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I am trying to prove the following inequality:

If A1, A2, A3, A4,.... is a sequence of events. Prove the following:

$$\Pr(lim_{n \to \infty}inf A_n) <= lim_{n \to \infty}inf P(A_n)$$

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. $$\Pr(lim_{n \to \infty}inf A_n) <= lim_{n \to \infty}\Pr\left(\bigcup\limits_{n=N}^{\infty} A_n\right)$$

Am I right till here?. Is my approach correct? Thank you.

1 Answers 1

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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).

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    `lim inf = lim` How this is implied ?. If it is so, then `lim sup=lim` will also hold, which suggests that the given sequence `A1,A2,A3...` converges also. But, we have no information about its convergence. – 2017-02-02
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    This is a basic analysis result. A sequence $(a_n)_{n\in\mathbb{N}}$ converges to $L\in \mathbb{R}$, that is $\lim_{n\to \infty} a_n=L$, if and only if $\liminf_{n\to \infty}a_n = \limsup_{n\to\infty} a_n = L$. We know that – 2017-02-02
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    $\mathbb{R}\ni P(\liminf_{n\to \infty} A_n) = \lim_{n\to\infty} P\left ( \bigcap_{k\geq n} A_k \right)$, so the the sequence of $(P\left ( \bigcap_{k\geq n} A_k \right))_{n\in \mathbb{N}}$ converges and the limit equals liminf. – 2017-02-02
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    Note that $P(\liminf A_n)\in [0,1]$ since it is the measure of a measurable set. Continuity from below yields that $(P\left ( \bigcap_{k\geq n} A_k \right))_{n\in \mathbb{N}}$ converges and its limit is given by $P(\liminf A_n)$. – 2017-02-02
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    Furthermore it does not entail that $A_1,A_2,...$ converges in measure. Simply take the sequence $A_1=[0,1]$, $A_2=[0,1/2]$, $A_3=[0,1],...$ and $P$ the Lebesgue measure on $[0,1]$. Now $P(A_n)$ does not converge since it oscillates between $1$ and $1/2$. However $P(\cap_{k\geq n} A_k) = P([0,1/2]$ so $(\cap_{k\geq n} A_k)$ converges. – 2017-02-02
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    Not sure if I can still reach anyone here but I have problems with @Martin example above. If you define your $A_i$ as above and furthermore $P(A_{2k+1})=\frac{1}{4}$ and $P(A_{2k})=\frac{3}{4}$. Then $P(\liminf(A_n))=P([0,1/2])=\frac{3}{4}$ but $\liminf(P(A_n))=\frac{1}{4}$ a contradiction to the inequality. Where is my mistake? – 2018-07-07
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    @Erdbeer0815 The problem with your counterexample is as follows. In my example we have that $A_{2k+1}= [0,1]$ and $A_{2k}=[0,1/2]$ for any $k\in \mathbb{N}$. Thus we can't define a probability measure $P$ such that $P(A_{2k+1}) = 1/4 \leq 3/4 = P(A_{2k})$. This is because [probability measures is monotone](https://math.stackexchange.com/questions/656219/monotonicity-of-measures) (all positive measures as a matter of fact). Meaning that since $A_{2k} \subset A_{2k+1}$ we must have that $P(A_{2k}) \leq P(A_{2k+1})$, which does not hold in your example. – 2018-07-09