Tightness of measures

Remarks

On $\R^d$, a family $\mathcal{M}$ is tight if and only if $\sup_{\mu \in \mathcal{M}} \mu(\set{x : \abs{x} > R}) \to 0$ as $R \to \infty$, because the compact sets are the closed bounded sets. Example: the normal laws $N(m, 1)$ with $\abs{m} \le 1$ form a tight family. Given $\varepsilon$, choose $r$ with $P(\abs{Z} > r) < \varepsilon$ for $Z$ standard normal; then $K = [-1 - r, 1 + r]$ works, since $N(m, 1)(\R \setminus K) \le P(\abs{Z} > r)$. Non-example: the point masses $\delta_n$, $n \in \N$, on $\R$. Each compact $K$ lies in some $[-R, R]$, and then $\delta_n(\R \setminus K) = 1$ for $n > R$. Also non-tight: the normal laws $N(0, n)$, $n \ge 1$, whose mass spreads out. A single Borel probability measure on a Polish space is tight, and so is every finite family. On a Polish space, Prokhorov's theorem says that $\mathcal{M}$ is tight if and only if it is relatively compact for weak convergence. Every sequence of real random variables that converges in distribution is tight.