If Haar measure is $\sigma$-finite, is the underlying topological space $\sigma$-compact?

Let $X$ be a locally compact Hausdorff group and $\lambda$ a left Haar measure on $X$. Assume that $\lambda$ is $\sigma$-finite. Is it true that $X$ is $\sigma$-compact?

Attempt: Write $X = \bigcup_n X_n$ here $X_n$ is a Borel set of $X$ for all $n$ with $\lambda(X_n) < \infty$. My idea was to approximate each $X_n$ by a compact subset $K_n$, and then $X=\bigcup_n X_n$ would be well approximated by $\bigcup_n K_n$, but we still would have to make up for the complement and I don't see how to do that. I will most likely need to use properties of the Haar measure.

Also, note that the converse is true. Any help is welcome!

Solution 1:

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.