Measure of countable union

Question

In my measure theory class, the professor prove that if $\mu$ is a finite measure on a space $X$ ($\mu(X) < \infty$) and $A_1 \subset A_2 \subset A_3 \subset \cdots$, then $\lim_{n\to\infty} \mu(A_n)=\mu(\bigcup_{n} A_n)$. My question is:

Is this false for infinite measures?

I've tried to play a bit with the Lebesgue measure in $\mathbb{R}$, but I could not produce any counter-examples. Any help? Thanks!

Answer 1

This is always true. Apply sigma additivity on

$B_n=A_n\backslash A_{n-1}$ where $B_1=A_1$. Note that the $B_n$ sets are disjoint, hence we can use sigma additivity in the 2nd equality $$\mu(A_n)=\mu(\bigcup_{i=1}^n B_i)=\sum_{i=1}^n \mu(B_i).$$ Taking limits on both sides gives $$\lim_{n\rightarrow \infty}\mu(A_n)=\sum_{i=1}^\infty \mu(B_i)=\mu(\bigcup_{i=1}^\infty B_n)=\mu(\bigcup_{i=1}^\infty A_n).$$

In the 2nd equality we used sigma-additivity "the other way", and the 3rd equality follows because of equality of the two sets.

Answer 2

Robert Bartle's The Elements of Integration (page 21 of the textbook) says:

3.4 Lemma: Let $\mu$ be a measure defined on a $\sigma$-algebra on $X$.

(a) If $(A_n)$ is an increasing sequence of sets in $X$, then $$\mu \left(\bigcup_{n=1}^\infty A_n \right)=\lim_{n \to \infty} \mu(A_n).$$

(b) If $(B_n)$ is an decreasing sequence of sets in $X$ and if $\mu(B_1) < \infty$, then $$\mu \left(\bigcup_{n=1}^\infty B_n \right)=\lim_{n \to \infty} \mu(B_n).$$

The increasing sequence of sets does not require that $\mu(X) < \infty$.

Answer 3

There isn't any problem when $\mu$ can take the value $+\infty$. If $\mu(A_n) =\infty$ for some $N$, then $$\mu\left(\bigcup_{n=1}^\infty A_n\right) \geqslant \mu(A_N)=\infty,$$ and $$\sum_{n=1}^\infty \mu(A_n)\geqslant \mu(A_N)=\infty,$$ so $$ \mu\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu(A_n).$$