Almost Everywhere Convergence Suppose $\Omega\subset\mathbb{R}^n$ is a bounded domain and $u_n:\Omega\rightarrow \mathbb{R}$ is a sequence of positive measurable functions, such that $u_n\geq u$ for some $u:\Omega\rightarrow\mathbb{R}$ positive and measurable. Suppose that no subsequence of $u_n$ converges almost everywhere to $u$. Can I conclude from it, that in some set of positive measure $$\liminf u_n(x)>u(x)\,?$$
Thanks.

The conclusion is false. Consider the following counter-example. Let $E\subset\Omega$ be a measurable set, such that both $E$ and $\Omega\setminus E$ are of positive measure. Let $u=0$ and for every $n\ge 1$, let $u_{2n-1}=\mathbf{1}_E$ and $u_{2n}=\mathbf{1}_{\Omega\setminus E}$.
However, there eixsts some measurable $E\subset\Omega$ of positive measure, such that $\limsup_{n\to\infty}u_n>u$ on $E$. To see this, note that $\Omega$ is bounded and no subsequence of $u_n$ converges a.e. to $u$ will imply that $u_n$ cannot converge in measure to $u$. That is to say, there exist $\epsilon>0$, $\delta>0$ and a sequence $n_k\to\infty$, such that for each $k\ge 1$, $E_k:=\{x\in\Omega: u_{n_k}(x)\ge u(x)+\epsilon\}$ has measure no less than $\delta$. Then for $E:=\limsup_{k\to\infty}E_k$, its measure is no less than $\delta$, and $\limsup_{n\to\infty}u_n\ge u+\epsilon$ on $E$.
