A critiria of measurability related to Lebesgue Density Theorem The statement is as follows.
Assume $E\subset\mathbb{R}$, $\forall x\in E$, $$\lim_{\delta\to0}\frac{m^*(E^c\cap (x-\delta,x+\delta))}{2\delta}=0$$
where $E^c$ denotes the complement of $E$. Can we conclude that $E$ is a Lebesgue measurable set? $m$ denotes the standard Lebesgue measure and $m^*$ denotes the outer Lebesgue measure on $\mathbb R$.
Any suggestion is welcomed. Thanks a lot!
 A: The answer is yes, $E$ is measurable. Here is a theorem that holds for $\mathbb{R}^n$

Theorem: Let $E\subset \mathbb{R}^n$. Then
$$\begin{align}
\lim_{\delta\rightarrow0}\frac{m^*(E\cap B(x;\delta))}{m(B(x;\delta))}=1\tag{1}\label{one}
\end{align}$$
for almost all $x\in E$. Furthermore, $E$ is Lebesgue measurable iff
$$\begin{align}
\lim_{\delta\rightarrow0}\frac{m^*(E\cap B(x;\delta))}{m(B(x;\delta))}=0\tag{2}\label{two}
\end{align}$$
for almost all $x\in E^c$.

A proof of this result can be found in Jones, F. Lebesgue Integral on Euclidean space, Revised edition, Jone & Bartlett, 2001. p. 664

Here is an outline to the proof: To simplify arguments, I consider only the case where $m^*(E)<\infty$. Let $A$ a Borel set such that $E\subset A$ and $m^*(E)=m(A)$. (Take for example open sets $E\subset A_n$ such that $m(A_n)<m^*(E)+2^{-n}$ and define $A=\bigcap_nA_n$).
Claim: for any Lebesgue measurable set $B$,
$$ m^*(E\cap B)=m(A\cap B)$$
Indeed, as $A$ is measurable
$$m^*(E\cap B)=m^*(E\cap B\cap A)+m^*(E\cap B\cap A^c)=m(A\cap B)$$
It follows that
$$\frac{m^*(E\cap B(x;\delta))}{m(B(x;\delta))}=\frac{m(A\cap B(x;\delta))}{m(B(x;\delta))}$$
Therefore, by Lebesgue's theorem
$$\begin{align}\lim_{\delta\rightarrow0}\frac{m^*(E\cap B(x;\delta))}{m(B(x;\delta))}=\lim_{\delta\rightarrow0}\frac{m(A\cap B(x;\delta))}{m(B(x;\delta))}=1_A(x)\qquad \text{a.a}\,x\in\mathbb{R}^n\tag{3}\label{three}
\end{align}$$
Since $E\subset A$, \eqref{one} follows, as well as \eqref{two} when $E$ is Lebesgue measurable.
Conversely, assume that \eqref{two} holds for all $x\in E^c$. By \eqref{three}, $A\cap E^c$ is a null set and so Lebesgue measurable; hence
$E=A\setminus (A\cap E^c)$ is Lebesgue measurable.
