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.
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.