Recently, I have come across this proof about weak convergence of measures. I'm trying to see why the usual hypothesis is "$X$ is a locally compact separable metric space". Then I come across below result, i.e.,
Theorem: Let $(E, d)$ be a locally compact separable metric space. Then $E$ is a Radon space.
Could you have a check on my attempt?
Proof: Since $E$ is locally compact and separable metric space, there exists a countable basis $(U_n)$ such that $\overline{U_n}$ is compact for every $n$. Let $$ K_m := \bigcup_{n=1}^m \overline{U_n} \quad \forall m \in \mathbb N. $$
Then $K_m$ is compact and $\bigcup_m K_m = E$.
Because $\mu$ is inner regular, it suffices to prove that $\mu$ is tight on every closed subset $B$ of $E$. The claim then follows immediately because $$ B = \bigcup_m (E \cap K_m), $$ and because $E \cap K_m$ is a compact subset of $E$. This completes the proof.
