91Ó°ÊÓ

Let $A_1,A_2,…$be a sequence of events such that for every$P\left(A_i\right)=1$, prove$P\left(\underset{k=1}{\overset{∞}{âˆ\copyright }}{A}_k\right)=1$.

This is Bonferroni's inequality,$P\left({âˆ\copyright }_{k=1}^n{A}_k\right)â‰\yen \underset{k=1}{\overset{n}{∑}}P\left(A_k\right)-(n-1)$

Directly from this inequality, it follows that for any $n$events with probabilityrole="math" localid="1649443762370" $1,P\left(\underset{k=1}{\overset{n}{âˆ\copyright }}{A}_k\right)â‰\yen 1$.

From Axioms probability of the outcome, space is$1$, and the probability of every event i.e. subset is less or equal to$1$.

$P\left(\underset{k=1}{\overset{n}{âˆ\copyright }}{A}_k\right)=1$

Prove this for$∞$.

We see that:

$A_1⊃\left(A_1âˆ\copyright A_2\right)⊃…⊃\underset{k=1}{\overset{n}{âˆ\copyright }}{A}_k$

In some ordering of the$A_1,A_2…$

Otherwise, some$A_k$is a subset of the intersection of all the sets, and therefore the probability of the intersection is greater than$P\left(A_k\right)=1$.

So ${\left({âˆ\copyright }_1^n{A}_k\right)}_n$is a descending sequence of events, and$\underset{k=1}{\overset{∞}{âˆ\copyright }}{A}_k=\underset{n→∞}{\lim }\underset{k=1}{\overset{n}{âˆ\copyright }}{A}_k$.

Proposition $6.l.$for this sequence means

$P\left(\underset{k=1}{\overset{∞}{âˆ\copyright }}{A}_k\right)=\underset{n→∞}{\lim }P\left(\underset{k=1}{\overset{n}{âˆ\copyright }}{A}_k\right)=1$

Since the right-hand side holds limes of a constant sequence with value$1$, according to the equation $\left(1\right)$above.