# weak convergence and CDF

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?

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