TITLE: $A = \liminf(A_n) = \limsup(A_n)$ then $P(A_n) \to P(A)$
QUESTION [3 upvotes]: Let $A = \liminf(A_n) = \limsup(A_n)$. Show that $P(A_n) \to P(A)$

$Def.: \limsup A_n = \bigcap\limits_{m\geq1}\left[\bigcup\limits_{n\geq m}A_n\right]$ and
    $\liminf A_n = \bigcup\limits_{m\geq1}\left[\bigcap\limits_{n\geq m}A_n\right]$
$P$ a probability measure on $(\Omega,\mathscr{F},P)$.


I know from a already proven theorem that, if $P$ is $\sigma$-additive (and it is), then if $A_n \uparrow A\implies P(A_n)\to P(A)$ and if $B_n \downarrow B \implies P(B_n) \to P(B)$.
Let $E_m = \bigcup\limits_{n\geq m}A_n\implies E_m \downarrow A \implies P(E_m) \to P(A)$
Let $F_m = \bigcap\limits_{n\geq m}A_n \implies F_m \uparrow A\implies P(F_m) \to P(A)$
Now, I don't believe this shows that $P(A_n) \to P(A)$. How can I finish this proof?

REPLY [4 votes]: For fixed $n$, $E_n \supset A_n\supset F_n$; therefore $P(E_n) \ge P(A_n)\ge P(F_n)$. Since both $P(E_n) \to P(A)$ and $P(F_n) \to P(A)$, by the squeeze theorem $P(A_n) \to P(A)$.
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