Let $A_1,A_2,\ldots$ be Lebesgue measurable subsets of $\mathbb{R}$. It is certainly true, by monotonicity of $\mu$, that $$\mu\left(\bigcup_{i=1}^nA_i\right)\leq \mu\left(\bigcup_{i=1}^\infty A_i\right)$$
Is it true that $$\lim_{n\rightarrow\infty}\mu\left(\bigcup_{i=1}^nA_i\right)= \mu\left(\bigcup_{i=1}^\infty A_i\right)?$$
I've tried to think about the case where $\mu\left(\bigcup_{i=1}^\infty A_i\right)$ is infinite, but still can't find any counterexample.
Best Answer
This is precisely continuity from below of measures:
Set $E_n = \bigcup_1^n A_i$, $E=\bigcup_1^\infty A_i$. Then $E_1 \subset E_2 \subset\ldots$, and $E = \bigcup_1^\infty E_n$. Therefore we may apply continuity from below to obtain $$ \lim_{n\to\infty}\mu\left(\bigcup_1^n A_i\right) = \lim_{n\to\infty}\mu(E_n) = \mu(E)=\mu\left(\bigcup_1^\infty A_i\right). $$
It follows from the countable additivity property of measures. Set $B_1 = A_1$, $B_i = A_i\setminus A_{i-1}$. Note that the $B_i$-s are pairwise disjoint, $\bigcup_1^n B_i = \bigcup_1^n A_i$, and $\bigcup_1^\infty B_i=\bigcup_1^\infty A_i$. Can you finish?