Let $X$ be a Lusin space. A subset $H \subset \mathcal{P}(X)$ is relatively compact in the topology of weak convergence if $H$ is *tight*: for every $\varepsilon > 0$ there exists a compact set $K \subset X$ such that $\mu(K^c) < \varepsilon$ for all $\mu \in H$. Moreover, when $X$ is Polish, tightness is also necessary for relative compactness.