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

haar-measuremeasure-theorytopological-groups

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!

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.