Let $E$ be a metric space, $\mathcal B$ the Borel $\sigma$-algebra on $E$ and $\mathcal M_1(E)$ the set of probability measures on $(E,\mathcal B)$. I'm interested in when and why the following implication holds.
If $\mathcal{M}_1(E)$ is compact, then $M \subset \mathcal{M}_1(E)$ is exponentially tight.
Is this always true? For what reason?
A script I'm studying says something along the lines of "we do not need to worry about exponential tightness (of $M \subset \mathcal{M}_1(E)$) because $\mathcal{M}_1(E)$ is compact". Here, $E$ is a bounded, open, connected subset of $\mathbb R^d$ with a smooth boundary and $M$ is the sequence of local times of a Brownian motion. And I see that $\mathcal{M}_1(E)$ is compact in this case, but not why exponential tightness follows.
It sounds to me like there's a simple argument that makes this statement true in a general setting, but I couldn't come up with a proof. So I wonder: why and when is it true?