# Limit of measure of finite unions

I am beginning to learn measure theory, and I have a basic doubt regarding to measure of union of sets, and limits.

Let $\mathcal{A}$ be a $\sigma$-algebra on $X$, and $\mu$ be a finite measure on $\mathcal{A}$. Suppose that $A_1, A_1, A_3, ...$ are elements of $\mathcal{A}$, and further we know $$\mu\left(\bigcup_{i=1}^{n}A_i\right)=\sum_{i=1}^{n} \mu(A_i)$$ for every positive integer $n$. How can we conclude the following? $$\mu\left(\bigcup_{i=1}^{\infty}A_i\right)=\sum_{i=1}^{\infty} \mu(A_i)$$

My idea: Let $E_k=\bigcup_{i=1}^{k} A_i$. Then, $E_1\subseteq E_2\subseteq E_3\subseteq\cdots$ Also, $\bigcup_{k=1}^{\infty}E_k=\bigcup_{i=1}^{\infty} A_i$. Then, using continuity from below argument (Theorem 1.8 (c) in Folland), we get $$\sum_{i=1}^{\infty}\mu(A_i)=\lim_{n\to\infty}\mu\left(\bigcup_{i=1}^{n} A_i\right)=\lim_{n\to\infty} \mu(E_n) = \mu\left(\bigcup_{i=1}^{\infty}E_n\right)=\mu\left(\bigcup_{i=1}^{\infty}A_n\right)$$

Assuming that my argument is correct (is it actually?), I would like to know whether this is the standard/canonical method for this basic problem?

• Do you need the $A_i$ to be disjoint? Sep 19, 2013 at 6:46
• @IanColey: Nope, $A_i$s need not be distinct. I am only assuming $\mu\left(\bigcup_{i=1}^{n}A_i\right)=\sum_{i=1}^{n} \mu(A_i)$ for every positive integer $n$. Sep 19, 2013 at 6:48
• @Quickbeam2k1: Doesn’t that just mean that the sequence does not fulfill the requirements of the lemma? Sep 23, 2013 at 20:11
• @Prism: Looks correct and pretty standard to me. Sep 23, 2013 at 20:13
• @EikeSchulte: Oh alright, I am glad to hear that :) Sep 23, 2013 at 20:18

The sets $\{A_i\}$ don't have to be disjoint. However, the intersection of any two sets must have measure $0$ as can be shown by induction:
And since $$\mu\left(\bigcup_{i=1}^{n-1} A_i\right) + \mu(A_n) = \mu\left(A_n \cup \bigcup_{i=1}^{n-1} A_i\right)$$
We have $$\mu\left(A_n \cap \bigcup_{i=1}^{n-1} A_i\right) = 0.$$
• Thanks! This is excellent! Just to make sure, in order to conclude that $\mu(A_i\cap A_j)=0$ for $i\neq j$, we will need to use monotonicity after distributing the "big" union in the last line. Sep 27, 2013 at 0:11
• Happy to help! Since $A_n \cap A_k \subset A_n \cap \bigcup_{i=1}^{n-1} A_i$ for $k \in \{1, \ldots, n-1\}$, then $\mu(A_n \cap A_k) \le \mu\left(A_n \cap \bigcup_{i=1}^{n-1} A_i\right) = 0$. Since this holds for all $n \in \mathbb N$, we conclude that $\mu(A_i \cap A_j) = 0$ for all $i \ne j$. Sep 27, 2013 at 9:14