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Let $B$ be a bounded Borel set of $\mathbb{R}$, Show that if $A$ is a finite union of disjoint intervals, the Lebesgue measure of $A\triangle B$ can be arbitrarily small. Also show that this remains true as long as $B$ has finite Lebesgue measure.

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Sometimes this is taken to be the definition of a measurable set. – user17794 Oct 19 '12 at 0:56

The second case is just as easy as the first. Let $\epsilon>0$ be given. Since $B$ is measurable, there is an open set $U\supseteq B$ such that $m(U-B)<\frac{\epsilon}{2}$. $U$ is a countable disjoint union of open intervals, say $U=\bigcup_{i=1}^{\infty}U_{i}$. Now since the measure of $U$ is finite, there must be an $N$ such that $m(\bigcup_{i=N+1}^{\infty}U_{i})<\frac{\epsilon}{2}$. Let $A=\bigcup_{i=1}^{N}U_{i}$. Then $A$ is a finite disjoint of open intervals, and:

$m(A\Delta B)=m(B-A)+m(A-B)\leq m(U-A)+m(U-B)\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

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Addition to the the answer above: "Since $B$ is measurable, there is an open set $U$ s.t. $m(U - B) < \frac{\epsilon}{2}$". Existence of a set (open or not) is since the Lebesgue measure is non atomic, then the fact follows by Sierpiński theorem. But why $U$ is open? – user90793 Jan 24 at 1:56

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