Let $\displaystyle \{E_n\}$ be a sequence of lebesgue measurable subsets of R such that

$\displaystyle \sum\limits_{n=1}^{\infty} m(E_n) < \infty $. If $\displaystyle E= \lim_{n} \sup E_{n} $, then show that m(E)=0

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- Apr 10th 2010, 01:44 AMChandru1Sequence of leesgue measurable sets
Let $\displaystyle \{E_n\}$ be a sequence of lebesgue measurable subsets of R such that

$\displaystyle \sum\limits_{n=1}^{\infty} m(E_n) < \infty $. If $\displaystyle E= \lim_{n} \sup E_{n} $, then show that m(E)=0 - Apr 10th 2010, 02:27 AMMoo
Hello,

Look at the Borel-Cantelli theorem for measure theory. - Apr 10th 2010, 09:15 AMsouthprkfan1
Since $\displaystyle \sum\limits_{n=1}^{\infty} m(E_n) < \infty $, then for all $\displaystyle \epsilon $ > 0, there exists an N such that if k > N, then $\displaystyle \sum\limits_{n=k}^{\infty} m(E_n) < \epsilon $. Thus, it must be the case that for each k > N we have $\displaystyle m(E_k) < \epsilon $. And finally, we have that sup{$\displaystyle m(E_k)$} < $\displaystyle \epsilon $ for all k > N.