I've been stuck at this problem for a while. I'm to show that the following two statements are equivalent.
$(i)$ $E$ is a Lebesgue measurable set with finite measure.
$(ii)$ For every $\epsilon > 0$, $\exists$ a bounded open set $U$ such that $m^*(E \Delta U) \leq \epsilon$.
To show $(i) \implies (ii)$, I've used the almost open criteria of Lebesgue measurability to conclude that for every $\epsilon > 0$, $\exists$ a bounded open set $V$ with finite measure, such that $m^*(E \Delta V) \leq \epsilon$. But I'm having difficulty with the boundedness part. I guess an open set with finite measure doesn't necessarily has to be bounded.
To show $(ii) \implies (i)$, again we can use the almost open criteria of Lebesgue measurability to conclude that $E$ is a Lebesgue measurable set. But somehow I need to use the boundedness condition to imply that $E$ has a finite measure.
Any help would be greatly appreciated!
$(i) \implies (ii)$ Since $E$ is measurable $m(E) = m^*(E)$ so $\forall \frac{\epsilon}{2}$ there exists $O=\cup_{i=1}^{\infty}I_i$ where $I_i$ are disjoint open intervals, such that $E\subset O$ and $m(E) + \frac{\epsilon}{2}>m(O)$. Since E has finite measure O has finite measure so $\lim_{n \to\infty} \sum_{i=n}^{\infty}m(I_i) = 0$. In particular there exists N such that $X=\bigcup_{i=N}^{\infty}I_i$, $ m(X) \leq \frac{\epsilon}{2}$. Let S=O\X
$$E\Delta S = (E\setminus S) \cup (S \setminus E)) = (E\setminus(O\setminus X)) \cup ((O\setminus X)\setminus E)\subset X \cup (O\setminus E) $$ So $m^*({E\Delta S}) \leq m^*({X}) + m^*(O\setminus E)\leq \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon $ . Note that S is bounded as required (as a finite union of intervals). Note also that I freely switch between outer measure and measure for sets with are clearly measurable.