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Let $E\subseteq \mathbb{R}^n$ be a set such that for every $\varepsilon>0$, there is an open set $U\subseteq \mathbb{R}^n$ such that $\lambda^*(E\triangle U)\leq \varepsilon$, where $\lambda^{*}$ is the outer measure.

Show that E is lebesgue-measurable.

I've tried several things that didn't work... I'd like to hear suggestions

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  • $\begingroup$ Ah, you're quite right. I thought I'd seen this problem before and went off memory. $\endgroup$ – Nobody Jul 29 '16 at 21:07
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Let $A \subseteq \mathbb{R}^{n}$. We need to show that $\lambda^{*}(A) = \lambda^{*}(A\cap E) + \lambda^{*}(A\cap E^{c})$, where $E^{c} = \mathbb{R}^{n} - E$. Since $\lambda^{*}$ is subadditive, this amounts to showing $$ \lambda^{*}(A) \geq \lambda^{*}(A\cap E) + \lambda^{*}(A\cap E^{c}). $$ SO, let $\epsilon > 0$ and choose an open $U \subseteq \mathbb{R}^{n}$ such that $\lambda^{*}(U\triangle E) < \epsilon$. Since $U$ is measurable, $$ \lambda^{*}(A\cap E) = \lambda^{*}(A\cap E \cap U) + \lambda^{*}(A\cap E\cap U^{c}), $$ and $$ \lambda^{*}(A\cap E^{c}) = \lambda^{*}(A\cap E^{c} \cap U) + \lambda^{*}(A\cap E^{c}\cap U^{c}). $$ Putting these together, we get $$ \lambda^{*}(A\cap E) + \lambda^{*}(A\cap E^{c}) \leq \lambda^{*}(A\cap U) + \lambda^{*}(A\cap U^{c}) + \lambda^{*}(E^{c} \cap U) + \lambda^{*}(E \cap U^{c}) \leq \lambda^{*}(A) + \epsilon,$$ where the last inequality comes from the fact that $\lambda^{*}(A\cap U) + \lambda^{*}(A\cap U^{c}) = \lambda^{*}(A)$ by the measurability of $U$.

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