$\lim_nP(X_n \in [x_n,u_n])=P(X \leq u)?$ 
Let $(x_n)_n$ and $(u_n)_n$ be 2 sequences of real numbers such that $\forall n \in \mathbb{N},x_n \leq u_n,$ $ \lim_nx_n=-\infty$ and $\lim_n u_n= u\in \mathbb{R}.$
Let $(X_n)_n$ be a sequence of real random variable converging in distribution to the random variable $X$. We suppose that $P(X=u)=0.$
Is it true that $\lim_n P(X_n \in [x_n,u_n])=P(X \leq u)?$

We have $P(X_n \in [x_n,u_n])=P(X_n -u_n\leq 0)-P(X_n <x_n).$ $(X_n-u_n)_n$ converges in distribution to $X-u$, the portmanteau theorem $(P_{X-u}(\partial(]-\infty,0]))=0)$ gives us $\lim_n P(X_n \leq u_n)=P(X \leq u).$
Now we have to prove that $\lim_n P(X_n<x_n)=0,$ but in this case the above trick won't work.
What should I do to prove the second case?
 A: Let $F_X$ be the cumulative density function of $X$. $F_X$ is monotonically increasing (in the sense of "non-decreasing"), and $\lim_{x\to-\infty} F_X(x)=0$. 
Let $\varepsilon > 0$. Then there exists a $B\in\Bbb R$ such that $F_X(x)<\frac\varepsilon2$ for all $x\le B$. By definition of convergence in distribution, we have $$\lim_{n\to\infty} F_{X_n}(x)=F_X(x) \text{ for all continuity points } x \text{ of }F_X.$$
Since $F_X$ is monotonic, it can have at most countably many points of discontinuity. Let $x\le B$ be a continuity point of $F_X$.
We can choose an $N\in\Bbb N$ such that $ |F_{X_n}(x)-F_X(x)|<\frac\varepsilon2$ for all $n\geq N$. 
Now we see that $|F_{X_n}(x)|\le |F_{X_n}(x)-F_X(x)|+|F_X(x)|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$ for all $n\geq N$. We can fix a $\tilde N \geq N$ such that $x_n \le x$ for all $n\geq \tilde N$ (since the $x_n$ tend to $-\infty$). 
For $n\geq\tilde N$ we then have, since the $F_{X_n}$ are increasing, $|F_{X_n}(x_n)|\le|F_{X_n}(x)|<\varepsilon$. Thus, $$\lim_{n\to\infty}\Bbb P(X_n < x_n)\overset{\text{Def.}}=\lim_{n\to\infty} F_{X_n}(x_n)=0.$$
