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.
 A: You could use the one-dimensional case and Lévy's continuity theorem: We have $$\varphi_n(x) := \int_{\mathbb{R}^d} \exp(i \langle x , y \rangle) \, \mathrm{d} F_n(y).$$
The one-dimensional distribution function
$$G_x(t):= F_n(xt)$$
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 $t \in \mathbb{R}$. Here, we need to check that any continuity point of $G_x$ is also one of $F$. This will be done at the end. (Since $F$ is multidimensional and we only have the continuity on a ray, it could be in general wrong.)
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. (For example one can use the portmanteau theorem, write an open set as countable union of open intervals, approximate the endpoints of the interval by continuity points of the distribution function and conclude the one-dimensional case.)
To finish the proof, we need to show that for a continuity point $t_0 \in \mathbb{R}$ of $G_x$ the point $t_0 x$ is a continuity point of $F$. $F$ is monotone in the sense that for $x = (x_1,\ldots,x_d) \leq y = (y_1,\ldots,y_d)$ (i.e. $x_i \leq y_i$ for all indices $i=1,\ldots,d$) one has $F(x) \leq F(y)$. Now $F$ is (by measure-continuity) right-continuity in the sense that $F(y_k) \rightarrow F(y)$ for monotone sequences $(y_k)_{k}$ with $y_k \downarrow y$. If $F(y_k) \rightarrow F(y)$ is also true for monotone sequences with $y_k \uparrow y$, then $y$ is already a continuity point.
In fact, for any sequence $(y_k)_{k}$ with $y_k \rightarrow y$ one has $y -|y_k-y| \leq y_k \leq y+ |y_k-y|$, where the absolute-value is to be taken component-wise. By taking monotone subsequences of $|y_k-y|$ we find monotone subsequences $(u_{k_j},v_{k_j})_j$ of $u_k := y-|y_k-y|$ and $v_k :=y+|y_k-y|$. Now we know that
$$F(y) \leftarrow F(u_k) \leq F(y_k) \leq F(v_k) \rightarrow  F(y)$$
showing the continuity.
If $F$ is not continuous at $t_0x$, then there exists a monotone sequence $(y_k)_k$ with $y_k \uparrow t_0 x$, but $F(y_k) + \varepsilon \leq F(t x_0)$ for some $\varepsilon >0$. Now take $t_k := \max_{i=1,\ldots,d} |y_{k,i} - t_0 x_i|$ and note that $x (t_0 - t_k) \leq y_k$ and therefore $G_x(t_0-t_k) \leq F(y_k) \leq G_x(t_0) - \varepsilon$. Since $t_k \rightarrow 0$, we get a contradiction!
