When a sigma-finite space is a sigma-compact space? $X$ 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.
 A: 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 $\mu(G_n)<\infty$.
