0

Let $X$ be some topological space, and $\mu$ finite Borel regular measure on $X$. If for some $x \in X$, it holds that for every open neighborhood $U$ of $x$ we have $\mu(U)=1$ , does it follow that $\mu$ is $\delta_x$?

I have seen that statement being used for compact Hausdorff space, and intuitively it seems to make sense (for Hausdorff spaces at least), but I can't find the rigorous way to prove that.

Delpins
  • 61
  • 2
    We don't need compactness or Hausdorff. If $S\ni x$ is an arbitrary Borel set, then outer regularity immediately implies $\mu(S)=1$. Taking complements we have $\mu(S)=0$ whenever $x\notin S$. – M W Sep 19 '23 at 18:19
  • @MW Thank you. It makes sense now. – Delpins Sep 19 '23 at 18:31

0 Answers0