Prove that $P(\bigcup_{i=1}^{\infty} A_i) = 1$

Question

$A_i$ $(i=1,2,...)$ are independent events $\sum_{i=1}^{\infty}P(A_i) = \infty.$ Prove that: $P(\bigcup_{i=1}^{\infty} A_i) = 1 $

Can someone please help me out with this question?

Answer 1

I believe this is related to the second Borel-Cantelli lemma. The proof goes as follows:

\begin{align}P\left( \bigcup_{i=1}^{\infty} A_i \right) &\geq P\left( \bigcup_{i=n}^{\infty} A_i \right)\\ &=\lim_{k \to \infty} P\left( \bigcup_{i=n}^{k} A_i \right)\\ &= 1-\lim_{k \to \infty} P\left( \bigcap_{i=n}^{k} A_i^C \right) \\ &= 1-\lim_{k \to \infty}\prod_{i=n}^k \left(1-P(A_i)\right)\\ &\geq 1-\lim_{k \to \infty} \exp\left(-\sum_{i=n}^k P(A_i)\right)\\ &= 1\\ \end{align}

Where the second-to-last inequality follows from the fact that $$-\ln(1-P(A_i))=\sum_{k=1}^{\infty}\frac{[P(A_i)]^k}{k}\geq P(A_i)$$

Answer 2

By virtue of the non-trivial part of Borel-Cantelli lemma, one is allowed to conclude that the event $\displaystyle \mathrm{A} = \{A_n, \mathrm{i.o.}\} = \bigcap_{k = 1}^\infty \bigcup_{n = k}^\infty A_n$ has total mass equal to one. Notice that $\mathrm{A}$ lies within the union of the $A_n$ and the exercise follows.

Answer 3

TL;DR BCL2 tells us $A_n$ occurs infinitely often. So obviously, at least one $A_n$ occurs.

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Pf:

$$\sum_{i=1}^{\infty}P(A_i) = \infty$$ and $A_n$ indp

By BCL2, we have $P(\limsup A_n) = 1$

Note that $$\limsup A_n \subseteq \bigcup_n A_n \tag{*}$$

QED

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Intuition:

$(*)$ should be not so difficult to prove and should be easily understood intuitively:

In the sequence

$$A_1, A_2, A_3, A_4, ...$$

we deduce by BCL2 that for any day $m$, (let's say $A_m$ corresponds to day $m$), there is some future day $n \ge m$ that will occur eg on day 1, there is some day $n \ge 1$ s.t. $A_n$ will occur. On day $m=100$, there is some $n \ge 100$ s.t. $A_n$ will occur. We say that $A_n$ occurs infinitely often.

So if $A_n$ occurs infinitely often, then obviously at least one of those $A_n$'s will occur i.e. $$P(\bigcup_{n=1}^{\infty} A_n) = 1$$

Answer 4

If $P\left( \bigcup_{n=1}^\infty A_n \right) < 1,$ then $P\left( \bigcap_{n=1}^\infty (\text{not } A_n) \right)>0.$ By independence, we have $$ \prod_{n=1}^\infty P(\text{not } A_n) > 0, \text{ so } \log\prod_{n=1}^\infty (1-P(A_n))>-\infty. $$ $$ \sum_{n=1}^\infty -\log (1-P(A_n)) <\infty. $$ If you can show that when $0\le p<1$ then $p\le -\log(1-p),$ then you get $$ \sum_{n=1}^\infty P(A_n) < \infty. $$ The inequality $p\le -\log(1-p)$ depends on the base of the logarithm being $\le e$ but $>1;$ however, this is not essential, because for any base $>1$ one can find a suitable positive constant and say $p\le -(\text{constant}\cdot\log(1-p)).$

Observe that if $\sum_{n=1}^\infty p_n = \infty$ then $\sum_{n=\mathbf N}^\infty p_n=\infty$ no matter how be $\mathbf N$ gets. Thus after any such index $\mathbf N$, one can find some $n$ for which $A_n$ occurs. Therefore, infinitely many of them occur. Consequently this whole argument amounts to a proof of one of the Borel–Cantelli lemmas.