Lebesgue measurable sets have the same Lebesgue measure as Borel sets

I have read that if $A$ is Lebesgue-measurable, then there exists Borel sets $B,C$, with $B\subset A\subset C$, such that $m(B) = m(C) = m(A)$. It is clear for me that such a set C exists, just by taking intersections of unions of rectangles covering $A$. But I don't see how to obtain the set $B$.

Thank you!

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2 Answers

Take complements of intersections of unions of rectangles covering the complement.

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That was simple! Thank you! –  Ferenc Nov 5 '12 at 23:04

This can depend on how you define Lebesgue measurable sets.

One common definition (or provably equivalent) is that $A$ is Lebesgue measurable if $A=B\cup N$ where $B$ is Borel and $N$ is a subset of a Borel set of measure zero.

In this case by definition if $A$ is Lebesgue measurable then there is some $B$ which is Borel and $A\setminus B$ is a subset of measure zero, and therefore has the same measure, and on the other hand $N\subseteq D$ where $D$ is Borel of measure zero, so $A\subseteq B\cup D$ and the same argument shows that the measure is preserved.

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