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If $\{B_n\}_{n=1}^{\infty}$ is a descending collection of sets and $m(B_1)<\infty$, then $m\left(\bigcap\limits_{n=1}^{\infty}B_k\right)=\lim\limits_{k\rightarrow \infty} m(B_k).$

Why is it necessary to have $m(B_1)<\infty$?

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    It isn’t necessary to have m(B_1)<\infty, $b$ut it is necessary to have m(B_k)<\infty for some $k$.2012-10-01

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Hint: Consider the collection $\{(n,\infty)\}_{n=1}^\infty$.

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    Yes, that sounds reasonable. I guess unless we're in the extended reals the expression is indeed undefined. Right?2012-10-01