Question: Prove or disprove the following: Let $E \subset \mathbb{R}^n$. If given any $\epsilon > 0$, there exists an open $U \supset E$ such that the outer measure of $E, |E|_e > |U| - \epsilon$ then $E$ is Lebesgue measurable set in $\mathbb{R}^n$.
I believe this statement is false since the definition of Lebesgue measurable set is slightly different. That is given any $\epsilon >0$ there exists open $U \supset E$ such that $$|U- E|_e < \epsilon.$$
However, I can't seem to find any non-measurable sets that satisfy the following hypothesis to show a counterexample. Any hints/reference will be appreciated.
( notation: Lebesgue outer measure $m^*( E )$ )
Let $E \subset \mathbb{ R }^1$ ( Lebesgue measurable or not Lebesgue measurable). Suppose $m^*( E )$ is finite.
We will show for any $\epsilon > 0$ there is an open set $U \supset E$ such that $m^*( E ) > m^*( U ) - \epsilon$.
By definition, $m^*( E ) = \inf \{ \sum_{ n = 1 }^{ \infty } L( J_n ) : E \subset \cup_{ n = 1 }^{ \infty } J_n \}$, where $J_n$ is a bounded open interval and $L$ is the length of the interval.
Let $\epsilon > 0$. Since $m^*( E )$ is finite, there is a sequence of open intervals $( J_n )$ such that $E \subset \cup_{ n = 1 }^{ \infty } J_n$ and $\sum_{ n = 1 }^{ \infty } L( J_n ) < m^*( E ) + \epsilon$. $\:$ $\cup_{ n = 1 }^{ \infty } J_n$ is an open set. $\:$ Let $U = \cup_{ n = 1 }^{ \infty } J_n$. $\:$ $E \subset U$.
$$m^*( U ) = m^*( \cup_{ n = 1 }^{ \infty } J_n ) \le \sum_{ n = 1 }^{ \infty } m^*( J_n ) = \sum_{ n = 1 }^{ \infty } L( J_n )$$
$$m^*( U ) \le \sum_{ n = 1 }^{ \infty } L( J_n ) < m^*( E ) + \epsilon$$
$$m^*( U ) - \epsilon < m^*( E )$$