# Continuity of measure — descending collection of sets

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

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