To prove the convergence of inverse of CDFs. Prove that if $F_n \Rightarrow F$ and $x$ is such that there is at most $a \in \mathbf R$ with $F(a)=x$,
then $F_n^{-1}(x) \rightarrow F^{-1}(x)$.
Hint:For any $\epsilon \gt 0$, you can choose a $y$ such that $F$ is continuous at $y$ and $F^{-1}(x)-\epsilon \lt y \lt F^{-1}(x) $.
$F_n \Rightarrow F$ means weakly convergence or convergence in distribution.
Any help please, thank you very much!
 A: For the inverse CDF $$F^{-1}(u):=\inf\{ t: F(t)> u\}$$ it is easy to see that
$$\text{if $F(a)>x$ then $F^{-1}(x)\le a$}\tag1$$
$$\text{if $F(a)\le x$ then $F^{-1}(x)\ge a$}.\tag2$$
Let $x\in(0,1)$ be such that there is at most one $a$ with $F(a)=x$. Establish the preliminary result:

Lemma: $F(y)<x$ for every $y<F^{-1}(x)$.
Proof: If there exists $y<F^{-1}(x)$ such that $F(y)\ge x$, then for every $a\in(y, F^{-1}(x))$ we have $a>y$ so $F(a)\ge F(y)\ge x$; but also $F^{-1}(x)> a$ so by (1) we have $F(a)\le x$. It follows that $F(a)=x$ for infinitely many $a$, violating the hypothesis.

Suppose $F_n\Rightarrow F$. To prove the convergence of $F_n^{-1}(x)$ to $F^{-1}(x)$, let $\epsilon>0$. Since $F$ has at most countably many discontinuities, we can find a continuity point $y$ of $F$ such that $F^{-1}(x)-\epsilon<y<F^{-1}(x)$.
By convergence in distribution, we have $F_n(y)\to F(y)$ as $n\to\infty$. From the Lemma we know that $F(y)<x$, so there exists $N$ such that $F_n(y)<x$ for all $n>N$. By (2), this implies $F_n^{-1}(x)\ge y$ for such $n$.
Next, let $y'$ be a continuity point of $F$ such that $F^{-1}(x)<y'<F^{-1}(x)+\epsilon$. Then $F_n(y')\to F(y')$, and by (2) we have $F(y')> x$ so there exists $N'$ such that $F_n(y')>x$ for all $n>N'$. By (1), this implies $F_n^{-1}(x)\le y'$ for such $n$.
For $n$ exceeding both $N$ and $N'$ we've shown
$$F^{-1}(x)-\epsilon<y\le F_n^{-1}(x)\qquad\text{and}\qquad F_n^{-1}(x)\le y'<F^{-1}(x)+\epsilon,$$
hence $|F_n^{-1}(x)-F^{-1}(x)|<\epsilon$.
