Intersection of countably many sets Here's a question I got as a homework assignment:

Let $\{A_i\}_{i=1,\ldots,\infty}$ a sequence of events in the probability
  space $(\Omega,F,\mathbb{P})$. Show that if $\mathbb{P}(A_i)=1$ for
  all $i$ then $\mathbb{P}(\bigcap_{i-1}^{\infty}A_i)=1$

So, as the equation is very obvious, I don't know how to prove it.
Any suggestions?
Thanks!
 A: Look at the sequence of sets $B_{k+1} = A_{k+1} \bigcap B_k$, where $B_1 = A_1$. Prove that $\mathbb{P}(B_k) = 1$ and note that $$B_1 \supseteq B_2 \supseteq B_3 \supseteq \cdots$$ and $$\bigcap_{k=1}^{\infty} A_k = \bigcap_{k=1}^{\infty} B_k$$
Recall the following result you should have proved for sequence of nested decreasing sets.
Continuity from above: If $C_k \downarrow C$, that is, $C_1 \supseteq C_2 \supseteq C_3 \cdots$ and $\displaystyle \bigcap_{k=1}^{\infty} C_k = C$, then $\mathbb{P}(C_k) \downarrow \mathbb{C}$.
A: $$
\bigcap_{n=1}^{+\infty}A_n=\Omega\setminus B
\quad
B=\bigcup_{n=1}^{+\infty}(\Omega\setminus A_n)
\quad
\mathbb P(\Omega\setminus A_n)=0
\quad
\mathbb P(B)\leqslant\sum_{n=1}^{+\infty}\mathbb P(\Omega\setminus A_n)
$$
A: That really depends on what you already know about probability. If you know that the countable union of events of probability zero has probability zero then this is merely a case of applying DeMorgan laws:
Let $B_i$ be the complement of $A_i$, then $\Bbb P(B_i)=0$, then $\Bbb P(\bigcup B_i)=0$ as well, and by DeMorgan laws this is the same as saying $\Bbb P(\bigcap A_i)=1$.
