# Lebesgue outer measure regularity

In $$\mathbb{R}$$, the outer regularity of Lebesgue measure gives $$m^*(A)=\inf\{m(A)\mid E \text{ is open and } A \subset E\}$$.

Can we replace $$E$$ as measurable sets instead of open sets? I am not able to prove it. Looking for some hints. Thanks in advance.

You can but the result is somewhat trivial. Since every open set is Lebesgue measurable, we have $$\inf\{m(E)\mid E \text{ is measurable},\; A \subseteq E\} \leq \inf\{m(E)\mid E \text{ is open}, \;A\subseteq E\} = m^*(A).$$ This follows because the left infimum has a greater range. By monotonicity, $$m^{*}(A) \leq m^{*}(E) = m(E)$$ holds for all measurable $$E$$.
Therefore, we have $$m^*(A) \leq \inf\{m(E)\mid E \text{ is measurable}, A \subseteq E\}$$