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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!

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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. 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.

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  • $\begingroup$ Thank you for your wonderful couterexample! In fact, in your example $\chi_{E_j}$ converge weakly to $\frac{1}{2}$ (constant function). $\endgroup$ Nov 1, 2021 at 15:39
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    $\begingroup$ @YongpanHuang : Yes, your comments about the convergence to $1/2$ and about the equality to $1/2$ are correct. So, are you satisfied with this answer? If so, please mark it accordingly. $\endgroup$ Nov 1, 2021 at 16:02
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    $\begingroup$ See also encyclopediaofmath.org/wiki/Rademacher_system $\endgroup$ Nov 1, 2021 at 16:11
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    $\begingroup$ @YongpanHuang : I am glad the answer was helpful. Just in case you forgot about these guidelines: mathoverflow.net/help/someone-answers and mathoverflow.net/help/accepted-answer $\endgroup$ Nov 1, 2021 at 16:12
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    $\begingroup$ @GeraldEdgar : Thank you for your comment. Yes, this is what I actually had in mind when writing this answer. :-) $\endgroup$ Nov 1, 2021 at 16:13

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