Let $X_{n}$ be a sequence of real random variables and let $X$ a real random variable defined over the same p.s. $(\Omega,\mathcal{A},\mathbb{P})$ and such that $X_{n}(\omega)\searrow X(\omega)$ for all $\omega\in\Omega$. Let $x_{n}$ be a sequence of reals such that $x_{n}\searrow_{n} x$ and such that $x_{n} \in Med(X_{n})$ for all $n\in\mathbb{N}$.
I wonder if $\lim\sup_{n} \{\omega\in\Omega : X_{n}(\omega)\leq x_{n}\} = \{\omega\in\Omega : X(\omega)\leq x\}$, where $\lim\sup_{n} A_{n} = \cup_{n=1}^{\infty}\cap_{k=n}^{\infty} A_{k}$.
The definition of median involves probabilities, so we can alter the random variables on a set of probability $0$ without altering the median. On a set $E$ of probability $0$ let $X_n(\omega)=\frac 1 n, X(\omega)=0$ and $x_n=\frac 1 {2n}, x=0$. On the complement of $E$ let $X_n(\omega)=\frac 1 n, X(\omega)=0$ and $x_n=\frac 1 {n}, x=0$. Then the left side does not include any point of $E$ but RHS includes every $\omega \in \Omega$.