This is not a repeat question because I am not asking for an answer. I would like to know if my reasoning is valid – which I could not find in the replies to previously asked instances of this question.
This is the problem:
Show that if $P(A_{i}) = 1$ for all $i \geq 1$ then $P(\ \bigcap_{i=1}^{\infty}A_i)=1.$
This is my proof:
Let S denote the universe/sample space which contain events $A_{i}$'s. Suppose $\exists j$ such that when $i\neq j$, $A_{i}\neq A_{j}$.
Then $P(A_{i}) + P(A_{j}) = 1 + 1 = 2 \gt 1 = P(S), \bot.$ Therefore $A_{i} = A_{j}$ for all i, j $\ge$ 1, so $P(\bigcap_{i=1}^{\infty}A_i) = P(\bigcap_{i=1}^{\infty}A_1) = P(A_1) = 1.$
Basically I am asserting that all $A_{i}$'s must be equal in order to be probability 1 which, while intuitive to me, I'm not sure is valid reasoning.
Best Answer
No, your reasoning is not valid. $A_i\neq A_j$ does not imply $P(A_i\cup A_j)=P(A_i)+P(A_j)=1+1$. In order to use $P(A\cup B)=P(A)+P(B)$, you need $A$ and $B$ to be disjoint, not just unequal.
There is no way to prove $A_i=A_j$. It could be that the difference between $A_i$ and $A_j$ is nonempty, but still has probability zero. For example, suppose $X$ is a random number chosen uniformly between $0$ and $1$, and let $$ A_1=\{X\neq 0.5\},\qquad A_2=\{X\neq 0.25\},\qquad A_3=\{X\neq 0.75\} $$ Then all the $A_i$ have probability one, but none of them are equal to each other. For example, $A_1\neq A_2$, because $A_1$ contains the outcome where $X=0.25$, but $A_2$ does not.