Let $(A_n)$ be a decreasing sequence of sets. I am looking for an example that if $\mu^*(A_1) = \infty$, $$\mu^*(\cap_{n \in N} A_n)\not= \lim_{n\to \infty}\mu^*(A_n).$$
I saw from another post that if we define sets to be $A_1 = [0, \infty)$ and $A_n =\emptyset$ for $n \ge 2$, this sequence of sets can be an example which satisfies the above conditions. But, I do not understand why.
I think that $\lim_{n\to \infty}\mu^*(A_n)= 0$ as $A_n$ is empty set for $n \ge 2$. But, I also think that $\cap_{n \in N} A_n = \emptyset$. That is, $\mu^*(\cap_{n \in N} A_n)=0$ as well.
Could you elaborate on this?
The classic example along these lines is $$ A_n = [n,\infty).$$ Note that each set has $\mu(A_n)=\infty,$ so $\lim_n \mu(A_n) = \infty$ but $\bigcap_n A_n = \emptyset.$
However, if the $A_n$ are decreasing and any of them have finite measure, then continuity holds.
I think copper.hat's example in the comments of the post you link refer to an earlier edit that didn't make clear that they were looking for an example with $\lim_n \mu(A_n) \ne \mu(\bigcap_n A_n).$