If $A_{n}\rightarrow A$, when can we say $\mu\left(A_{n}\right)\rightarrow\mu\left(A\right)$, where $\mu$ is a measure on $\left(X,\mathcal{A},\mu\right)$ and $A_{n},A\in\mathcal{A}$ for all $n$?
As far as I know, if $A_{n}\nearrow A$, then $\mu\left(A_{n}\right)\nearrow\mu\left(A\right)$. I also think if $A_{n}\searrow A$, then $\mu\left(A_{n}\right)\searrow\mu\left(A\right)$.
Is it true if $A_{n}\rightarrow A$, then $\mu\left(A_{n}\right)\rightarrow\mu\left(A\right)$ as well? (I don't think so...)
I see in a lot of proofs they use the $$A_{n}\searrow A,\;\;\;\text{then}\;\;\;\mu\left(A_{n}\right)\searrow\mu\left(A\right)$$ ”trick”, but they always say if $\mu\left(X\right)<\infty$, than the previous trick holds. More precisely, I always see: if $\mu\left(X\right)<\infty$ and $A_{n}\searrow\emptyset$, then $\mu\left(A_{n}\right)\rightarrow0$, but I don't really get why we need the $\mu\left(X\right)<\infty$ criterion.
Let $X$ be a covering of all $A_n$. By dominated convergence theorem, $$\lim_{n\rightarrow\infty} \int_{X} i_{A_n} d\mu=\int_{X}i_{A}d\mu$$ (We assume the measure of all sets involved is finite so that the indicators are integrable)