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 $\mu(G_n)<\infty$.

The MO page here addresses your question (answer and comments).

By the way, your characterization of Haar measure is missing regularity conditions. Of course the main technical property of Haar measure that we care about is its invariance under left (or right) multiplication, but you need further properties to define a Haar measure. For example, according to your definition the counting measure on every topological group is a Haar measure! It is a common "error" to describe Haar measure in such a way that counting measure fits the description even though counting measure is not a Haar measure on $G$ when $G$ is not discrete.

What you left out are the following two conditions on $\mu$:

(i) for each Borel subset $A \subset G$, $\mu(A) = \inf \mu(U)$ where $U$ runs over open subsets of $G$ *containing* $A$,

(ii) for each open or $\sigma$-finite subset $A \subset G$, $\mu(A) = \sup \mu(K)$ where $K$ runs over compact subsets of $G$ *contained in* $A$.

These conditions taken together are called $\sigma$-*regularity*, and a Haar measure on $G$ is a $\sigma$-regular Borel measure on $G$ fitting the conditions you gave. Such conditions should be found in a proof that Haar measure is unique up to a scaling factor. Something like these conditions had better be used in a proof of uniqueness, because without them counting measure would be a Haar measure for $\mathbf R$ with its usual topology, which is false.

A measure that satisfies (i) and (ii) for all Borel subsets $A \subset G$ is called *regular*. Regularity is a stronger condition than $\sigma$-regularity and Haar measure on some locally compact groups is not regular. For example, let $G = S^1 \times \mathbf R_d$ where $\mathbf R_d$ is $\mathbf R$ with the discrete topology. Topologically this is like a cylinder of "independent circles".
Counting measure is a Haar measure on $\mathbf R_d$ and on $S^1$ there is a standard normalized Haar measure $d\theta/2\pi$. The product of these is a Haar measure on $G$ and it is not regular: the subset $A = \{1\} \times [0,1]$ of $G$ is closed (and not open), so $A$ is a Borel subset of $G$. Check $\mu(K) = 0$ for all compact $K \subset A$ and $\mu(U) = \infty$ for all open $U \supset A$. Thus condition (ii) for a Haar measure does not apply to $A$ (which is neither open nor $\sigma$-finite, so there is no contradiction).

## Best Answer

This is true.

Notice that if $K$ is a compact subset of a locally compact space, we can find an open $\sigma$-compact $U\supseteq K$. Indeed, by compactness, there is a relatively compact open $U_1\supseteq K$ (a finite union of relatively compact neighbourhoods of elements of $K$). Put $K_1=\overline U_1$, then iterate. $U=\bigcup U_n=\bigcup_n K_n$ is open and $\sigma$-compact.

Using this, you can show that if $X$ is a locally compact space with a $\sigma$-finite Radon measure (locally finite, outer regular, inner regular with respect to compact sets), by slightly refining the argument you provided, there is an open, $\sigma$-compact $U\subseteq X$ of full measure.

Now, if $X$ is a group, then $G=\langle U\rangle$ is also $\sigma$-compact (if $U=\bigcup_n K_n$, then $G$ is the union of finite products of $K_n$-s and their inverses) and open, and hence closed. Since it still has full measure, it follows that $G=X$, so $X$ is $\sigma$-compact.

Edit: As a mildly interesting observation, this last step also works if $U$ has cofinite measure, i.e. $X\setminus U$ has finite measure. In this case, $G=\langle U\rangle$ either equals $X$ (so we are done), or $X\setminus G$ is a nonempty union of cosets of $G$ of finite measure. It follows that $G$ has finite measure, and thus so does $X$, which must be compact in this case (a locally compact group of finite Haar measure is necessarily compact).You don't even need $U$ to be open for this argument, although you do need to use the fact in the parenthesis at the end of the previous paragraph.