# Problem on Lebesgue differentiation theorem

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.

• If $m$ denotes the Lebesgue measure then it is not true. E.g. let $E$ be a singleton. Nov 16, 2015 at 14:17
• Is the question correct, Take$E=\Q$ ,its measure is 0 hence numerator is always 0 Nov 16, 2015 at 14:19
• You need m(E) not 0 too.
– Paul
Nov 16, 2015 at 14:20
• 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 $m(E)$ Nov 16, 2015 at 14:21
• The statement is true for almost every $x \in E$. Nov 16, 2015 at 14:44

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.