# counterexample of weak convergence for quantile function

As we know, if a sequence of distribution functions $\{F_n\}$converge weakly to $F$, i.e. $\lim_{n \to \infty}F_n(x)=F(x)$ for all continuous points $x\in\mathbb{R}$ of $F$, then for the corresponding quantile functions, we have $\lim_{n\to\infty}Q_n(u)=Q(u)$ for all continuous points $u\in(0,1)$ of Q.

So my question is if $u$ is not a continuous point of Q, can we find an example that $Q_n(x)$ does not converge?

The discontinuous points (jumps) in $Q$ corresponds to flat points in $F$, but I don't know how to construct such a $\{F_n\}$.

Let $F_n(x)=0$ for $x\le 0$, $F_n(x)=1$ for $x\ge 3$, $F_n(x)=\frac12+\frac{(-1)^n}{n}$ for $1\le x\le 2$. Extend $F_n$ to the intervals $(0,1)$ and $(1,2)$, for example affinely: as $F(x)=F(1)x$ for $0<x<1$ and $F_n(x)=(1-F_n(2))(x-2)+1$ for $2<x<3$.
As $n\to \infty$, the sequence $F_n$ converges uniformly. The median of the random variable described by $F_n$ is less than $1$ when $n$ is even, and greater than $2$ when $n$ is odd. Hence, the median does not have a limit as $n \to\infty$.