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?

  • $\begingroup$ Do you need the $A_i$ to be disjoint? $\endgroup$
    – Ian Coley
    Sep 19, 2013 at 6:46
  • $\begingroup$ @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 $n$. $\endgroup$
    – Prism
    Sep 19, 2013 at 6:48
  • $\begingroup$ @Quickbeam2k1: Doesn’t that just mean that the sequence does not fulfill the requirements of the lemma? $\endgroup$ Sep 23, 2013 at 20:11
  • $\begingroup$ @Prism: Looks correct and pretty standard to me. $\endgroup$ Sep 23, 2013 at 20:13
  • $\begingroup$ @EikeSchulte: Oh alright, I am glad to hear that :) $\endgroup$
    – Prism
    Sep 23, 2013 at 20:18

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

  • $\begingroup$ 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. $\endgroup$
    – Prism
    Sep 27, 2013 at 0:11
  • $\begingroup$ 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 $i \ne j$. $\endgroup$ Sep 27, 2013 at 9:14

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