Showing that the intersection of countable events equals 1?

Question

How would I show that if $P(A_{i}) = 1$ for all $i \geq 1$, then $P\displaystyle\left(\bigcap_{i=1}^{\infty }A_{i}\right)=1$?

I am rather stuck on how to prove this out. I could rewrite $\bigcap_{i=1}^{\infty }A_{i}$ as an intersection of the complements of $A_{i}$, and then try to show that the probability equals zero? Because if the probability of the intersections of the complements is zero, then I can conclude that it's 1 for the unions.

Any suggestions on the first steps, or a way to approach this would be great!

Answer 1

You're basically already there, now use countable sub-additivity and squeeze the value out:

$$1\ge P\left(\bigcap_{i=1}^\infty A_i\right)^c = 1 - P\left(\bigcup_{i=1}^\infty A_i^c\right)\ge 1-\sum_{i=1}^\infty P(A_i^c)=1$$

Answer 2

A different approach: you can use induction to show $\bigcap_{r=1}^{n}A_1$ is a decreasing sequence of sets, which is $\bigcap\limits_{r=1}^nA_r\subseteq A_{n-1}\subseteq\cdots\subseteq A_1$.

When $n=2$, we have $A_1\bigcap A_2\subseteq A_1$ which is true. Assume that for all $n\geq0$ and $1\leq r\leq n$, $\bigcap\limits_{r=1}^nA_r\subseteq A_{n-1}\subseteq\cdots\subseteq A_1$. In order to show the case $n+1$ is true, we do the followings: $$\bigcap\limits_{r=1}^{n+1}A_r=\left(\bigcap\limits_{r=1}^nA_r\right)\bigcap A_{n+1}\subseteq\bigcap\limits_{r=1}^nA_r\subseteq A_{n-1}\subseteq\cdots\subseteq A_1$$ by hypothesis. By mathematical principal of induction, $\bigcap\limits_{r=1}^nA_r$ is a decreasing sequence of sets. If $n$ approaches to infinity, we will have $\bigcap\limits_{r=1}^{\infty}A_r$ contains in previous sets by the previous result. Since $P(A_r)=1$ for all $r\geq1$, we can obtain $P(\bigcap\limits_{r=1}^{\infty}A_r)=P\left(\lim\limits_{n\to\infty}\bigcap\limits_{r=1}^nA_r\right)=\lim\limits_{n\to\infty}P(A_n)=1$ by continuous probability.