Is a Lie group $\sigma$-compact? Is a Lie group $\sigma$-compact?
I want to use the Riesz representation theorem for show there is $\mu$ a bi-invariant Haar measure on a Lie group $L$  such that
$$I(f)=\int_L f d \mu$$
where $I$ is a linear functional on a Lie group $L$.
 A: I suppose that you are interested only in connected Lie groups. Then, yes, it is $\sigma$-compact. In fact, if $U$ is a compact neighborhood of the identity element, then $\{U^n\,|\,n\in\mathbb{N}\}$ is a countable collection of compact sets whose union is the whole group.
A: Yes. Of course, it all depends on your definitions - but it seems that most people require differential manifolds (and hence Lie groups) to be second countable. Since they are also locally compact, this means they are $\sigma$-compact.
However, one should be careful. The book Structure and Geometry of Lie Groups by J. Hilgert and K. Neeb doesn't require manifolds or Lie groups to be second countable, so for example $\mathbb{R}$ with the discrete topology is considered a Lie group (and it's not $\sigma$-compact). But this is pretty nonconventional.
EDIT: It is also important to note that, as José pointed out, every connected Lie group is automatically $\sigma$-compact (and in fact second countable). I'll copy the proof from here:
Suppose $G$ is a connected Lie Group. Let $K$ be any compact neighbourhood of the identity. Then the interior $\text{int}(K)$ is an open neighbourhood of the identity as well. Take $V = \text{int}(K) \cap \text{int}(K)^{-1}$; $V$ is still an open neighbourhood of the identity and $V \subset K$.
Consider $H = \bigcup_{n \in \mathbb{N}}V^n$. This is an open subgroup of $G$. Therefore it is closed, since its complement $H^C = \bigcup_{g \neq e \in G} gH$ is open. Because $G$ is connected and $H$ is nonempty, $H = G$. In particular,
$$ G = \bigcup_{n \in \mathbb{N}}V^n = \bigcup_{n \in \mathbb{N}}K^n.$$
