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$X$ is a topological space, $m$ is a $\sigma-$finite measure on $B(X)$, and what condition can make $X$ be a $\sigma-$compact space?

This question is from topological groups (for me). Locally compact Hausdorff groups which are $σ-$compact are $σ-$finite under Haar measure. For example, all connected, locally compact groups $G$ are σ-compact.

So in fact, I just want to know that what condition can make a locally compact Hausdorff group be $\sigma-$compact.

Or what condition can make a locally compact Hausdorff group with $\sigma-$finite measure be a $\sigma-$compact space.

Or a locally compact Hausdorff group with $\sigma-$finite Haar measure must be a $\sigma-$compact space? Is this right?

I know that,"A locally compact, Hausdorff, second countable space is $σ-$compact."

Thanks a lot.

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  • $\begingroup$ If the measure is finite (in the sense that compact sets have finite measure) and it is inner regular, then the two appear to be equivalent $\endgroup$ – Prahlad Vaidyanathan May 18 '15 at 16:44
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It seems the following.

A topological group $G$ is called $\omega$-precompact or $\omega$-bounded, if for each neighborhood $U$ of the unit there exist a countable subset $C$ of the group $G$ such that $G=CU=UC$ (or, equivalently, if for each neighborhood $U$ of the unit there exist a countable subset $C$ of the group $G$ such that $G=CU$).

It is obvious, that a locally compact topological group $G$ is $\sigma$-compact iff $G$ is $\omega$-precompact.

Proposition. Let $G$ be a locally compact group with a left invariant $\sigma$-finite and $\sigma$-additive measure $\mu$ on the Borel algebra $B(G)$ on the group $G$ such that $\mu(U)>0$ for each non-empty open subset $U$ of a space $G$ and there exists a non-empty open subset $U_0$ such that $\mu(U_0)<\infty$. Then the group $G$ is $\sigma$-compact.

Proof . Assume the converse. Then the group $G$ is not $\omega$-precompact. This easily implies that there exists a non-empty open subset $V$ of the group $G$ and an uncountable subset $A$ of the group $G$ such that the family $\{aV: a\in A\}$ is disjoint.$^*$ Let $v_0\in V$, $u_0\in U_0$ be an arbitrary points. Put $U=V\cap v_0u_0^{-1}U_0$. Then $0<\mu(U)<\infty$ and the family $\{aU: a\in A\}$ is disjoint. Let $\{G_n\}$ be a family of $\mu$-measurable subsets of the group $G$ such that $\mu(G_n)<\infty$ for each $n$ and $G=\bigcup G_n$. Then for each $n$ and each $a\in A$ the set $G_n\cap aU$ is $\mu$-measurable and has finite measure $\mu$. The additivity of the measure $\mu$ implies that for each $n$ and each $\varepsilon>0$ the set $\{a\in A:\mu(G_n\cap aU)>\varepsilon\}$ is finite.$^{**}$ Therefore the set $\{(n,a): \mu(G_n\cap aU)>0\}$ is countable. So if the set $A$ is uncountable then there exists an element $a\in A$ such that $\mu(G_n\cap aU) =0$ for each $n$. Since the measure $\mu$ is $\sigma$-additive, we have $\mu(aU)=\mu(U)=0$, a contradiction. $\square$


$^*$ This is a standard fact in the theory of topological groups. Since the group $G$ is not $\omega$-precompact, there exists a neighborhood $U$ of the unit of $G$ such that $CU\ne G$ for each countable subset $C$ of the group $G$. Pick a neighborhood $V$ of the unit of $G$ such that $V=V^{-1}$ and $V^2\subset U$. From here we can construct the required set $A$ by any of two ways.

On the first way we pick as $A$ a maximal subset of $G$ such that the family $\{aV: a\in A\}$ is disjoint. Such a subset $A$ exists by Zorn Lemma. We claim that $AU=G$. Indeed, assume the converse, there exists a element $g\in G\setminus AU$. If $aV\cap gV\ne\varnothing$ for some $a\in A$, then $g\subset aVV^{-1}\subset aU\subset AU$, a contradiction. Thus $aV\cap gV=\varnothing$ for each $a\in A$, and a family $\{aV: a\in A\cup\{g\}\}$ is disjoint, which contradicts the maximality of $A$. Since $AU=G$, by the previuous paragraph, $A$ is uncountable.

On the first way, using the claim from the prepreviuous paragraph, by transfinite recursion we can construct $A$ as a sequence $A=\{a_\alpha:\alpha<\omega_1\}$ of elements of $G$ such that $a_\alpha\not\in a_\beta U$ for each $\alpha<\omega_1$ and each $\beta<\alpha$. This also implies that $a_\alpha V\cap a_\beta V=\varnothing$, which can be shown using the ideas from the previous paragraph.

$^{**}$ Indeed, assume that for some $n$ and $\varepsilon>0$ the set $\{a\in A:\mu(G_n\cap aU)>\varepsilon\}$ is infinite. Pick any sequence $\{a_m\}$ of distinct points of this set. Then $$\mu(G_n)\ge \mu(G_n\cap\{a_1,\dots, a_m\}U)=\sum_{i=1}^m \mu(G_n\cap a_iU)\ge m\varepsilon$$ for each $m$, which contradicts to $\mu(G_n)<\infty$.

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  • $\begingroup$ @Brock Assume that $G=\bigcup K_n$, where each $K_n$ is compact. Let $U$ be an arbitrary neighborhood of the unit. Then $\{xU:x\in G\}$ is an open cover of the set $G$, and, therefore, of each of the sets $K_n$. Therefore for each $n$ there exists a finite subset $C_n$ of $G$ such that $C_nU\supset K_n$. Put $C=\bigcup C_n$. Then $CU=\bigcup C_nU\supset\bigcup K_n=G$. $\endgroup$ – Alex Ravsky May 23 '17 at 5:07
  • $\begingroup$ Now assume that $G$ is $\omega$-precompact. Since the group $G$ is locally compact, there exists a compact neighborhood $U$ of the unit of $G$. Since the group $G$ is $\omega$-precompact, there exists a countable subset $C=\{c_n\}$ of the group $G$ such that $G=CU$. Thus $G$ is a union $\bigcup c_nU$ of countably many compact sets. $\endgroup$ – Alex Ravsky May 23 '17 at 5:07
  • $\begingroup$ @Brock I added the proof. $\endgroup$ – Alex Ravsky May 27 '17 at 13:27

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