I have $u_{n}$ sequence of $H^{1}_{0}(\Omega)$ where $\Omega$ is open bounded and connected domain in $\mathbb{R}^{n}$ with $n>1$.
$u_{n}\rightarrow u$ in $H^{1}_{0}(\Omega)$ norm. Let $\Omega_{K}=\{x\in \Omega : u(x)>K a.e.\}$ with $K>0$.
It's true or false that there exist $n_{0}$ such that $\forall n\geq n_{0}$ $$u_{n}(x)>\dfrac{K}{2}$$ almost everywhere in $\Omega_{K}$ and uniformly respect to $x$.
No, it's not true. Let $\Omega$ be the unit ball. Pick a function $v\in H_0^1(\Omega)$ that is unbounded from above near $0$, say $v(x)=\max(0,\sqrt{-\log |x|}-1)$. Define $u_n(x)=1-v(x)/n$. The sequence converges to $u\equiv 1$ in the $H^1$ norm. The set $\{u>2/3\}$ is all of $\Omega$, but the set $\{u_n>1/3\}$ is only a spherical shell, for every $n$.
As defined, $u_n$ is not in $H_0^1$, but you can multiply it by a cutoff function such as $\max(1,1-2|x|)$. The effect remains: the set $\{u_n>1/3\}$ has a "hole" near $0$.