In my probability class, my professor talked something about the Outer continuity. It is the following:
If $A_{i+1} \subseteq A_i$ for all $i \in \mathbb{N}$, and $A = \bigcap_{i \in \mathbb{N}A_i}$, then $P(A) = \lim_iP(A_i)$.
I saw and proved this before. But this time he asked us to find a counterexample which shows that this one might fail in $\sigma$-finite measure, instead of finite measure.
In office hour, he actually gave an answer to us, which is like this:
Let $P$ be the Lebesgue measure on the real line. Denote $A_n = (n, +\infty)$, then $P(A) = 0$ since $A = \emptyset$. However, $\lim_{n\to\infty}P(A_n) = +\infty$, thus $P(A) \not = \lim_{n\to\infty}P(A_n)$.
I have not really learned measure theory thoroughly, so I have some troubles to get the idea why $\lim_{n\to\infty}P(A_n) = +\infty$, but $P(A) = 0$ here. Can anyone explain this a little bit? Thanks so much!
$P(A_n)$ is the length of the interval $(n,\infty)$, which is infinity for each $n$. Thus $\lim P(A_n)=\infty$.
Suppose $x\in A$. There exists $N$ such that $x<N$. So $x\notin (N,\infty)=A_N$. This contradicts $x\in\bigcap A_n$. So in fact there is no element in $A$, i.e. $A$ is the empty set.