Let $\seq{\mu_n}$ be a family of [[Probability|probability]] measures on $\real^d$. Suppose that 1. There exists a probability measure $\mu$ such that $\mu_n \to \mu$ [[Weak Convergence of Measures|weakly]], then $\seq{\mu_n}$ is [[Tight|tight]]. 2. If $\seq{\mu_n}$ is tight, then there exists a subsequence $\seq{\mu_{n_k}}$ such that $\mu_{n_k} \to \mu$ weakly. *Proof*. Suppose that $\mu_n \to \mu$ weakly, but $\seq{\mu_n}$ is not tight. Then there exists $\varepsilon > 0$ such that for every $K \subset \real^n$, $\sup_{n \in \nat}\mu_n(\real^n \setminus K) > \varepsilon$. In particular, for every $k \ge 1$, there exists $n_k \in \nat$ such that $\mu_{n_k}(\overline{B(0, k)}^c) > \varepsilon$. As $\mu_n \to \mu$ weakly, $\mu_{n_k} \to \mu$ weakly, and for any $R > 0$, $ \mu(B(0, R)) \le \liminf_{n \to \infty}\mu_{n_k}(B(0, R)) \le \liminf_{n \to \infty}\mu_{n_k}(B(0, k)) \le 1 - \varepsilon $ This implies that $\mu(\real^n) < 1$, which contradicts the assumption that $\mu$ is a probability measure. Task 3: other direction.