# Verifying $P(\lim_{n \to \infty}\inf A_n) \leq \lim_{n \to \infty}\inf P(A_n) \leq \lim_{n \to \infty}\sup P(A_n) \leq P(\lim_{n \to \infty}\sup A_n)$

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..

• Please avoid math-only titles. Commented Mar 30, 2020 at 16:38
• Is $lim_{n\to\infty}inf$ suppose to be $\liminf_{n\to\infty}$? (Also \lim and \inf and \liminf will make it look better, same holds for \sup and \limsup.) Commented Mar 30, 2020 at 16:39
• By the way, this is a special case of Fatou's lemma. If a sequence of real numbers $(a_n)_{n\in\mathbb N}$ converges, then $$\lim_{n\to\infty}a_n=\limsup_{n\to\infty}a_n=\liminf_{n\to\infty} a_n$$ Commented Mar 31, 2020 at 8:29
• But why will the sequence {$P(B_n)$} will be converging?It will be converging in the case if there exist a $k\in N$ such that $B_k=B_{k+i}$ for all $i \in \mathbb{N}$ .and this is possible if $A_k=A_{k+i}$ for all $i \in \mathbb{N}$.But how can this be possible if the sample space is infinite Commented Apr 1, 2020 at 10:48

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?