Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

1 Answer 1

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:

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.