Proof that every random variable has a median, proof check? I already checked the question Does a median always exist?. But I am not convinced that the limits exist.
Say I want to show $m:=\inf\{x\in\mathbb{R}\mid\mathbb{P}(X\leq x)\geq \frac{1}{2}\}$. Then for $x_n$ decreasing sequence with $x_n \to m$ and $x_n>m$, we have:
$$\mathbb{P}(X\leq m)=\lim_{n\rightarrow \infty} \mathbb{P}(X\leq x_n)=F(m^+)\geq\frac{1}{2}.$$
Since $(-\infty,m]=\cap_n(-\infty,x_n)$.
Now for the other inequality:
For $x<m$, $\mathbb{P}(X\leq x) < \frac{1}{2}$ since $m$ is the infimum. Now $x_n \to m$ increasing sequence and $\cup_n(-\infty,x_n]=(-\infty,m)$ since $m$ is in none of the sets
$$\lim_{n\to< \infty}\mathbb{P}(X\leq x_n)=\mathbb{P}(X<m )<\frac{1}{2} \quad \text{or should it be $\leq$?}$$
and $\mathbb{P}(X\geq m)\geq \frac{1}{2}$

Setting $a=\inf\left\{ x\in\mathbb{R}\mid P\left(X\leq x\right)\geq0.5\right\} $
and $b=\sup\left\{ x\in\mathbb{R}\mid P\left(X\geq x\right)\geq0.5\right\} $
we find: $$P\left(X\leq a\right)\geq0.5\text{ and }P\left(X\geq b\right)\geq0.5$$
Further the assumption $b<x<a$ leads to $P\left(X\leq x\right)<0.5\wedge P\left(X\geq x\right)<0.5$
contradicting that $P\left(X\leq x\right)+P\left(X\geq x\right)\geq1$.
So we are allowed to conclude that $a\leq b$ and that: $$m\in\left[a,b\right]\iff P\left(X\leq m\right)\geq0.5\wedge P\left(X\geq m\right)\geq0.5$$
The RHS states exactly that $m$ is a median of the distribution of
$X$.
