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In the first chapter of Probability wih Martingales (Willams) I came across the following example. Book says it's wrong, I don't understand what is wrong in that. Could somebody please explain why it's wrong ?

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It illustrates that the assumption $\mu(A_1)<\infty$ in the following is crucial:

Let $(X,\mathcal{E},\mu)$ be a measure space. If $(A_n)_{n\geq 1}$ is a sequence of sets from $\mathcal{E}$ such that $A_1\supseteq A_2\supseteq \cdots$ and $\mu(A_1)<\infty$, then $$ \mu\left(\bigcap_{n=1}^\infty A_n\right)=\lim_{n\to \infty}\mu(A_n). $$

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  • $\begingroup$ Thank you. So from the example we have $\operatorname{Leb}(\cap_n H_n) = \lim_{n \rightarrow \infty} \operatorname{Leb}(H_n)=\operatorname{Leb}(\emptyset)= \infty$. Could you please explain what is wrong with that? As I see, it doesn't violate the theorem. Is there a reason $\operatorname{Leb}(\emptyset)=\infty$ can't be? $\endgroup$ – triomphe Aug 21 '13 at 16:03
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    $\begingroup$ How can any measure of empty set be infinity? $\endgroup$ – Did Aug 21 '13 at 16:41
  • $\begingroup$ For any measurable set $H$: $\operatorname{Leb}(H) = \operatorname{Leb}(\emptyset \cup H) = \operatorname{Leb}(\emptyset) + \operatorname{Leb}(H)= \infty + \operatorname{Leb}(H)$? $\endgroup$ – RghtHndSd Aug 21 '13 at 18:40
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What is wrong is the idea that for $H_1 \supset H_2 \supset \dots$, we have $$\operatorname{Leb}(\cap_n H_n) = \lim_{n \rightarrow \infty} \operatorname{Leb}(H_n).$$

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  • $\begingroup$ The fact that this idea is wrong in general. $\endgroup$ – Did Aug 22 '13 at 13:40

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