In my measure theory class, the professor prove that if $\mu$ is a finite measure on a space $X$ ($\mu(X) < \infty$) and $A_1 \subset A_2 \subset A_3 \subset \cdots$, then $\lim_{n\to\infty} \mu(A_n)=\mu(\bigcup_{n} A_n)$. My question is:
Is this false for infinite measures?
I've tried to play a bit with the Lebesgue measure in $\mathbb{R}$, but I could not produce any counter-examples. Any help? Thanks!
Best Answer
This is always true. Apply sigma additivity on
$B_n=A_n\backslash A_{n-1}$ where $B_1=A_1$. Note that the $B_n$ sets are disjoint, hence we can use sigma additivity in the 2nd equality $$\mu(A_n)=\mu(\bigcup_{i=1}^n B_i)=\sum_{i=1}^n \mu(B_i).$$ Taking limits on both sides gives $$\lim_{n\rightarrow \infty}\mu(A_n)=\sum_{i=1}^\infty \mu(B_i)=\mu(\bigcup_{i=1}^\infty B_n)=\mu(\bigcup_{i=1}^\infty A_n).$$
In the 2nd equality we used sigma-additivity "the other way", and the 3rd equality follows because of equality of the two sets.