Let $\{E\_j\}$ be measurable subsets in $B\_1\subset\mathbb{R}^n$ and $\exists$ $A>0$, such that
$\int\_{B\_1}\chi\_{E\_j}(x)dx\geq A$ for any $j=1,2,3,...$. Can we select a subsequence of functions $\{\chi\_{E\_j}(x)\}$ such that $\chi\_{E\_j}(x)\rightarrow\chi\_E(x)$ a.e. $x\in B\_1$, for some measurable set $E$ in $B\_1$.


Note that $\chi\_{E\_j}\in L^2(B\_1)$, then $\exists$ subsequence (we still use $E\_j$) $\chi\_{E\_j}$ converge weakly to $f\in L^2(B\_1)$. From the weakly convergence, it not hard to see $0\leq f(x)\leq 1$ for $a.e. x\in B\_1$ and $\int\_{B\_1}f(x)dx\geq A$.


The main question is that whether $f$ is a characteristic function for some measurable set in $B\_1$. Maybe there exists a counterexample. I don't know how to solve this problem.


Thanks!



a.e. convergence of characteristic functions
The answer is no. E.g., let $n=1$ and let $B\_1$ be the set of all irrational numbers in the interval $[0,1]$. For each $x\in B\_1$, let $b\_j(x)$ be the $j$th binary digit of $x$. For each natural $j$, let $E\_j:=\{x\in B\_1\colon b\_j(x)=1\}$, so that $\chi\_{E\_j}=b\_j$ on $B\_1$.


Then no subsequence of the sequence $(\chi\_{E\_j})$ [converges in measure](https://en.wikipedia.org/wiki/Convergence_in_measure). Indeed, otherwise there would exist an increasing sequence $(j\_k)$ of natural numbers such that the sequence $(b\_{j\_k})$ converges in measure and hence $b\_{j\_{k+1}}-b\_{j\_k}\to0$ in measure (as $k\to\infty$), so that
$$\tfrac12=\lambda(\{x\in B\_1\colon b\_{j\_{k+1}}(x)\ne b\_{j\_k}(x)\}) \\
=\lambda(\{x\in B\_1\colon|b\_{j\_{k+1}}(x)-b\_{j\_k}(x)|\ge1\})\to0,$$ where $\lambda$ is the Lebesgue measure -- a contradiction.


Thus, no subsequence of the sequence $(\chi\_{E\_j})$ converges almost everywhere.


