Properties of the Probability Measure 2 Let $(\Omega, F, \mathbb{P})$ be a Probability space and $A_n \in F$ for $n \in \mathbb{N}$. I need to show:
(i) Let $\sum\limits_{n=1}^\infty \mathbb{P}(A_n)<\infty$, then $\mathbb{P}[\bigcap\limits_{m\in \mathbb{N}}\bigcup\limits_{n\geq m} A_n]=0 $
(ii) Find an example with $\sum\limits_{n=1}^\infty \mathbb{P}(A_n)=\infty$ and $\mathbb{P}[\bigcap\limits_{m\in \mathbb{N}}\bigcup\limits_{n\geq m} A_n]<1 $.
For (i) I tought this: We know that $\mathbb{P}(\emptyset)=0 $. Since $\sum\limits_{n=1}^\infty \mathbb{P}(A_n)<\infty$ is a sum with infinitive summands, the sum is not inifity itself and $\mathbb{P}$ is always positive, only finite summands can be bigger than zero. This means that there is an index $N$ with with $\mathbb{P}(A_k)=0$ for all $k\geq N$. The next step would be looking at $$\mathbb{P}[\bigcap\limits_{m\in \mathbb{N}}\bigcup\limits_{n\geq m} A_n]= \mathbb{P}[(A_1\cup A_2 \cup ... A_N \cup \emptyset\, \cup ...)\cap(A_2\cup A_3 \cup ... A_N \cup \emptyset\, \cup ...)\cap...(A_N \cup \emptyset\, \cup ...)\cap((\emptyset \cup \emptyset \cup ...  \cup \emptyset\, \cup ...)]= \mathbb{P}(\emptyset)=0$$ Is this ok?
(ii)Here unfortunatly I have no idea and would appreciate help.
 A: For the first part, it unfortunately does not work, because there are a lot of convergent series which have infinitely many non-zero terms, like $\sum_{n\geqslant 1}n^{-2}$ or $\sum_{n\geqslant 1}2^{-n}$. Instead, the key point is to use $\sigma$-sub-additivity of a probability measure to bound $\Pr\left(\bigcup_{n\geqslant m}A_m \right)$  remainders of the series $\sum_{m\geqslant 1}\Pr\left(A_m\right)$.
For the second part, let $A_n=A$ where $\Pr\left(A\right) \in (0,1)$.
A: (i) 
We can have $P(A_n)=2^{-n}>0$ for every $n$ so that $\sum_{n=1}^{\infty}P(A_n)=\sum_{n=1}^{\infty}2^{-n}=1<\infty$.
This shows that your reasoning is not correct.
Note that for every $k\in\mathbb N$ we have: $$\bigcap_{m\in\mathbb N}\bigcup_{n\geq m}A_n\subseteq\bigcup_{n\geq k}A_n$$and consequently by finite $\sum_{n\geq 1}P(A_n)$ for every $k\in\mathbb N$ we have:$$0\leq P(\bigcap_{m\in\mathbb N}\bigcup_{n\geq m}A_n)\leq P(\bigcup_{n\geq k}A_n)\leq\sum_{n\geq k}P(A_n)=\sum_{n\geq 1}P(A_n)-\sum_{n<k}P(A_n)\tag1$$
For every $\epsilon>0$ we can choose $k$ in such a way that the RHS does not exceed $\epsilon$ so this proves that $P(\bigcap_{m\in\mathbb N}\bigcup_{n\geq m}A_n)$ must be $0$.
(ii) See the answer of Davide.
A: Here is another try for (i), maybe this time? I would use the $\sigma$-continuity:
We know that for $B_n\uparrow B$ which means $B_1\supset B_2 \supset ... $, if $\mathbb{P}(B_1)<\infty$ and $\bigcap\limits_{n=1}^\infty B_n \in F$ we have 
$$
\lim_{m\rightarrow\infty} \mathbb{P}(B_n) = \mathbb{P}(\bigcap\limits_{n=1}^\infty B_n)
$$
Now for $B_n:=\bigcup\limits_{n\geq m} A_n$ this is true since $\bigcup\limits_{n+1\geq m} A_{n+1}\subset\bigcup\limits_{n\geq m} A_n$. Furthermore $P(\bigcup_{n\geq k}A_n)\leq\sum_{n\geq k}P(A_n)<\infty$ and since $A_n\in F$ and $F$ is a $\sigma$-Algebra, we also have $\bigcap\limits_{n=1}^\infty A_n \in F$. So we get:
$$
\mathbb{P}(\bigcap\limits_{m=1}^\infty B_n)=\mathbb{P}(\bigcap\limits_{m=1}^\infty \bigcup\limits_{n\geq m} A_n)= \lim_{m\rightarrow\infty} \mathbb{P}(\bigcup\limits_{n\geq m} A_n)=0
$$
To be honest I just kept thinking about this because I know that your answers are right, but I don't really understand them trough and trough. Is this idea ok or: Where did I make the mistake?
