Let us pretend we have a field called $\mathscr{R}_0$ and a probability measure $\mathbb{P}$ which is finitely addivitive on it. Suppose that there is a sequence of events $(A_n)_{n\in\mathbb{N}}$ such that $A_n\in\mathscr{R}_0\text{ all }n$.
I quote Billingsley (1995):
We would like to show that $\mathbb{P}$ is also countably additive on $\mathscr{R}_0$ and this will follow if, starting from a sequence $(A_n)_{n\in\mathbb{N}}$, $A_n\in\mathscr{R}_0$ and $A_n\downarrow \emptyset$ together imply $\mathbb{P}(A_n)\downarrow 0$.
Suppose that $A_1\supset A_2\supset\cdots$ and that $\mathbb{P}(A_n)\geq\varepsilon>0$ for all $n$.
$\color{red}{\text{The problem is to show that }\bigcap_{n} A_n \text{ must be nonempty.}}$
My question revolves around the $\color{red}{\text{ red part}}$ immediately above.
Since $(A_n)_{n\in\mathbb{N}}$ is a decreasing sequence of events, I know that:
$$\lim\limits_{n\to\infty}A_n=\bigcap_{n=1}^{\infty}A_n\tag{1}$$
Since, as stated in my reference, we know that:
$$A_n\in\mathscr{R}_0\hspace{0.4cm}\text{ and }\hspace{0.4cm} A_n\downarrow \emptyset\implies \mathbb{P}(A_n)\downarrow0\tag{2}$$
I would like to have that:
- $A_n\in\mathscr{R}_0$;
- $\lim\limits_{n\to\infty}A_n=\emptyset$, that is, equivalently, $A_n\downarrow\emptyset$;
Condition $1.$ is satisfied by assumption, while for condition $2.$ to be fulfilled I would expect that I have to show that:
$$\lim\limits_{n\to\infty}A_n=\bigcap_{n=1}^{\infty}A_n=\emptyset\tag{3}$$
and not that "$\bigcap_n A_n$ is nonempty".
So, why does it suffice to show that "$\bigcap_n A_n$ is nonempty" in order for $(2)$ to hold true, if one of the condition required for that result is precisely that $A_n\downarrow\emptyset$?
Best Answer
A fintely additive probability measure $P$ is countably additive if and only if $P(A_n) \to 0$ whenever $A_n$ decreases to empty set. Suppose there is sequence $A_n$ decreasing to empty set such that $P(A_n)$ does not tend to $0$. Since $P(A_n)$ is decreasing it follows that $\epsilon =\inf P(A_n) >0$. So we have $P(A_n) \geq \epsilon$ for all $n$. We can get a contradiction by showing that $\bigcap A_n$ must be empty.