I am trying to prove the following inequality:
If $A_1, A_2, A_3, A_4,\ldots$ is a sequence of events. Prove the following:
$$\Pr\left(\lim_{n \to \infty}\inf A_n\right) \le \lim_{n \to \infty}\inf P(A_n)$$
I am doing the following:
$$\ \bigcup \limits_{N=1}^{\infty} \bigcap\limits_{n=N}^{\infty}A_n \subseteq\bigcup\limits_{n=N}^{\infty} A_n$$
Using the monotonocity of measures :
$$\Pr\left(\bigcup \limits_{N=1}^{\infty} \bigcap\limits_{n=N}^{\infty}A_n \right) \le \Pr\left(\bigcup\limits_{n=N}^{\infty} A_n\right)$$
Now the left-hand side is a lower bound for the term on the right-hand side.
$$\Pr\left(\lim_{n \to \infty}\inf A_n\right) \le \lim_{n \to \infty}\Pr\left(\bigcup\limits_{n=N}^{\infty} A_n\right)$$
Am I right till here?. Is my approach correct?
Thank you.
Best Answer
No your inclusion does not make sense, what is $N$ on the RHS? Also the RHS on the last equality does not equal $\liminf P(A_n)$. Here is how you prove it.
Assume that $(\Omega,\mathbb{F},P)$ is a probability space and that $A_n\in \mathbb{F}$ for all $n\geq 1$. Now note that $$ \liminf_{n\to \infty} A_n := \bigcup_{n\geq 1} \bigcap_{k\geq n} A_k\in \mathbb{F}, $$ and that $ \bigcap_{k\geq n} A_k \subset \bigcap_{k\geq n+1} A_k$ for any $n\geq 1$. Thus by continuity from below of the probability measure $P$, and $\cap_{k\geq n}A_k \subset A_n$, $$ P(\liminf_{n\to \infty} A_n) = P\left( \bigcup_{n\geq 1} \bigcap_{k\geq n} A_k\right) = \lim_{n\to\infty} P\left ( \bigcap_{k\geq n} A_k \right) $$ $$ =\liminf_{n\to\infty}P\left ( \bigcap_{k\geq n} A_k \right) \leq \liminf_{n\to\infty}P\left( A_n \right) $$ where we used the monotonicity of $P$ and that $\liminf=\lim$ whenever the limit exists (which is does in our case).