mu\left(E\setminus\bigcup_{j\le N}E_j\right) +\sum_{j\le N}\mu(E_j\setminus F_j)<\varepsilon/2. \tag{1.6}$$ Consequently $\mu(U\setminus F)<\varepsilon$. Thus $\mathcal R$ is a sigma-algebra containing all open sets, hence contains every Borel set.

Apply (1.4) with gap below $\varepsilon/2$. Choose $n$ such that $\mu(X\setminus D_n)<\varepsilon/2$, and put $K=F\cap D_n$. This set is compact, lies in $E$, and $\mu(U\setminus K)\le\mu(U\setminus F)+\mu(X\setminus D_n)<\varepsilon$. This proves (1.2). Local finiteness follows from global finiteness.

If $E$ is completion-measurable, there are Borel $E_0\subset E\subset E_1$ with $\mu(E_1\setminus E_0)=0$. Apply compact inner approximation to $E_0$. Its compact subsets already lie in the actual set $E$, and their measures approach $\mu(E)$. $\square$

This proves, rather than assumes, the finite-Borel-to-Radon step under the stated topological hypothesis. In that scope the complete source Exercise 5(a) follows from the one-element separation and finite-cover proof in the earlier lesson, Lemma 2.1.

2. Ergodicity supplies a compact carrier without regularity

We now drop every open-set sigma-compactness assumption on $X$, and do not assume that $\mu$ is Radon. Suppose $X\ne\varnothing$ is locally compact Hausdorff, $G$ is a separable locally compact Hausdorff group, and $G\times X\to X$ is jointly continuous. Let $\mu$ be a nonzero finite Borel measure with full support. Ergodicity means that every exactly $G$-invariant Borel set is null or conull.

Theorem 2.1 (a conull sigma-compact carrier). There are increasing compact $C_n\subset X$ whose union $S$ is exactly invariant and conull: $$S=\bigcup_n C_n,\qquad GS=S,\qquad \mu(X\setminus S)=0. \tag{2.1}$$

Proof. The group is sigma-compact even though it has not been assumed second countable. Choose a relatively compact open identity neighborhood $W$ and a countable dense set $D\subset G$. Every $gW^{-1}$ meets $D$, so $g\in dW$ for some $d\in D$. Thus the compact sets $d\overline W$ cover $G$. Finite unions give increasing compact $G_n\uparrow G$.

Choose a nonempty relatively compact open set $V\subset X$, and put $C=\overline V$. The set $GV=\bigcup_{g\in G}gV$ is open and exactly invariant. It has positive measure because full support gives $\mu(V)>0$. Ergodicity therefore makes $GV$ conull.

Set $C_n=G_nC$. It is the continuous image of the compact product $G_n\times C$, hence compact and closed. The set $$S=GC=\bigcup_n G_nC \tag{2.2}$$ is Borel, contains $GV$, and is conull. Multiplication by any fixed group element permutes the full set of parameters $G$, so $GS=S$ exactly. This proves (2.1). $\square$

Notice what this does and does not assert. It makes the ambient carrier sigma-compact. It has not proved that a positive Borel subset of that carrier contains a positive compact subset. We will not need this latter conclusion for measurable wandering freeness.

Lemma 2.2 (a support point on a compact carrier). Let $C\subset X$ be compact, $E\subset X$ Borel, and $0<\mu(E\cap C)<\infty$. There is $x_0\in C$ such that $$\mu(E\cap C\cap U)>0 \quad\text{for every open neighborhood }U\text{ of }x_0. \tag{2.3}$$ The point need not belong to $E$.

Proof. If no such point existed, for each $x\in C$ choose an open neighborhood $U_x$ with $\mu(E\cap C\cap U_x)=0$. Finitely many $U_x$'s cover the compact set $C$. Finite subadditivity would then give $\mu(E\cap C)=0$, a contradiction. The proof uses neither inner nor outer regularity. $\square$

For example $E=(0,1)$, $C=[0,1]$ and Lebesgue measure have $x_0=0$ satisfying (2.3), although $0\notin E$. The global pointwise freeness hypothesis in the next section applies to this point as well.

3. The full measurable compact wandering conclusion

Lemma 3.1 (uniform separation at one point). Let a topological group act jointly continuously on a Hausdorff space. Suppose $K\subset G$ is compact and $x_0\notin Kx_0$. There is an open neighborhood $U_0$ of $x_0$ such that $$U_0\cap KU_0=\varnothing. \tag{3.1}$$

Proof. The orbit fragment $Kx_0$ is compact. A point and a disjoint compact set have disjoint open neighborhoods: separate the point from each member of the compact set, choose a finite cover on that set, intersect the point neighborhoods and unite the other neighborhoods. Let these resulting neighborhoods be $U_*$ of $x_0$ and $V$ of $Kx_0$.

For each $k\in K$, joint continuity gives neighborhoods $W_k$ of $k$ in $G$ and $U_k$ of $x_0$ with $W_kU_k\subset V$. Choose a finite set of parameters whose $W_k$'s cover $K$. Intersect their $U_k$'s with $U_*$, obtaining $U_0$. For every $k\in K$, at least one selected $W_j$ contains $k$, so $kU_0\subset W_jU_j\subset V$. Therefore $KU_0\subset V$, whereas $U_0\subset U_*$ and $V\cap U_*=\varnothing$. This proves (3.1). $\square$

Theorem 3.2 (source Exercise 5(c), without Radon regularity). Under Section 2's hypotheses, assume every point of $X$ has trivial stabilizer. For every compact $K\subset G\setminus\{e\}$ and every positive-measure Borel $E\subset X$, there is a Borel $F\subset E$ with $$\mu(F)>0,\qquad kF\cap F=\varnothing\quad(k\in K). \tag{3.2}$$ Consequently the action on $L^\infty(X,\mu)$ satisfies the compact projection condition.

Proof. If $K=\varnothing$, take $F=E$. Otherwise Theorem 2.1 gives increasing compact $C_n$ with conull union. Some $n$ satisfies $\mu(E\cap C_n)>0$. Finiteness of $\mu$ makes this measure finite. Lemma 2.2 supplies $x_0\in C_n$ satisfying (2.3). Trivial stabilizers and $e\notin K$ give $x_0\notin Kx_0$. Lemma 3.1 provides $U_0$ satisfying (3.1).

Take the actual Borel set $$F=E\cap C_n\cap U_0. \tag{3.3}$$ It has positive measure by (2.3). Since $F\subset U_0$, every $kF\subset KU_0$ is disjoint from $F$. This is (3.2), simultaneously for the entire compact family. Compactness of $F$ has not been asserted.

For the algebraic statement, let $p=\mathbf1_E$ be a nonzero projection and $q=\mathbf1_F$. Quasi-invariance makes $\alpha_k(f)(x)=f(k^{-1}x)$ an automorphism on measurable classes. We have $$0\ne q\le p,\qquad q\alpha_k(q)=\mathbf1_{F\cap kF}=0 \quad(k\in K). \tag{3.4}$$ Thus the exact compact projection condition holds. No union of individual exceptional null sets has been used. $\square$

The standing separable group condition, the finite Borel measure, full support, quasi-invariance and ergodicity are the printed assumptions relevant here. The proof places no additional countability or regularity condition on $X$. Quasi-invariance is used only to interpret (3.4) as an action on the measure algebra. The set conclusion precedes it.

Corollary 3.3 (the carrier is the only measured input). Suppose a jointly continuous pointwise free action on a Hausdorff $X$ has a sigma-finite Borel measure with a conull countable union of compact sets. Then (3.2) holds. Neither ergodicity nor full support is required if that carrier is already given.

Proof. Write $X=\bigcup_j H_j$ up to null sets with $\mu(H_j)<\infty$, and let the compact carrier be $\bigcup_n C_n$. A positive $E$ has $0<\mu(E\cap H_j\cap C_n)<\infty$ for some $j,n$. Apply Lemma 2.2 to the Borel set $E\cap H_j$ and compact $C_n$. Lemma 3.1 and the set $F=E\cap H_j\cap C_n\cap U_0$ finish the proof. $\square$

When a finite Radon measure is given instead, the earlier lesson supplies a positive compact $L\subset E$, and Lemma 2.2 applied to $E=L,C=L$ gives a support point in $L$. Lemma 3.1 proves the source's part (b) for every prescribed compact $K$ avoiding the identity. The printed part (b) does not introduce $K$; Definition XIII.1.3 supplies the intended parameter, which is explicit here. Under Theorem 1.2's hypotheses, the earlier compact extraction completes part (a) as well.

Lemma 3.4 (an atom is a point on the standard conull subspace). Let $(X,\Sigma,\mu)$ be a finite measure space with a measurable conull subset $S$ whose relative sigma-algebra admits a countable family $B_n$ separating its points. If $A\in\Sigma$ is a measure atom, meaning $$\mu(A)>0,\qquad \mu(A\cap H)\in\{0,\mu(A)\}\quad(H\in\Sigma), \tag{3.5}$$ then there is $x\in A\cap S$ with $\mu(\{x\})=\mu(A)$. Thus $A$ agrees with that singleton modulo null sets.

Proof. Each $B_n$ is measurable in $X$ because it is measurable in the measurable subspace $S$. The set $A\cap S$ has measure $\mu(A)$. Choose $C_n=B_n$ when $\mu(A\cap B_n)=\mu(A)$, and otherwise choose $C_n=S\setminus B_n$. In the latter case the atom condition gives $\mu(A\cap B_n)=0$, hence again $\mu(A\cap C_n)=\mu(A)$. Countable subadditivity gives $$F=A\cap S\cap\bigcap_{n\ge1}C_n,\qquad \mu(A\setminus F)=0,\qquad \mu(F)=\mu(A)>0. \tag{3.6}$$ Any two distinct points of $S$ differ on some $B_n$, whereas all points of $F$ have the same membership in every $B_n$. Thus $F$ contains at most one point. Its positive measure makes it nonempty, so $F=\{x\}$, proving the assertion. No continuity of the point code or change of topology has been used. $\square$

A standard Borel subspace supplies such a family: realize it as a Polish Borel space and pull back a countable base, which separates distinct points. Takesaki I, Appendix defines a standard measure by a null deletion leaving a standard Borel point space. Only this measurable implication is used here.

Theorem 3.5 (the full atomic branch of source Exercise 5). Retain the finite quasi-invariant ergodic measure and pointwise free action of Section 3, and suppose the measure is standard in the preceding sense. If $\mu$ has a measure atom, then $G$ is countable and one orbit $O=Gx_0$ is conull and consists of positive point atoms. Every positive Borel $E\subset X$ contains $x\in O$ such that $$L=\{x\}\subset E,\qquad \mu(L)>0,\qquad gL\cap L=\varnothing\quad(g\ne e). \tag{3.7}$$ This proves part (a) simultaneously for every nonidentity element; parts (b),(c) follow with the same point. Compact inner regularity is not an additional premise.

Proof. Lemma 3.4 supplies one point of positive mass. Let $$D=\{x\in X:\mu(\{x\})>0\},\qquad D_n=\{x:\mu(\{x\})\ge 1/n\}. \tag{3.8}$$ Each $D_n$ is finite: an $m$-point subset has measure at least $m/n$, so $m\le n\mu(X)$. Thus $D=\bigcup_nD_n$ is countable. Hausdorffness makes singletons closed, so $D$ is Borel. Quasi-invariance gives $\mu(\{gx\})>0$ if and only if $\mu(\{x\})>0$. Consequently $D$ is exactly invariant. It is nonnull, so ergodicity makes it conull.

Fix $x_0\in D$. The orbit map $g\mapsto gx_0$ takes values in the countable set $D$ and is injective by freeness. Hence $G$ is countable. The orbit $O$ is then a countable Borel set, exactly invariant and nonnull, so ergodicity makes it conull. A positive $E$ has $E\cap O\ne\varnothing$. Choose $x$ there. It has positive mass; its singleton is compact in the original topology, and freeness gives $gx\ne x$ for every $g\ne e$. This proves (3.7).

Every neighborhood of $x$ has positive intersection with $L=\{x\}$, so $x$ is the density point for part (b). Lemma 3.1 separates it uniformly from any prescribed compact $K\subset G\setminus\{e\}$. For part (c), $q=\mathbf1_{\{x\}}$ is a nonzero projection below $\mathbf1_E$, with $q\alpha_g(q)=0$ for every nonidentity $g$, hence for every such compact family. $\square$

Corollary 3.6 (the unresolved branch is atomless). At the full standard-measure source scope, either the measure is atomless or Theorem 3.5 proves all three clauses. In particular an uncountable $G$ forces the measure to be atomless.

Proof. A finite measure that is not atomless has a positive measure atom by definition. Lemma 3.4 and Theorem 3.5 apply, and their countability conclusion excludes this alternative when $G$ is uncountable. The atomless alternative is retained even for countable groups; its compact-extraction step is not proved by this dichotomy. $\square$

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Figure 1. The exact atomic route is Lemma 3.4, Theorem 3.5 and Corollary 3.6. The countable point code is used only on the original conull measurable subspace $S$; its intersection retains the whole mass of the atom and contains exactly one point. The conull orbit consists of positive point atoms. Choosing one actually in $E$ gives compact $L=\{x\}$, disjoint from every nonidentity translate. Compactness is never transported through a Borel isomorphism. Source hypothesis: Takesaki I, Appendix standard-measure definition; source conclusion: Takesaki III, Exercise XIII.1.5(a)–(c).

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Figure 2. The two paths distinguish compact extraction from measurable freeness. The upper path is Theorem 1.2 and the earlier one-element lemma. The middle path is Theorems 2.1 and 3.2, using a conull carrier $GC$, finite compact coverage, a support point and uniform compact-parameter separation. The lower interval example uses exact $K=[-3,-2]\cup[2,3]$, $U=(-2/5,2/5)$ and translates $KU=(-17/5,-8/5)\cup(8/5,17/5)$. The ordinal panel illustrates the failure of bare finite Borel regularity; it is outside the exercise's countability and full-support assumptions. Sources: Takesaki III, XIII.1, Definition 1.3 and Exercise 5; Example 4.2 below.

Lemma 3.7 (a maximal compact remainder). Let $\mu$ be a finite Borel measure on a Hausdorff space and let $E$ be Borel. Put $$r(E)=\sup\{\mu(C):C\subset E,\ C\text{ compact}\}. \tag{3.9}$$ There is a Borel $F\subset E$ such that $$\mu(F)=\mu(E)-r(E),\qquad \mu(C)=0\quad\text{for every compact }C\subset F. \tag{3.10}$$ Consequently, the assertion that every positive Borel subset of $E$ contains a positive compact subset is equivalent to compact inner regularity on every Borel subset of $E$.

Proof. Choose compact $C_n\subset E$ with $\mu(C_n)\to r(E)$; if $r(E)=0$, take $C_n=\varnothing$. The finite unions $D_n=\bigcup_{j\le n}C_j$ are compact and contained in $E$. Their masses are at most $r(E)$ and at least $\mu(C_n)$, so $\mu(D_n)\to r(E)$. Thus $D=\bigcup_nD_n$ is Borel with $\mu(D)=r(E)$. Set $F=E\setminus D$. If $C\subset F$ is compact, then $C\cap D_n=\varnothing$ and $C\cup D_n$ is a compact subset of $E$. Hence $$\mu(C)+\mu(D_n)\le r(E). \tag{3.11}$$ Letting $n\to\infty$ gives $\mu(C)=0$ and proves (3.10).

For the consequence, apply this construction to any Borel $H\subset E$. If $r(H)<\mu(H)$, its positive remainder contradicts the asserted positive compact extraction. Therefore $r(H)=\mu(H)$ for every such $H$. The reverse implication follows by choosing a compact approximant with positive mass. The construction produces an actual Borel subset in the original space. $\square$

Call a Borel $A\subset X$ a Radon part of $\mu$ if every Borel $E\subset A$ is compact inner regular. Equivalently, the measure $\mu_A(H)=\mu(H\cap A)$ is a finite Radon measure on $X$. Indeed, apply inner regularity to $H\cap A$ to get compacts contained in that actual set; conversely, for $E\subset A$, compact approximants for $\mu_A(E)$ lie in $E$, where $\mu_A=\mu$. Global finiteness supplies local finiteness.

Proposition 3.8 (the largest Radon part). For a finite Borel measure on a locally compact Hausdorff space, the Radon parts form an ideal under Borel subsets and countable unions. Their membership is unchanged by a Borel null modification. There is a Borel Radon part $R$, unique modulo null sets, such that $$\mu(A\setminus R)=0 \quad\text{for every Borel Radon part }A. \tag{3.12}$$ The set $R$ is a measure carrier, not a claimed closed support.

Proof. Restriction to a Borel subset preserves the defining approximation. If $A\triangle B$ is null and $A$ is a Radon part, each Borel $E\subset B$ has $\mu(E\cap A)=\mu(E)$. Compact approximants contained in $E\cap A$ therefore prove that $B$ is also a Radon part.

Suppose $A_j$ are Radon parts and $A=\bigcup_jA_j$. For Borel $E\subset A$, disjointize the pieces: $E_j=E\cap(A_j\setminus\bigcup_{i<j}A_i)$. Finiteness chooses $N$ such that $\sum_{j>N}\mu(E_j)<\varepsilon/2$. For $j\le N$, choose compact $K_j\subset E_j$ with $\mu(E_j\setminus K_j)<\varepsilon/(2N)$. Their finite union is compact, lies in the actual $E$, and misses mass below $\varepsilon$. Thus $A$ is a Radon part.

Let $m$ be the supremum of the masses of all Borel Radon parts. Choose parts $A_n$ whose masses tend to $m$, taking empty sets if $m=0$, and put $R=\bigcup_nA_n$. By the union assertion, $R$ is a Radon part; its mass equals $m$. For any Radon part $A$, the union $R\cup A$ is again a part, so $\mu(R\cup A)\le m=\mu(R)$. This proves (3.12). Applying it in both directions proves uniqueness modulo null sets. $\square$

Lemma 3.9 (transport of a Radon part). Suppose $g:X\to X$ is a homeomorphism and both $g$ and $g^{-1}$ preserve $\mu$-null sets. Then $A$ is a Radon part if and only if $gA$ is a Radon part.

Proof. The finite measure $\eta(H)=\mu(gH)$ is absolutely continuous with respect to $\mu$. We need its uniform finite-measure consequence: for every $\varepsilon>0$ there is $\delta>0$ such that $$\mu(H)<\delta\ \Longrightarrow\ \eta(H)<\varepsilon. \tag{3.13}$$ If this failed, choose $H_n$ with $\mu(H_n)<2^{-n}$ and $\eta(H_n)\ge\varepsilon$. The tail unions $V_m=\bigcup_{n\ge m}H_n$ decrease to a $\mu$-null set. But $\eta(V_m)\ge\varepsilon$, so finiteness and continuity from above make their intersection have $\eta$-mass at least $\varepsilon$, contradicting absolute continuity. This proves (3.13).

For Borel $E\subset gA$, choose compact $C\subset g^{-1}E\subset A$ with $\mu(g^{-1}E\setminus C)<\delta$. Then $gC\subset E$ is compact in the original topology and $\mu(E\setminus gC)=\eta(g^{-1}E\setminus C)<\varepsilon$. Thus $gA$ is a Radon part. Apply the same argument to $g^{-1}$ for the converse. Neither invariance of the measure nor a bounded Radon–Nikodym derivative was assumed. $\square$

Theorem 3.10 (ergodic regularity dichotomy). Let $G$ act on a locally compact Hausdorff $X$ by homeomorphisms, preserving the null sets of a nonzero finite Borel measure $\mu$. Suppose the measure-algebra action is ergodic: a Borel $B$ with $\mu(gB\triangle B)=0$ for every $g$ is null or conull. Then exactly one of the following holds:

1. $\mu$ is Radon on the original $X$.
2. Every positive Borel $E\subset X$ contains a positive Borel $F\subset E$ for which every compact $C\subset F$ has measure zero.

Moreover, one positive Borel Radon part suffices for the first alternative. Under the chapter's jointly Borel standard-measure action convention, exactly invariant-set ergodicity suffices: Free actions and the crossed-product diagonal, Proposition 4.7, proves the required invariant-class equivalence for a separable locally compact group.

Proof. Take the maximal Radon part $R$ from Proposition 3.8. Lemma 3.9 makes $gR$ a Radon part, so $\mu(gR\setminus R)=0$. Applying this to $g^{-1}$, and then transporting its null exceptional set by $g$, gives $\mu(R\setminus gR)=0$. Therefore $$\mu(gR\triangle R)=0\qquad(g\in G). \tag{3.14}$$ This is invariance of one measure class for every fixed $g$; no intersection of uncountably many conull sets is taken. Ergodicity makes $R$ null or conull.

If $R$ is conull, the null-modification assertion of Proposition 3.8 makes $X$ itself a Radon part. Hence $\mu$ is Radon. Any positive Radon part forces $\mu(R)>0$, so forces this case.

If $R$ is null, no positive Borel set is a Radon part. Given positive $E$, there is therefore a Borel $H\subset E$ with $r(H)<\mu(H)$. Lemma 3.7 supplies the positive Borel $F\subset H\subset E$ in alternative 2. The alternatives are disjoint because Radon regularity would give a positive compact subset of this $F$. $\square$

Corollary 3.11 (the exact compact-extraction question). In the finite, pointwise free source setting, fix any $g\ne e$. The assertion $$\text{for every positive Borel }E,\ \exists\,L\subset E\text{ compact}: \mu(L)>0,\quad gL\cap L=\varnothing \tag{3.15}$$ is equivalent to $\mu$ being Radon in the original topology. If the action is ergodic in Theorem 3.10's sense, it is enough to prove (3.15) for all positive Borel subsets of one positive Borel part $A$.

Proof. Assertion (3.15) supplies positive compact extraction after forgetting the disjointness condition. Lemma 3.7 with $E=X$ proves compact inner regularity on all Borel sets. Conversely, the original one-element separation and finite-cover proof in Compact fixed sets and free Polish models, Lemma 2.1, gives (3.15) for a finite Radon measure. For the localized assertion, Lemma 3.7 makes $A$ a positive Radon part, and Theorem 3.10 makes $\mu$ Radon everywhere. The quantifier ranges over every positive Borel subset; one positive compact set alone does not suffice. $\square$

Corollary 3.12 (the possible atomless obstruction). Retain the full source setting and its standing jointly Borel standard-measure convention. If the second alternative of Theorem 3.10 occurs, then $\mu$ is atomless. For every positive Borel $E$ and every compact $K\subset G\setminus\{e\}$, there is a positive Borel $F\subset E$ such that $$F\cap kF=\varnothing\quad(k\in K),\qquad \mu(C)=0\quad(C\subset F\text{ compact}). \tag{3.16}$$

Proof. A positive measure atom would, by Lemma 3.4, give a positive singleton. That singleton is a positive Radon part, contradicting the second alternative. Thus the measure is atomless.

Theorem 3.10 first gives a positive Borel $H\subset E$ with no positive compact subset. Apply the complete measurable wandering proof, Theorem 3.2, to $H$ and the prescribed $K$. It supplies a positive Borel $F\subset H$ missing all its $k$-translates at once. Every compact subset of $F$ is a compact subset of $H$, hence null. This proves (3.16). The statement is conditional on alternative 2 occurring; no example satisfying all the standing standard-measure dynamical hypotheses is asserted. $\square$

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Figure 3. Proposition 3.8 selects the largest Radon part $R$ modulo null sets; Lemma 3.9 and ergodicity force its mass to be zero or the full mass. In the right branch, Lemma 3.7 removes a countable increasing union of compact approximants and leaves an actual positive Borel remainder containing no positive compact subset. Theorem 3.2 then gives the same measurable wandering conclusion there. The figure records an exhaustive conditional dichotomy, not existence of its right branch at the standing standard-measure scope. Proof locators: Lemmas 3.7 and 3.9, Proposition 3.8, Theorem 3.10 and Corollaries 3.11–3.12. Source target: Takesaki III, XIII.1.5(a)–(c), printed11–12/PDF31–32.

These reductions identify the remaining source decision precisely: do the full standing atomless standard-measure, free, ergodic, quasi-invariant, full-support hypotheses force the first alternative? A measure-only nonregular example does not answer that question. A compact conull carrier, or even a positive compact set, does not supply a Radon restriction on that carrier.

Lemma 3.13 (support and nonsingular transport). For a finite Borel measure on a Hausdorff space, define $$S_\mu=\{x:\mu(U)>0\text{ for every open }U\ni x\}. \tag{3.17}$$ This set is closed. If $X$ is compact and $\mu(X)>0$, it is nonempty. Equivalent Borel measures have the same support. If $h:X\to X$ is a homeomorphism, then $S_{h_*\mu}=hS_\mu$. In particular every nonsingular homeomorphism preserves $S_\mu$ exactly.

Proof. The complement of $S_\mu$ is the union of all open sets of measure zero, and is therefore open. If the support were empty on a compact $X$, those zero-measure open sets would cover $X$; a finite subcover would force $\mu(X)=0$. Equivalence of measures gives the same collection of zero-measure open sets. An open neighborhood $U$ of $hx$ has $h^{-1}U$ as an open neighborhood of $x$, and $(h_*\mu)(U)=\mu(h^{-1}U)$. This proves the transport formula. Combining it with equivalence proves exact invariance. No uncountable intersection of conull sets is involved. $\square$

Nonemptiness of the support has not asserted that it carries any mass. The next proposition gives a precise test for that failure.

Proposition 3.14 (a locally countable conull carrier has null support). Let $X$ be compact Hausdorff, $\mu$ a Borel probability with $\mu(\{x\})=0$ for every $x$, $D\subset X$ closed, and $Y\subset D$ Borel with $\mu(Y)=1$. Suppose that for every $y\in Y$ there is an ambient open neighborhood $U_y$ such that $U_y\cap D$ is countable. Then $$\varnothing\ne S_\mu\subset D\setminus Y,\qquad \mu(S_\mu)=0,\qquad \mu(X\setminus S_\mu)=1. \tag{3.18}$$ The open set $X\setminus S_\mu$ is a union of zero-measure open sets but has measure one. Every compact subset of $Y$ is countable and null.

Proof. Since $Y$ is conull, $\mu(X\setminus D)=0$. Countable additivity and the singleton hypothesis give $\mu(U_y\cap D)=0$; hence $\mu(U_y)=0$. Thus no $y\in Y$ belongs to the support. No point outside the closed set $D$ belongs to it either, since $X\setminus D$ is a zero-measure open neighborhood. This proves $S_\mu\subset D\setminus Y$. Lemma 3.13 supplies nonemptiness and closedness, and the displayed containment gives its zero mass.

By the definition of support, $X\setminus S_\mu$ is the union of the zero-measure open neighborhoods just described, including any others. It is Borel and has measure one. Countable additivity makes every countable subunion null; it does not make this arbitrary union null.

If $C\subset Y$ is compact, the $U_y$'s cover $C$. Finitely many suffice, and $C$ is contained in their finite union of countable intersections with $D$. Therefore $C$ is countable and has measure zero. $\square$

Corollary 3.15 (the two proposed repairs). Under Proposition 3.14's hypotheses, replacing $\mu$ by any equivalent finite measure leaves the support unchanged and null. Restricting $\mu$ to its support gives the zero measure. Suppose additionally that $X$ has a countable dense subset $Q=\{q_n:n\ge1\}$ of distinct points. Choose $c_n>0$ with $\sum_n c_n=1$, set $\eta=\sum_n c_n\delta_{q_n}$, and fix $0<t<1$. The probability $$\lambda=(1-t)\mu+t\eta \tag{3.19}$$ has full support, but cannot be ergodic under any group of nonsingular homeomorphisms. It still has a positive Borel subset with no positive compact subset.

Proof. Lemma 3.13 proves the equivalent-measure assertion; equivalence also preserves the nullity of $S_\mu$. Its restriction is zero because every Borel subset of it is null.

Every nonempty open set meets $Q$, so its $\lambda$-measure is positive. The set of positive point masses of $\lambda$ is exactly $Q$, and $\lambda(Q)=t$. A nonsingular bijection preserves the property that a singleton is null and hence permutes $Q$. Thus $Q$ is an exactly invariant Borel set of intermediate measure for every proposed nonsingular action. This contradicts ergodicity.

Finally $Y\setminus Q$ is Borel with $\lambda(Y\setminus Q)=1-t$. Every compact $C\subset Y\setminus Q$ is countable by Proposition 3.14, has $\mu(C)=0$, and misses the atomic measure's carrier. Therefore $\lambda(C)=0$. This proves the last assertion in the actual original topology. $\square$

Primary-source comparison. Wage's CH construction uses $I=[0,1]$ and $X_0=I\times\{0,1\}$; write $a^i=(a,i)$. In $Z=(X_0,\sigma)\times(X_0,\tau)$, the double diagonal $D=\{(a^i,a^j):a\in I,\ i,j\in\{0,1\}\}$ is closed and $Y=\{(a^1,a^1):a\in I\}$. He identifies $Y$'s Borel structure with that of $I$, and gives each $y$ a neighborhood meeting $D$ in a countable subset of $Y$. These reported properties put the transported Lebesgue probability under Proposition 3.14 and exclude full support. This is a comparison of those properties, not an import of the transfinite construction or a counterexample to all the source dynamical hypotheses.

Open diagram at full size

Figure 4. This is a schematic of the hypotheses and exact conclusions, not a geometric model of Wage's spaces. Proposition 3.14 puts the nonempty closed support inside the null set $D\setminus Y$; the conull open complement is an arbitrary union of null open sets. Lemma 3.13 proves invariance under nonsingular homeomorphisms. Corollary 3.15 gives the exact mixed masses $t$ and $1-t$: adding dense positive point masses supplies full support and also an invariant Borel set of intermediate measure. Sources: Takesaki III, XIII.1.5, for the full-support and ergodicity obligations; Wage for the compared diagonal properties. The standing atomless source question remains open.

Atomless orbit mixtures: support can be repaired while returns decide ergodicity

The dense-atomic repair in Corollary 3.15 is not the only way to supply full support. One can instead sum translates of an atomless measure. The following results prove what this operation preserves and identify its exact ergodicity requirement. They are conditional on the specified seed and action; no omitted part of Wage's construction is assumed proved here.

Let a countable group $G$ act on a compact Hausdorff space $X$ by homeomorphisms. The group has the discrete topology, so this is a jointly continuous action. Let $\nu$ be a Borel probability, choose $c_g>0$ with $\sum_{g\in G}c_g=1$, and define $$\nu_g(B)=\nu(g^{-1}B),\qquad \mu(B)=\sum_{g\in G}c_g\nu_g(B),\qquad [B]_G=\bigcup_{g\in G}gB. \tag{3.20}$$ All saturations in this subsection are countable unions of Borel sets. We write $S(\eta)=\operatorname{supp}_X\eta$ for the original-topology support of a measure.

Lemma 3.16 (null ideal and support of an orbit mixture). Formula (3.20) defines a quasi-invariant Borel probability, and for every Borel $B$ $$\mu(B)=0\ \Longleftrightarrow\ \nu(g^{-1}B)=0\text{ for every }g\in G \ \Longleftrightarrow\ \nu([B]_G)=0. \tag{3.21}$$ Its support is $$S(\mu)=\overline{\bigcup_{g\in G}gS(\nu)}. \tag{3.22}$$ In particular a minimal action, meaning that every orbit is dense, makes $\mu$ fully supported.

Proof. Each $\nu_g$ is a Borel probability. Interchanging two sums of nonnegative numbers proves countable additivity of $\mu$; its total mass is $\sum_gc_g=1$. Positivity of all weights proves the first equivalence in (3.21). A countable union has zero $\nu$-measure exactly when each of its members does. Inversion permutes the group indices, proving the second equivalence.

For $h\in G$, $$\nu_g(hB)=\nu_{h^{-1}g}(B). \tag{3.23}$$ Thus the first criterion in (3.21) is unchanged by $B\mapsto hB$. Applying this also to $h^{-1}$ proves quasi-invariance. This does not assert that $\nu$ itself is quasi-invariant.

Put $F=\overline{\bigcup_g gS(\nu)}$. If $x\in F$, each neighborhood of $x$ contains a point of $gS(\nu)$ for some $g$. That open neighborhood has positive $\nu_g$-measure and hence positive $\mu$-measure. Thus $F\subset S(\mu)$. Conversely, if $x\notin F$, a compact neighborhood $C$ of $x$ can be chosen inside $X\setminus F$. For each fixed $g$, every point of $C$ lies outside $S(\nu_g)=gS(\nu)$, so a finite cover by $\nu_g$-null open sets gives $\nu_g(C)=0$. The interior of $C$ is therefore $\mu$-null and contains $x$. This gives $S(\mu)\subset F$. The local compact shrinking used here is Lemma 1.1's first paragraph and does not require a countable base. Lemma 3.13 gives $S(\nu_g)=gS(\nu)$ and the nonemptiness of $S(\nu)$. Under minimality the orbit of any point of that support is dense, so (3.22) is all of $X$. $\square$

A positive open set need not meet the support of a nonregular measure: Proposition 3.14 supplies exactly such a conull open set. Formula (3.22) follows by compact neighborhoods outside the closure, and does not use that false inference.

Lemma 3.17 (strong standardness and atomlessness survive). Suppose $Y\subset X$ is Borel, $\nu(Y)=1$, and its relative Borel sigma-algebra makes it a standard Borel space. Then $$Y_\infty=[Y]_G \tag{3.24}$$ is an invariant Borel conull standard Borel subspace for $\mu$. If every singleton is $\nu$-null, then $\mu$ is atomless as a measure, not just zero on singletons.

Proof. A homeomorphism restricts to a Borel isomorphism $Y\to gY$, so each $gY$ is a Borel standard Borel subspace. Enumerate the group and disjointize these sets: $$E_n=g_nY\setminus\bigcup_{j<n}g_jY. \tag{3.25}$$ Each $E_n$ is a Borel subset of the standard Borel space $g_nY$, and is standard Borel. A countable disjoint union of standard Borel spaces is standard Borel. Here that assertion applies to the actual relative sigma-algebra: a subset of $Y_\infty$ is relatively Borel exactly when its intersection with every $E_n$ is Borel. One direction follows by restriction; the other follows by taking the countable union of the ambient Borel representatives in these Borel pieces. Thus the disjoint-union identification is a Borel isomorphism. The standard Borel subset and countable-disjoint-union results are the usual descriptive-set-theoretic prerequisites; no topological regularity theorem is being inferred from them.

The set $Y_\infty$ is invariant. Every $\nu_g$ gives it mass one, so $\mu(Y_\infty)=1$. Also $\mu(\{x\})=\sum_gc_g\nu(\{g^{-1}x\})=0$. Standard Borel spaces admit countable point separation. If $\mu$ had a positive measure atom, Lemma 3.4 on this conull subspace would make it a positive singleton, a contradiction. This proves measure atomlessness. $\square$

Proposition 3.18 (a purely non-Radon seed remains purely non-Radon). Suppose no Borel $A$ of positive $\nu$-measure is a Radon part for $\nu$. Then no positive Borel set is a Radon part for $\mu$. One sufficient seed condition is a conull Borel $Y$ for which $$\nu(C)=0\quad\text{for every compact }C\subset Y. \tag{3.26}$$

Proof. First suppose (3.26). If $A$ were a positive Radon part for $\nu$, its positive Borel subset $A\cap Y$ would have a positive compact subset, contradicting (3.26). Thus the sufficient condition has the stated consequence.

Now suppose $A$ is a positive Radon part for $\mu$. Some $g$ has $\nu_g(A)>0$. For every Borel $B$, $$\nu_g(B)\le c_g^{-1}\mu(B). \tag{3.27}$$ Given Borel $E\subset A$ and $\varepsilon>0$, inner regularity of $\mu$ on $A$ supplies compact $C\subset E$ with $\mu(E\setminus C)<c_g\varepsilon$. Inequality (3.27) gives $\nu_g(E\setminus C)<\varepsilon$. Hence $A$ is a Radon part for $\nu_g$. Pulling compact approximants back by the homeomorphism $g$ shows that $g^{-1}A$ is a positive Radon part for $\nu$, a contradiction. No equivalence of the seed and mixture, or bounded density of the whole mixture relative to the seed, is required. $\square$

Theorem 3.19 (the exact return criterion). Assume the conull Borel carrier $Y$ of Lemma 3.17. Call Borel $A\subset Y$ exactly return-invariant when $$gA\cap Y\subset A\quad\text{for every }g\in G. \tag{3.28}$$ Equivalently $[A]_G\cap Y=A$; equivalently, if $y,gy\in Y$, then $y\in A$ exactly when $gy\in A$. The mixture is ergodic if and only if every such $A$ has $\nu(A)\in\{0,1\}$. For these sets the exact mass is $$\mu([A]_G)=\nu(A). \tag{3.29}$$ In particular, if $G$ is countable, $\nu$ is atomless, and $$\nu(Y\cap gY)=0\quad(g\ne e), \tag{3.30}$$ then the mixture is not ergodic, even when it has full support.

Proof. The identity element gives $A\subset[A]_G\cap Y$. Condition (3.28) gives the reverse inclusion. For the pointwise formulation, (3.28) gives the forward implication and its application to $g^{-1}$ gives the reverse one.

For any exactly invariant Borel $B\subset X$, each $g^{-1}B=B$, so $$\mu(B)=\sum_gc_g\nu(B)=\nu(B)=\nu(B\cap Y). \tag{3.31}$$ The saturation $[A]_G$ is exactly invariant, and its intersection with $Y$ is $A$. This proves (3.29). If $\mu$ is ergodic, (3.29) proves the return condition's necessity. Conversely take exactly invariant $B$. Its set $A=B\cap Y$ satisfies (3.28), and (3.31) makes $\mu(B)$ zero or one.

This suffices for measure-algebra ergodicity as well. If $B$ is invariant modulo $\mu$-null sets, put $B_*=\bigcap_{g\in G}gB$. This countable intersection is exactly invariant. Since $e\in G$, $B_*\subset B$, and $B\setminus B_*$ is contained in the countable union of the null sets $B\setminus gB$. Thus $B_*$ agrees with $B$ modulo $\mu$. A completed-measurable invariant event can first be replaced by a Borel representative, since quasi-invariance preserves the null differences. This proves the equivalence under either convention.

Finally, under (3.30), choose Borel $A\subset Y$ with $0<\nu(A)<1$. Such a set exists because the probability is atomless; this last assertion needs only the defining fact that $Y$, of mass one, is not a measure atom. For $g\ne e$, the set $gA\cap Y$ is null by (3.30). Countability now gives $\nu([A]_G\cap Y)=\nu(A)$. The saturation is exactly invariant, so (3.31), which does not require (3.28), yields $\mu([A]_G)=\nu(A)\in(0,1)$. This is an explicit invariant event of intermediate mass. Pairwise disjoint translates of $Y$ are a sufficient special case of (3.30). $\square$

The exact return condition is deliberate. A seed-null set can have a positive saturation when the seed itself is not quasi-invariant. Exactifying invariance using $\nu$-null sets alone would therefore be invalid. The countable exactification above uses the already quasi-invariant mixture $\mu$. Equation (3.21) specifies the null ideal which is safe to saturate.

Proposition 3.20 (a complete conditional torus repair, with the remaining failure exposed). Let $I$ have cardinality at most the continuum. Suppose $D\subset\mathbb T^I$ is compact, $Y\subset D$ is Borel and standard Borel, and $\nu$ is a Borel probability on $D$ with $\nu(Y)=1$, null singletons, and (3.26) in $D$. There is a separable compact Hausdorff space $X$, a free minimal continuous $\mathbb Z$-action, and an atomless strongly standard, fully supported, quasi-invariant Borel probability $\mu$, with a conull Borel $E$ containing no positive compact subset. The constructed action is not ergodic.

Proof. Take a new distinguished index $*\notin I$. Choose real numbers $\alpha_j$, for $j\in\{*\}\sqcup I$, so that $1$ and every finite subfamily of the $\alpha_j$'s are rationally independent. Here is the cardinality argument even when $I$ is uncountable. Well-order $\{*\}\sqcup I$ with order type its initial cardinal ordinal, at most $\mathfrak c$. Before any stage its chosen real numbers, together with $1$, have rational span of cardinality at most the maximum of $\aleph_0$ and the cardinality of that preceding stage, which is strictly below $\mathfrak c$. Choose outside that span. A finite dependence would contradict its last choice.

Set $$X=\mathbb T^{\{*\}\sqcup I},\qquad a=(\alpha_j+\mathbb Z)_j,\qquad \tau_n(x)=x+na,\qquad j(d)=(0+\mathbb Z,d). \tag{3.32}$$ The ultrafilter proof of compactness and Hausdorffness in Separable compact actions with nonregular measures, Section 1, applies to this index set. Its Lemma 1.1 proves finite-torus density, so every basic open cylinder meets $\{na:n\in\mathbb Z\}$. Thus every orbit is dense and $X$ is separable. If $na=0$, the distinguished coordinate gives $n\alpha_*\in\mathbb Z$, which forces $n=0$. The action is free and jointly continuous because each discrete-group slice is a translation homeomorphism. The map $j$ is a continuous embedding with compact closed image.