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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.

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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 $P_n \Rightarrow P$.

The one-dimensional case should be treated in any book on probability theory. (Usually one uses the portmanteau theorem and write an open set as countable union of open intervals ...)

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