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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.

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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 $$m^*(A) \leq \inf\{m(E)\mid E \text{ is measurable}, A \subseteq E\}$$

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