Show that if $P\left(A_i\right)=1$for all$i\ge 1$, then$P\left(\underset{i=1}{\overset{\infty }{\cap }}{A}_i\right)=1$.

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

This is Bonferroni's inequality,$P\left({\cap }_{k=1}^n{A}_k\right)\ge \sum _{k=1}^{n}P\left(A_k\right)-(n-1)$

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

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

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

Prove this for$\infty$.

We see that:

$A_1\supset \left(A_1\cap A_2\right)\supset \dots \supset \underset{k=1}{\overset{n}{\cap }}{A}_k$

In some ordering of the$A_1,A_2\dots$

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 ${\left({\cap }_1^n{A}_k\right)}_n$is a descending sequence of events, and$\underset{k=1}{\overset{\infty }{\cap }}{A}_k=\underset{n\to \infty }{\lim }\underset{k=1}{\overset{n}{\cap }}{A}_k$.

Proposition $6.l.$for this sequence means

$P\left(\underset{k=1}{\overset{\infty }{\cap }}{A}_k\right)=\underset{n\to \infty }{\lim }P\left(\underset{k=1}{\overset{n}{\cap }}{A}_k\right)=1$

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