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we know that if $(X_n)_n$ is a sequence of real random variables, then it converges in distribution to a random variable $X$ if and only if $\lim_nF_{X_n}(x)=F(x)$ at every point $x \in \mathbb{R}$ where $F_X$ is continuous.

I think the result is true if the variable $X_n$ takes values in $\mathbb{R^d}$, I mean $(X_n)_n=((X_n^1,...,X_n^d))_n$ converges in distribution to a random variable $X=(X_1,...,X_d)$ if and only if $\lim_nF_{X_n}(x_1,...,x_d)=P(X_n^1 \leq x_1,...,X_n^d \leq x_d)=F(x)$ at every point $(x_1,...,x_d) \in \mathbb{R^d}$ where $F_X$ is continuous.

But how can we prove it?

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  • $\begingroup$ Which is the definition you have for convergence in distribution of a $d$-dimensional random vector? $\endgroup$ – Alejandro Nasif Salum Apr 17 at 19:28
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    $\begingroup$ No, By DEFINITION $(X_n)_n$ converges in distribution (weakly) to $X$ if for all continuous and bounded functions $f$, we have $\lim_n\int_{\mathbb{R^d}}f(x)dP_{X_n}(x)=\int_{\mathbb{R^d}}f(x)dP_X(x)$ $\endgroup$ – mathex Apr 17 at 19:34

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