> [!theorem] > > Let $\seq{\mu_n}$ be a family of $\real^d$-valued [[Probability|probability]] measures, and there exists $f: \real^d \to \complex$ such that the [[Characteristic Function|characteristic functions]] $\hat \mu_n \to f$ pointwise. If $f$ is [[Continuity|continuous]] at $0$, then there exists a probability measure $\mu$ such that $\mu_n \to \mu$ [[Weak Convergence of Measures|weakly]] with $f = \hat \mu$. > > *Proof*. It's sufficient to show that the family $\seq{\mu_n}$ is tight. Firstly, $f(0) = 1$. Let $\varepsilon > 0$, then there exists $\delta > 0$ such that $\abs{f(\xi)} < \varepsilon$ for all $\xi$ with $\abs{\xi} < \delta$. Let $R = 2/\delta$, $r = \delta$, and $e_j$ be the $j$-th standard coordinate vector. By the above lemma, > $ > \mu_n(\bracs{y \in \real^d: \abs{\angles{e_j, y}} > 2/\delta}) \le \braks{\frac{1}{\delta}\int_0^\delta \abs{1 - \hat \mu_n(se_j)}ds} \cdot \frac{1}{m(2)} > $ > where $m(2) \ge 1/2$, so > $ > \mu_n(\bracs{y \in \real^d: \abs{\angles{e_j, y}} > 2/\delta}) \le \frac{2}{\delta}\int_0^\delta \abs{1 - \hat \mu_n(se_j)}ds > $ > which converges to > $ > \frac{2}{\delta}\int_0^\delta \abs{1 - f(se_j)}ds \le 2\varepsilon > $ > Now, there exists $N \in \nat$ such that > $ > \mu_n(\bracs{y \in \real^d: \abs{\angles{e_j, y}} > 2/\delta}) \le 3\varepsilon > $ > for all $n \ge N$. This bounds the mass outside the box of radius $2/\delta$ to be less than $3d\varepsilon$.