Convergence of integral of quantiles I am trying to prove the following.
The following holds for $X, X_k \in L^{2}(\Omega)$.
Let $\alpha \in (0, 1)$ be fixed, and let the $\alpha$-quantile $\inf \{c \in \mathbb{R} \ | \ P(X\leq c)\geq \alpha \}$ be denoted $q_{\alpha}(X)$. Let $X_k$ be a sequence of random variables such that
\begin{align}
\int_{\alpha}^{1} q_{\beta}(X_k) \ d\beta \leq 0
\end{align}
for each $k$. If we have for a random variable $X$ that $\lim_{k \rightarrow \infty}\|X_k -X\|_{2} = 0$, then
\begin{align}
\int_{\alpha}^{1}q_{\beta}(X)\ d\beta \leq 0,
\end{align}
I can't seem to find the right way to go about this. I was hoping use a measure-theoretic argument, to show that as $k \rightarrow \infty$, $X_k$ and $X$ can disagree at most on a set with measure $0$, but I can't seem to set it up correctly. I had the following in mind.
Assume that $\lim_{k \rightarrow \infty}\|X_k -X\|_{2}$. Then, for any $\epsilon > 0$ we can find an $N \in \mathbb{N}$ such that
\begin{align}
P(\{ \omega \in \Omega \ | \ X_k(\omega) \neq X(\omega)\}) < \epsilon.
\end{align}
Unfortunately, I can't seem to make use of this in the above integrals.
Could I use a transformation like
\begin{align}
\int_{\alpha}^{1}q_{\beta}(X)\ d\beta  = \frac{1}{1-\alpha}\int X \chi_{\{X\geq q_{\alpha}\}} \ dP
\end{align}
to apply an measure theoretic argument? And even so, even if I show that $X_k, X$ disagree on a set of measure $0$, could not the values where they disagree tend to infinity?
Any ideas would be very welcome!
 A: I don't know of this solves your problem, but if the distributions were continuous, i.e. $_X(x-)=F_X(x)$, then the argument may not be complicated since $(\cdot;_)=_$ in law and then through a subsequence $(\cdot;_)$
converges weakly to $(\cdot;)$,  which is the same as  in law. The function $=\mathbb{1}_{[\alpha,1]}(\beta)=\mathbb{1}(q(\beta;)\geq q(\alpha;))$
has discontinuity only at $\alpha$ which under the law of $q(\cdot;)$
is of measure zero.
You also need to use handle by uniform integrability. The fact that $X_n$ converges to $$ in $L_2$ (and so in $L_1$)
imlplies that $\{X_n,X\}$
is a uniform integrable family, and that $_n$
also converges weakly to $X$. Uniform integrability implies that the measures $\nu_n():=\int_{X_n^{-1}(A)} X\,dP$
converge weakly to $\nu():=\int_{X^{−1}()}\,$.
From this and the continuity mapping theorem you may be able to get what you need.

Here is a short proof that $\nu_n$ converges weakly to $\nu$:
Let  $f\in\mathcal{C}_b(\mathbb{R})$, the continuity of
$x\mapsto xf(x)$ implies that
$ X_nf( X_n)\Rightarrow X f( X)$.
Since $|f( X_n) X_n|\leq \|f\|_\infty| X_n|$ and $X_n$ converges to $X$ in $L_1$,  $\{f( X_n) X_n\}$ is uniformly integrable. Hence
$\int f( X_n) X_n\,d\mu\rightarrow\int f( X) X\,d\mu$.
