Probability of sequence of events Question : Let ${A_n}$ be a sequence of events with P($A_n$)=1 $\forall n$ $\geq$1. Find P($\cup A_n$ ) and P($\cap A_n $)

What I did:
  Since it is given that P($A_n$)=$1$ we can easily say that all $A_n$ are sure events for all n $\geq 1$.So as all are sure events so each $A_n$ contains every other events so basically, we can think $\cup A_n$ as $A_k$ for any k $\in {1,2,....n}$ thus
P( $\cup A_n$ )=P( $A_k$)=$1$
Similarly for ( $\cap A_n $) we get it as $ 1$.

Am I doing any mistake ? Any help is appreciated.
 A: *

*For the first part: $P(\bigcup_{n \ge 1} A_n) \ge P(A_1)$

*For the second part: $P(\bigcap_{n \ge 1} A_n) = 1-P(\bigcup_{n\ge 1} A_n^c) = 1-0$. (You may need to prove that the countable union of zero-probability events has zero probability.)

A: The idea for union is okay and is an application of the monotonicity of measures:
$A_{1}\subseteq\bigcup_{n=1}^{\infty}A_{n}$ so that $P\left(\bigcup_{n=1}^{\infty}A_{n}\right)\geq P\left(A_{1}\right)=1$.
For intersection you did the same...? (I really wonder what that is here)
This can be solved like this:
$\bigcap_{n=1}^{\infty}A_{n}=\left(\bigcup_{n=1}^{\infty}A_{n}^{\complement}\right)^{\complement}$
so that: $$P\left(\bigcap_{n=1}^{\infty}A_{n}\right)=1-P\left(\bigcup_{n=1}^{\infty}A_{n}^{\complement}\right)\geq1-\sum_{n=1}^{\infty}P\left(A_{n}^{\complement}\right)=1$$
A: You need a proof for the intersection. Correct proof:
 $1\geq P(\cup_n A_n )\geq P(A_1)=1$ so $P(\cup_n A_n )=1$. Now $P(A_n^{c})=0$ for all $n$ so $0\leq P(\cup_n A_n^{c}) \leq \sum P(A_n^{c})=\sum 0=0$ so $P((\cup_n A_n^{c})=0$. This implies that $P(\cap_n A_n)=1$. 
