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Recently in my class there was talk of a sequence of sets $A_1 \supseteq A_2 \supseteq \dots $, where each set has infinite measure. Although I found an example that violated the equality (which was the point of the exercise): $$\mu(\bigcap_{n=1}^\infty A_n) = \lim_{n \rightarrow \infty}\mu(A_n),$$ namely, $$ A_n = [n, \infty[, \text{ where } n \in \mathbb{N}^+,$$ I am still a bit puzzled at how I did it. I can see that the measure of the intersection of the sets is $0$, as can be shown be contradiction (assume it weren't empty, containing a positive real number $r$; but then $r \notin [\lceil{r}\rceil, \infty[$, so it cannot be an intersection of all the sets).
Regarding $\lim_{n\rightarrow \infty}\mu(A_n)$, I'm somewhat conflicted:

  • On the one hand, $\forall \ n \ \mu(A_n) = \infty$. Hence, $\lim_{n \rightarrow \infty} \infty = \infty$
  • On the other hand, $\lim_{n \rightarrow \infty}A_n = \varnothing$. Hence, $\mu(\lim_{n \rightarrow \infty}A_n) = 0$.

So am I right to conclude that in general $\mu(\lim_{n \rightarrow \infty}A_n) \neq \lim_{n\rightarrow \infty}\mu(A_n)$? Why? Thanks for clarifying these issues.

Mitch Baker
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    You are right. Indeed, this is a basic theorem in measure theory (look up continuity from above), but with the additional hypothesis that $\mu(A_1) < \infty$, and your example is a typical one illustrating why that hypothesis is needed. –  Nov 10 '17 at 21:30

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Yes, you are right to conclude that, and you're right to conclude that because you've found a counterexample.

Duncan Ramage
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