Let $X_n$ ~ $U(0, 1/n)$ and $X = 0$ (almost surely equal to $0$, actually).
Does $X_n$ converge in distribution to $X$?
I know that $$F_X{_n} = \begin{cases} 0, & x < 0,\\ nx, & 0 \leq x \leq 1/n\\ 1, & x>1/n. \end{cases} $$ I need to prove that:
$\forall \epsilon>0, \forall x \neq 0, \exists N \in \mathbb{N}, \forall n \geq N,|F_X{_n} - F_x|<\epsilon$ ($1$)
My question relies in the 2nd case, when $0 \leq x \leq (1/n)$. I know that when n tends to infinity, this interval will degenerate at $0$. I also know that I don't need to prove ($1$) if x=$0$ because the CDF of $X$ is discontinuous at that point. But how can I put forward that argument formally?
$\forall \epsilon>0, \forall x >0$, set $N=\lceil \frac1x \rceil \in \mathbb{N}$. Then $\forall n \geq N$, $F_{X_n}(x)=1$ and $F_X(x)=1$.
Therefore, $$\forall \epsilon>0, \forall x >0, \exists N=\lceil \frac1x \rceil \in \mathbb{N}, \forall n \geq N, |F_{X_n}(x)-F_X(x)|=0<\epsilon.$$