While solving questions in probability, I found the following question.
Let $(u_n)_{n \in \mathbb{N}}$ and $(v_n)_{n \in \mathbb{N}}$ two sequence of real numbers such as $\forall n \in \mathbb{N},u_n\leq v_n,$ and $\lim_nu_n=\alpha,\lim_n{v_n}=\beta.$ Let $(X_n)_n$ a sequence of real random variables which converges in distribution to random variable X whose function of $F_X$ distribution is continue. Show that $\lim_n\mathbb{P}(X_n \in [u_n,v_n])= \mathbb{P}(X \in [\alpha,\beta])$.
In order to solve the question, I tried to use distribution functions but, I failed, because here we have a sequence of interval, so what should I do to solve this problem?
Fix a positive $\varepsilon$. There exists a $n_0$ such that for all $n\geqslant n_0$, $\left\lvert u_n-\alpha\right\rvert+\left\lvert v_n-\beta\right\rvert\lt \varepsilon$. Therefore, for $n\geqslant n_0$, $$\mathbb{P}(X_n \in [u_n,v_n])\leqslant \mathbb{P}(X_n \in [\alpha-\varepsilon ,\beta+\varepsilon])$$ from which we infer, by continuity of $F_X$, that $$ \limsup_{n\to +\infty}\mathbb{P}(X_n \in [u_n,v_n])\leqslant \mathbb{P}(X \in [\alpha-\varepsilon ,\beta+\varepsilon]). $$ Since this is true for all $\varepsilon$, using again continuity of $F_X$, we get $$ \limsup_{n\to +\infty}\mathbb{P}(X_n \in [u_n,v_n])\leqslant \mathbb{P}(X \in [\alpha ,\beta]). $$ An inequality with $\liminf$ can be inferred as well using the inclusion $[u_n,v_n]\supset [\alpha+\varepsilon,\beta-\varepsilon]$.