Suppose $X$ is normal space and $\mu$ is a Radon measure, then the union of open null sets is again a null set.

Question

I tried to prove this statement by contradiction and basic topology tricks, but failed.

Answer 1

Let $\{U_\alpha\}$ be your collection of open null sets, and let $U = \bigcup_{\alpha} U_\alpha$.

Suppose $K$ is any closed set contained in $U$. Then $K$ is compact and $\{U_\alpha\}$ is an open cover of $K$. By choosing a finite subcover, conclude that $\mu(K) = 0$.

So every closed set contained in $U$ is null. Now use the fact that $\mu$ is inner regular.