I'm reading some results from Measure Theory Volume 4 by D.H. Fremlin, and I'm stuck on something.
This is pulled out of one of his lemmas (stated more generally for topological groups):
A Haar measure $\mu$ on a locally compact group $G$ is strictly positive.
The proof is quite short:
It is sufficient to show that $\mu$ is positive on non-empty open sets containing 1.
Suppose for a contradiction that $U$ is an open set containing $1$ and $\mu(U) = 0$.
Then $G = \bigcup\limits_{x\in G}xU$, and since $G$ is $\tau$-additive, $\mu(G) = 0$, a contradiction.
I assume that $\tau$-additive means that $\mu(\bigcup_{\alpha\in J}E_{\alpha}) \leq \sum\limits_{\alpha\in J}\mu(E_{\alpha})$ is this correct?
If so, then why do Haar measures have this property. If not what does it really mean?
I don't know what $\tau$-additive means, but if $G$ is a locally compact Hausdorff abelian group and $\mu$ is a left Haar measure, then $\mu(U)\neq 0$ for $U$ a non-empty open subset of $G$, and the proof is almost what you've written (except, again, I don't know what is meant by $\tau$-additive). Remember that Haar measures are non-zero by definition. They are also inner regular on open sets. So suppose $\mu(U)=0$. Then, by left invariance, $\mu(gU)=0$ for all $g\in G$. Now, if $K$ is a compact subset of $G$, then $K$ is covered by finitely many translates of $gU$ of $U$ (here we use that $U$ is non-empty). So $\mu(K)=0$ by sub-additivity. Finally, by inner regularity on opens, $\mu(G)$ is equal to the supremum of $\mu(K)$ as $K$ ranges over the compact subsets of $G$. But this supremum is $0$ by what we've just shown, contrary to the assumption that $\mu\neq 0$.
EDIT: Assuming the set $J$ in your inequality is supposed to be countable, then this is countable sub-additivity (or I guess just sub-additivity) of $\mu$, and every measure has this property. I guess now that I think about it, the way ``$\tau$-additivity" is being invoked seems to be some sort of extension of countable sub-additivity to collections of sets which might not be countable (since the union in the proof cited might not be a countable one). I don't know why this property holds for Haar measures. So I guess my answer doesn't really address the question, although the proof I've given does show that one need not resort to $\tau$-additivity, whatever it is, so I will leave it for now, but will delete it if the OP requests it.