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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.

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Are you sure you mean $\liminf$ rathar than $\limsup$? –  23rd Nov 13 '12 at 13:00
    
Yes @Richard. What do u think? Is this true? –  Tomás Nov 13 '12 at 13:06
    
I think for $\liminf$, it is false. Consider the following counter-example. Let $E\subset\Omega$ be such that both $E$ and $\Omega\setminus E$ are of positive measure. Let $u=0$, $u_{2n}=\mathbf{1}_E$ and $u_{2n+1}=\mathbf{1}_{\Omega\setminus E}$. My suggestion is to replace $\liminf$ with $\limsup$ in your statement. –  23rd Nov 13 '12 at 13:14
    
well ok, thank you @Richard. –  Tomás Nov 13 '12 at 13:25
    
Please post your comment as an answer @Richard –  Tomás Nov 13 '12 at 14:02

1 Answer 1

up vote 2 down vote accepted

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 $E$.

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