From Probability Through Problems By Marek Capinski,Tomasz Jerzy Zastawnaik
Verify that
$P(\lim_{n \to \infty}\inf A_n) \leq \lim_{n \to \infty}\inf P(A_n) \leq \lim_{n \to \infty}\sup P(A_n) \leq P(\lim_{n \to \infty}\sup A_n)$
Solution as given : Consider
$B_n=\cap_{k=n}^{\infty}A_k$
then
\begin{eqnarray*}P(\lim_{n \to \infty}\inf A_n) &=& P(\cup_{n=1}^{\infty}B_n)…since \space \lim_{n \to \infty}\inf A_n =\cup_{n=1}^{\infty}\cap_{k=n}^{\infty}A_k \\ &=& \lim_{n \to \infty}P(B_n)…since \space B_1\subset B_2\subset…\\ &=& \lim_{n \to \infty} \inf P(B_n) ..since \space P(B_1)\leq P(B_2) \leq…\\ &\leq& \lim_{n \to \infty}\inf P(A_n)…since\space B_n\subset A_n \\ \end{eqnarray*}
What I am not getting is the $3^{rd}$ step,I mean how $\lim_{n \to \infty}P(B_n)=\lim_{n \to \infty} \inf P(B_n)$?Please explain..
Thanks in advance..
Best Answer
I think there may be some mistake in the book.We can write proof as
$B_n=\cap_{k=n}^{\infty}A_k$
for all $k \geq n$,$B_n \subset A_k$
so we can say $P(B_n) \leq P(A_k)$ for all $k \geq n$
So,we can say $P(B_n) \leq \inf_{k \geq n} P(A_k)$....(1)
So we proceed as
\begin{eqnarray*}P(\lim_{n \to \infty}\inf A_n) &=& P(\cup_{n=1}^{\infty}B_n)...since \space \lim_{n \to \infty}\inf A_n =\cup_{n=1}^{\infty}\cap_{k=n}^{\infty}A_k \\ &=& \lim_{n \to \infty}P(B_n)...since \space B_1\subset B_2\subset...\\ &\leq & \lim_{n \to \infty} \inf_{k \geq n} P(A_k) ..from (1) \\ &=& \lim_{n \to \infty}\inf P(A_n) \\ \end{eqnarray*}
is it correct way?