Limit of measure of finite unions

Question

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?

Answer 1

Your method is correct. This approach is standard. This is how I'd approach the problem.

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:

\begin{align} \mu\left(\bigcup_{i=1}^{n-1} A_i\right) + \mu(A_n) &= \mu\left(\bigcup_{i=1}^{n-1} A_i\right) + \mu\left(A_n - \bigcup_{i=1}^{n-1} A_i\right) + \mu\left(A_n \cap \bigcup_{i=1}^{n-1} A_i\right) \\ &= \mu\left(A_n \cup \bigcup_{i=1}^{n-1} A_i\right) + \mu\left(A_n \cap \bigcup_{i=1}^{n-1} A_i\right) \end{align}

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 $$ \mu\left(A_n \cap \bigcup_{i=1}^{n-1} A_i\right) = 0. $$