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I have no idea where to start for the following problem:

$E$ is a Lebesgue measurable set on $\mathbb{R}$. Show that for all $x \in E$, we have $$ \lim_{h\searrow 0} \frac{m(E\cap [x-h,x+h])}{2h} = 1$$ What if $x \notin E$?

I think it has something to do with Vitali Covering Lemma.

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    $\begingroup$ If $m$ denotes the Lebesgue measure then it is not true. E.g. let $E$ be a singleton. $\endgroup$
    – drhab
    Nov 16, 2015 at 14:17
  • $\begingroup$ Is the question correct, Take$E=\Q $ ,its measure is 0 hence numerator is always 0 $\endgroup$
    – Varun
    Nov 16, 2015 at 14:19
  • $\begingroup$ You need m(E) not 0 too. $\endgroup$
    – Paul
    Nov 16, 2015 at 14:20
  • $\begingroup$ I think the question will be $m(E) > 0 $ and for some $x$ in $E$ actually the $ m(R) $ $ R=\{x: $satisying the condition $\}$ is $m(E) $ $\endgroup$
    – Varun
    Nov 16, 2015 at 14:21
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    $\begingroup$ The statement is true for almost every $x \in E$. $\endgroup$ Nov 16, 2015 at 14:44

1 Answer 1

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What you are trying to prove is just Lebesgue's density theorem in $\mathbb R$. By choosing $f=\chi_E$ in the statement of Lebesgue's differentiation theorem, we get that, for almost every $x \in E$, $$ \lim_{h\to 0} \frac{1}{m(B(x,h))} \int_{B(x,h)}\chi_E(y)\,dy=\chi_E(x). $$ since $B(x,h)=(x-h,x+h)$, $\int_{B(x,h)}\chi_E(y)\,dy= m((x-h,x+h) \cap E)$ and $\chi_E(x)=1$ for every $x \in E$, we can rewrite this as $$ \lim_{h\to 0} \frac{m((x-h,x+h) \cap E)}{m(B(x,h))} =1, $$

which is what we wanted. If $x \notin E$, apply the same reasoning and remember that $\chi_E(x)=0$ in that case.

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