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A set $E\subset \mathbb{R}$ is measurable if given $\epsilon >0$ there is an open set $G$ and a closed set $F$ such that $F\subseteq E \subseteq G$ and $m(G-F)<\epsilon$, where $m$ is the outer measure. I'm trying to prove that if $E$ is a measurable set then $E\cap K$ is measurable for all compact $K$ (in $\mathbb{R}$). This is what I have so far:

Let $E$ be a measurable set and $K$ be a compact set. Since $K$ is compact it must be closed and bounded and all closed sets are measurable. Now given $\epsilon > 0$, there are open sets $G_K, G_E$ and closed sets $F_K, F_E$ such that $F_K\subseteq K \subseteq G_K$, $F_E\subseteq E \subseteq G_E$, $m(G_K-F_K)<\epsilon$ and $m(G_E-F_E)<\epsilon$. We then have $(F_E\cap F_K)\subseteq (E\cap K)\subseteq (G_E\cap G_K)$, where $(F_E\cap F_K)$ is closed and $(G_E\cap G_K)$ is open. Now I'm having trouble showing that $m((G_E\cap G_K)-(F_E\cap F_K))< \epsilon$. Any suggestions? Is there a way to write $(G_E\cap G_K)-(F_E\cap F_K)$ as the union of (nice) disjoint sets? I realize that I should probably use $\frac{\epsilon}{2}$ up above so everything works correctly.

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But all compact sets are measurable, so $E\cap K$ is the intersection of two measurable sets which we should know is measurable?

Or are you only trying to establish that the collection of measurable sets is an algebra?

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  • $\begingroup$ yep that slipped my mind....I made this much harder than it was supposed to be... thank you $\endgroup$ – MAM Apr 12 '15 at 5:32
  • $\begingroup$ @MAM Of course this assumes you are allowed to use that fact at your current "level". If you have not yet proved this about the measurable, but are in the process of doing it, the argument could be circular. $\endgroup$ – Jeppe Stig Nielsen Apr 12 '15 at 5:38

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