# Monotone convergence example

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 ? 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 $$\mu\left(\bigcap_{n=1}^\infty A_n\right)=\lim_{n\to \infty}\mu(A_n).$$
• 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? – triomphe Aug 21 '13 at 16:03
• For any measurable set $H$: $\operatorname{Leb}(H) = \operatorname{Leb}(\emptyset \cup H) = \operatorname{Leb}(\emptyset) + \operatorname{Leb}(H)= \infty + \operatorname{Leb}(H)$? – RghtHndSd Aug 21 '13 at 18:40
What is wrong is the idea that for $H_1 \supset H_2 \supset \dots$, we have $$\operatorname{Leb}(\cap_n H_n) = \lim_{n \rightarrow \infty} \operatorname{Leb}(H_n).$$