Convergence of median and convergence in distribution Let $X_n$ be a sequence of real random variables such that $X_n \rightarrow^{d} X$, that is, $X_n$ converges in distribution to $X$. Suppose that $X$ has a cummulative distribution function, $F$, continuous at every point. Then, $\lim_n F_{X_n} (x) =F_{X} (x) $ for every real number $x$.
Let us suppose that $X$ has a unique median, $m(X) $ and let $\{m(X_n) \} $ be a sequence of medians of $\{X_n\} $ (maybe for some $n$ they are not unique).
Prove that $m(X_n) \rightarrow m(X) $.
I think that, if we denote by $[a_n, b_n] $ the set of medians of $X_n$ for every $n$, if we prove that $a_n\rightarrow m(X) $ and the same with $b_n$ it suffices. But I don't know how to use that $X_n$ converges in distribution to X.
 A: Set $\ x^- = \liminf \ m(X_n)$.
It turns out that $\ x^- > - \infty$.
Indeed there exists $c$ such that $F_X(c)=\frac{1}{3}$. By the convergence in distribution $\exists \ N \ \text{such that}\ \forall \ n>N \implies F_{X_n}(c) < \frac{1}{3}+0.1 \ <0.5 $.
So $\ m(X_n) > c$.
Similar reasoning achieves that $x^- < + \infty$.
Now, assume by contradiction that $\ x^- < \ m(X)$.
Set $y=\frac{x^- + m(X)}{2}$.$\ $ As X has a unique median, $\ F_X(y)<\frac{1}{2}$.
Again the convergence in distribution implies that
$\exists \ N \ \text{such that}\ \forall \ n>N \implies F_{X_n}(y) < \frac{1}{2}\ $.  Therefore, $m(X_n) \geq y > x^-,\ $ contradicting $x^-$ being liminf of $m(X_n)$.
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Finally assume $\ x^- > \ m(X)$.
Set $y=\frac{x^- + m(X)}{2}$.$\ $ As X has a unique median, $\ P(X \geq y) < \frac{1}{2}$.
Again the convergence in distribution implies that
$\exists \ N \ \text{such that}\ \forall \ n>N \implies P(X_n \geq y) < \frac{1}{2}\ $.  Therefore, $m(X_n) \leq y < x^-$.
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For the limsup the same logic applies, therefore $\lim m(X_n) = m(X)$ .
