$\lim_{k \rightarrow \infty} \mu \left(\bigcup_{n=1}^k A_n \right) = \mu \left(\bigcup_{n=1}^\infty A_n \right)$

Question

The problem I'm working on is: Prove that for a family of measurable sets $A_k$ in $[a,b]$ the following is true $$\lim_{k \rightarrow \infty} \mu \left(\bigcup_{n=1}^k A_n \right) = \mu \left(\bigcup_{n=1}^\infty A_n \right)$$

This post was the most similar to my problem, however, the solution doesn't include the union(since it's an increasing sequence?)

Also, the answer to this question includes exactly what I am looking for , but says "there is no reason to think that" the equality always holds.

This got me confused. Please help!

$\mu$ is a measure and all I know about ${A_n}_{n=1}^{\infty}$ that it is a family of measurable sets in $I_0 =[a,b]$ — John Lennon, May 10 '15 at 07:38

score 3 · Answer 1 · answered May 10 '15 at 07:42

The continuity of the measure $\mu$ implies $$\mu(B) = \lim_{k \to \infty} \mu(B_k)$$ for any sequence $(B_k)_{k \in \mathbb{N}}$ of measurable sets satisfying $B_1 \subseteq B_2 \subseteq \ldots$ and $B = \bigcup_{n \in \mathbb{N}} B_n$. Apply this to

$$B_k := \bigcup_{n=1}^k A_n \qquad \text{and} \qquad B := \bigcup_{n=1}^{\infty} A_n.$$

$\lim_{k \rightarrow \infty} \mu \left(\bigcup_{n=1}^k A_n \right) = \mu \left(\bigcup_{n=1}^\infty A_n \right)$

1 Answers1