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



\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..

  • 2
    $\begingroup$ Please avoid math-only titles. $\endgroup$
    – Asaf Karagila
    Mar 30 '20 at 16:38
  • $\begingroup$ 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.) $\endgroup$
    – Asaf Karagila
    Mar 30 '20 at 16:39
  • $\begingroup$ 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 $$\lim_{n\to\infty}a_n=\limsup_{n\to\infty}a_n=\liminf_{n\to\infty} a_n$$ $\endgroup$ Mar 31 '20 at 8:29
  • $\begingroup$ 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 $\endgroup$ Apr 1 '20 at 10:48

I think there may be some mistake in the book.We can write proof as


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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.