# Convergence of distribution functions implies weak convergence of associated probability measures

I'm trying to prove the following:

Let $P,P_n$ be probability measures on $\mathbb{R}^d$ with distribution functions $F$ and $F_n$ respectively. Show that $F_n(t)\to F(t)$ for all continuity points $t$ of $F$ implies that $P_n$ converges weakly to $P$.

I'm using the following notion of weak convergence of measures.

$P_n$ is said to converge weakly to $P$ on a metric space $S$, if $\displaystyle\lim_{n\to\infty}\int_{S}f\,dP_n=\int_{S}f\,dP$ for all bounded and continuous functions $f:S\to\mathbb{R}$.

Related question are Equivalence of definition for weak convergence

Convergence of Probability Measures and Respective Distribution Functions

However, I'm still not getting any further. Any help would be appreciated.

You could use the one-dimensional case and the Lévy's continuity theorem: We have $$\varphi_n(x) := \int_{\mathbb{R}^d} \exp(i \langle x , y \rangle) \, \mathrm{d} F_n(y).$$ Now $$G_x(t):= F_n(xt)$$ induces a one-dimensional measure, say $\mu_{x,n}$. By assumptation we have $$G_x(t) \rightarrow G_x(t) := F(xt)$$ in every continuity point $t \in \mathbb{R}$. Thus, the one-dimensional case implies $\mu_{x,n} \Rightarrow \mu_x$, where $\mu_x$ is the measure induced by $G_x$. By Lévy's continuity theorem $$\varphi_n(x) = \widehat{\mu}_{x,n}(1) \rightarrow \widehat{\mu}_x(1) = \varphi(x),$$ where $\varphi(x) := \int_{\mathbb{R}^d} \exp(i \langle x , y \rangle) \, \mathrm{d} F(y)$. Thus $P_n \Rightarrow P$.