## Lemma on Measure of Increasing Sequences in X

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

(a) If ($E_n$) is an increasing sequence in X, then

$\mu (\bigcup_{n=1}^\infty E_n )=\lim \mu (E_n)$

(b) If $F_n$) is a decreasing sequence in X and if $\mu (F_1)<\infty$, then

$\mu (\bigcap_{n=1}^\infty F_n )=\lim \mu (F_n )$

Note: An increasing sequence of sets ($E_n$) means that for all natural numbers n, $E_n \subseteq E_{n+1}$. A decreasing sequence means the opposite, i.e. $E_n \supseteq E_{n+1}$.

Proof: (Elaboration of the proof given in Bartle’s book)

(a) First we note that if $\mu (E_n) = \infty$ for some n, then both sides of the equation are $\infty$, and inequality holds. Henceforth, we can just consider the case $\mu (E_n)<\infty$ for all n.

Let $A_1 = E_1$ and $A_n=E_n \setminus E_{n-1}$ for n>1. Then $(A_n)$ is a disjoint sequence of sets in X such that

$E_n=\bigcup_{j=1}^n A_j$, $\bigcup_{n=1}^\infty E_n = \bigcup_{n=1}^\infty A_n$

Since $\mu$ is countably additive,

$\mu (\bigcap_{n=1}^\infty E_n) = \sum_{n=1}^\infty \mu (A_n)$ (since ($A_n$) is a disjoint sequence of sets)

$=\lim_{m\to\infty} \sum_{n=1}^m \mu (A_n)$

By an earlier lemma $\mu (F\setminus E)=\mu (F)-\mu (E)$, we have that $\mu (A_n)=\mu (E_n)-\mu (E_{n-1})$ for n>1, so the finite series on the right side telescopes to become

$\sum_{n=1}^m \mu (A_n)=\mu (E_m)$

Thus, we indeed have proved (a).

For part (b),  let $E_n=F_1 \setminus F_n$, so that $(E_n)$ is an increasing sequence of sets in X.

We can then apply the results of part (a).

\begin{aligned} \mu (\bigcup_{n=1}^\infty E_n) &=\lim \mu (E_n)\\ &=\lim [\mu (F_1)-\mu (F_n)]\\ &=\mu (F_1) -\lim \mu (F_n) \end{aligned}

Since we have $\bigcup_{n=1}^\infty E_n =F_1 \setminus \bigcap_{n=1}^{\infty} F_n$, it follows that

$\mu (\bigcup_{n=1}^\infty E_n) =\mu (F_1)-\mu (\bigcap_{n=1}^\infty F_n)$

Comparing the above two equations, we get our desired result, i.e. $\mu (\bigcap_{n=1}^\infty F_n) = \lim \mu (F_n)$.