Let $(X,A)$ a measurable space.
$\mu: A\to [0,\infty]$ satisfying:
$\mu(X)<\infty$
$\mu(\emptyset)=0$
$\forall{E,F}\in{A}$ disjoint sets,
$\mu(E\cup F)=\mu(E)+\mu(F)$
Prove:
$\mu$ is a measure on $A$ iff
for every deceasing sequence $E_n$ in $A$ $\mu(\cap_{n=1}^{\infty} E_n)= \lim_{n\to \infty} \mu(E_n)$.
First: $\mu (\emptyset)=0$ (given).
The forward case is obvious since $\mu$ is a measure by assuming, and $\mu(X) < \infty$ implies that $\mu(E) <\infty$ for all $E\in{A}$.
So using a theorem we finish it.
The second condition is to show that for a sequence of disjoint sets $F_n \in{A}$ , $\mu(\cup_{n=1}^{\infty} F_n) =\sum_{n=1}^{\infty} \mu(F_n)$.
Let's define a new decreasing sequence $E_n$.
$E_n=X\setminus (\cup_{i=1}^{n} F_i)=\cap_{i=1}^{n} F_i
^{c}$.
$\mu(X)=\mu(E_n\cup E_n^{c})=(by 2)=\mu(E_n)+\mu(E_n^{c})$.
$\mu(E_n^{c})=\mu(\cup_{i=1}^{n} F_i)=\mu(X)-\mu(\cap_{i=1}^{n} F_i^{c})$
Now let n approaches infinity so:
$\mu(\cup_{i=1}^{\infty} F_i)=\mu(X)-lim_{n\to \infty} \mu(\cap_{i=1}^{n} F_i^{c})=\mu(X)-lim_{n\to \infty} \mu(X\setminus \cup_{i=1}^{n} F_i)$
Then can I use that $\mu(lim)=lim(\mu)$ ?
(Then I get
$\mu(\cup_{i=1}^{\infty} F_i)=\mu(X)-\mu(lim_{n\to \infty} X\setminus {\cap_{i=1}^{n} F_i)}=\mu(X)-\mu(lim_{n\to \infty} \cup_{i=1}^{n} F_i)$ .
So I can reduce $\mu(X)$ with its limit since $\mu(X)$ is finite (by 2), and use 3 after it.
Can someone help in this.
Best Answer
Hint
Let $\{F_i\}_{ i\in\mathbb N}$ a collection of disjoint sets. \begin{align*} \mu\left(\bigcup_{i=1}^\infty F_i\right)&=\mu(X)-\mu\left(\bigcap_{i=1}^\infty F_i^c\right)\\ &=\mu(X)-\lim_{n\to \infty }\mu\left(\bigcap_{i=1}^n F_i^c\right)\\ &=\mu(X)-\lim_{n\to \infty }\left(\mu(X)-\mu\left(\bigcup_{i=1}^n F_i\right)\right). \end{align*} I let you justify all steps and conclude.