Let $\{F_{n}\}_{n}$ be a sequence of cumulative distribution functions such that converge to $F$, in the sense that $F_{n}(x)\rightarrow F(x)$ for all $x\in\mathbb{R}$. We define the function infimum of the median by $MEi(F) = \inf\{x\in\mathbb{R} : F(x) = 1/2\}$, for all cumulative distribution function $F$.
My question is, whether or not, we have that $\lim\inf_{n} MEi(F_{n}) = \lim\sup_{n} MEi(F_{n})$.
I think so, but i'm not able to prove it.
Thanks in advance.
This is definitely not true.
Let $$ G_n(x) =\begin{cases} 0 & x<0 \\ 2^{-1}-2^{-n} & x\in [0,2^n) \\ \frac{1}{2} & x\in [2^n,4^n) \\ 1 & x\geq 4^n \end{cases} $$ and $$ H_n(x) =\begin{cases} 0 & x<-4^{-n} \\ \frac{1}{2} & x\in [-4^{-n},4^n) \\ 1 & x\geq 4^n \end{cases} $$ Then, clearly, both $H_n$ and $G_n$ converge pointwise to $$ F(x)=\begin{cases} 0 & x<0 \\ \frac{1}{2} & x\geq 0 \end{cases} $$ but if we pick $F_n$ to be $G_n$ for odd $n$ and $H_n$ for even $n$, we get that \begin{align} \liminf_{n\to\infty}MEi F_n=\lim_{n\to\infty} ME_i H_n=0\neq \infty=\limsup_{n\to\infty} ME_i G_n=\limsup_{n\to\infty} ME_i F_n \end{align}
