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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 $m(B_1)<\infty$, but it is necessary to have $m(B_k)<\infty$ for some $k$. – Brian M. Scott Oct 1 '12 at 8:18
up vote 6 down vote accepted

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

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What is $\lim_{n \to \infty} (n,\infty)$? $(\infty, \infty)$? I'm not sure what that is. Is it undefined? Or is it the empty set? – Rudy the Reindeer Oct 1 '12 at 8:17
@Matt the intersection of all of the sets is empty, but the measure of each set is infinite. So the limit of the measures is infinity, while the measure of the "limit" set is zero. – JSchlather Oct 1 '12 at 8:27
Sorry, does this mean "yes, $(\infty, \infty) = \varnothing$"? – Rudy the Reindeer Oct 1 '12 at 8:54
@MattN. I suppose that typically for any $a \in \mathbb R$ the convention is that $(a,a)$ is empty, so in the extended real number line we'd still have $(\infty,\infty)$ is empty. – JSchlather Oct 1 '12 at 9:29
Yes, that sounds reasonable. I guess unless we're in the extended reals the expression is indeed undefined. Right? – Rudy the Reindeer Oct 1 '12 at 11:41

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