Relation between $m^{***}$ with $m^*$? For any set $A$, define $m^{***} (A)\in [0, \infty]$ by
$$m^{***}(A)=\sup \{m^*(F) \mid F \subseteq A, F \text{  is closed } \}$$
So how is $m^{***}$ related to outer measure $m^*$?
There is a similar question already exists how are $m^{**}(E)$ and $m^{***}(E)$ related to $m^{*}(E)$?
said that $m^{***}=m^{**}=m^*$
However, the hint "Regularity theorem for Lebesgue measure" is kind of new to me. I have already proved $m^{**}=m^*$ in that question, and also in my problem, it's easy to see that $m^{***} \leq m^*$. However, I am not sure how to process the other direction. Any help? Thanks!
 A: There's a theorem from Tao's measure theory book that states the following to be equivalent for $E\subseteq \mathbb{R}^d$:


*

*$E$ is Lebesgue measurable.

*For every $\epsilon > 0$, there is an open set $U\supset E$ such that $m^*(U\setminus E) < \epsilon$.

*For every $\epsilon > 0$, there is an open set $U\subset \mathbb{R}^d$ such that $m^*(E\Delta U) < \epsilon$.

*For every $\epsilon > 0$, there is a closed set $F\subset E$ such that $m^*(E\setminus F) < \epsilon$.

*For every $\epsilon > 0$, there is a closed set $F\subset \mathbb{R}^d$ such that $m^*(E\Delta F) < \epsilon$.

*For every $\epsilon > 0$, there is a Lebesgue measurable set $E_{\epsilon}$ such that $m^*(E_{\epsilon}\Delta E) < \epsilon$.


This is essentially the regularity theorem for the Lebesgue measure. For a proof, look at Exercise 1.2.7 here. As you have already shown, $m^*$ and $m^{**}$ are equivalent definitions of the Lebesgue outer measure on $\mathbb{R}$ ($m^{**}$ is the more general definition for arbitrary Euclidean spaces). It is easy to show using the above theorem that the Lebesgue inner measure and Lebesgue outer measure are equal for Lebesgue measurable sets; namely, for any $\epsilon > 0$, we can choose a closed set $F\subseteq E$ and an open set $U\supseteq E$ such that $$m^{***}(E)\geq m^*(F)\geq m(E)-\frac{\epsilon}{2}$$ and $$m^{**}(E)\leq m^*(U)\leq m(E)+\frac{\epsilon}{2}$$ As this is true for all $\epsilon > 0$, we will have $m^{**}(E)\leq m(E)\leq m^{***}(E)$. Combined with the more obvious inequality $m^{***}(E)\leq m(E)\leq m^{**}(E)$, we will have $m^{**} = m^{***} = m$ for Lebesgue measurable sets. When $E$ is not Lebesgue measurable, however, the inner and outer measure will not agree.
