Compactness of the set of finite Borel measures

Question

Suppose $X$ is a compact subset of $\mathbb{R}^n$ for some $n \in \mathbb N$. Let $\mathcal M(X)$ denote the space of all finite Borel measures on $X$. Is $\mathcal M(X)$ compact under some commonly used topology?

I have very little knowledge of functional analysis to answer this on my own. Hence posting here.

Answer 1

The answer is obviously not. Consider a sequence of measures $\mu_n = n \mathbb{1}_{x=0}$, $n \in \mathbb N$. This has no convergent subsequence.