Can weak convergence of distribution function imply the convergence of left limit?

Question

If $F_n$ and $F$ are distribution functions, and $F_n$ weakly converges to $F$. Then we know that $F_n(x) \rightarrow F(x)$ when $x$ is the continuous point of $F$.

I want to ask: can we deduce that $F_n(x-) \rightarrow F(x-)$ for every $x$? And $F_n(x+) \rightarrow F(x+)$ ? It's easy when the $x$ is the continuous point of $F$, but if x is a discontinuous point?

Answer 1

No.

Let $F_n$ be the cdf of a uniformly distributed random variable on $[-1/n, 1/n]$. Then $F_n(0) = 1/2$ for all $n$. But the $F_n$ converge weakly to the cdf of the point mass at $0$, that is, to $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\ge0$. So $\lim_{n\to\infty} F_n(0)$ matches neither the right nor left hand limit of $F(x)$ at $x=0$.