I just started to read about some basic things in probability. Would be happy if someone could tell about the validity of the proof I wrote.
So, let $ \left(\varOmega,\mathcal{F},\mathbb{P}\right) $ be a probability measure.
Let $ \left(A_{n}\right)_{n} $ be a sequence of events. We define the sets:
$ \limsup(A_{n})=\bigcap_{k}\bigcup_{n\geq k}A_{n} $
$ \liminf(A_{n})=\bigcup_{k}\bigcap_{n\geq k}A_{n} $
If $ \liminf(A_{n})=\limsup(A_{n}) $ then we say that $ lim\left(A_{n}\right) $ exists and $ \liminf(A_{n})=\limsup(A_{n})=\lim A_{n} $.
Now, I already proved that for a sequence of events $ \left(A_{n}\right)_{n} $ such that $ A_{1}\subseteq A_{2}…\subseteq A_{n}… $ it follows that $ \lim A_{n}=\bigcup_{n}A_{n} $ and that
$ \mathbb{P}\left(\lim A_{n}\right)=\underset{n}{\lim}\mathbb{P}\left(A_{n}\right) $
So when you read my proof, assume I already proved this ^.
Now enough with the introduction:
let $ \left(A_{n}\right)_{n} $ be some sequence of events in the probability space such that $\lim{A_n} $ exists. I'll try to prove that $ \mathbb{P}\left(\lim A_{n}\right)=\underset{n}{\lim}\mathbb{P}\left(A_{n}\right) $.
First, denote for any $ k \in \mathbb{N} $ :
$ \bigcap_{n\geq k}A_{n}=B_{k} $
So for any $ k $, it follows that $ B_{k}=B_{k+1}\cap A_{k} $ and thus $ B_{k}\subseteq B_{k+1} $.
And now using what I wrote earlier,
$ \underset{k}{lim}\mathbb{P}\left(B_{k}\right)=\mathbb{P}\left(\lim B_{k}\right)=\mathbb{\mathbb{P}}\left(\bigcup_{k}B_{k}\right)=\mathbb{P}\left(\bigcup_{k}\bigcap_{n\geq k}A_{n}\right)=\mathbb{P}\left(\limsup A_{n}=\liminf A_{n}\right)=\mathbb{P}\left(\underset{k}{\lim}A_{k}\right)=\underset{k}{\lim}\mathbb{P}\left(\bigcap_{n\geq k}A_{n}\right) $
Now, For any k:
$ \bigcap_{n\geq k}A_{n}\subseteq A_{k} $ and thus:
$ \mathbb{P}\left(\bigcap_{n\geq k}A_{n}\right)\leq\mathbb{P}\left(A_{k}\right) $
which implies that
$ \lim_{k}\mathbb{P}\left(\bigcap_{n\geq k}A_{n}\right)\leq\lim_{k}\mathbb{P}\left(A_{k}\right) $
In addition:
Im not sure with my arguments here
for any $ k$ it follows that $ A_{k}\subseteq\bigcup_{n\geq k}A_{n} $ and therefore $ A_{k}\subseteq\bigcap_{k}\bigcup_{n\geq k}A_{n} $
Thus:
$ \mathbb{P}\left(A_{k}\right)\leq\mathbb{P}\left(\bigcap_{k}\bigcup_{n\geq k}A_{n}\right) $
Now $ \mathbb{P}\left(\bigcap_{k}\bigcup_{n\geq k}A_{n}\right) $ = $ \mathbb{P}\left(\limsup A_{n}=\lim_{k}A_{k}\right) $ does not depend on k, so $ \lim_{k}\mathbb{P}\left(A_{k}\right)\leq\lim_{k}\mathbb{P}\left(\limsup A_{n}=\lim_{k}A_{k}\right)=\mathbb{P}\left(\lim_{k}A_{k}\right)$
And all in all we get that
$ \lim_{k}\mathbb{P}\left(A_{k}\right)\leq\mathbb{P}\left(\lim_{k}A_{k}\right)\leq\lim_{k}\mathbb{P}\left(A_{k}\right) $
Thus $ \mathbb{P}\left(\lim_{k}A_{k}\right)=\lim_{k}\mathbb{P}\left(A_{k}\right) $
We'll happy to hear if that's fine or maybe I messed up one of those steps. Thanks in advance
Best Answer
If we state $B_k=\bigcup_{n=k}^{\infty}A_n$ and $C_k=\bigcap_{n=k}^{\infty}A_n$ then for every $n$:$$C_n\subseteq A_n\subseteq B_n\text{ and consequently }P(C_n)\leq P(A_n)\leq P(B_n)$$
This with $P\left(C_{n}\right)\nearrow P\left(\liminf A_{n}\right)$ and $P\left(B_{n}\right)\searrow P\left(\limsup A_{n}\right)$.
This indicates that every limit point of sequence $(P(A_n))_n$ must be an element of interval: $$[P(\liminf A_n),P(\limsup A_n)]$$
We could also express this as:$$P(\liminf A_n)\leq \liminf P(A_n)\leq\limsup P(A_n)\leq P(\limsup A_n)$$
This tells us that sequence $(P(A_n))_n$ will converge if $P(\liminf A_n)=P(\limsup A_n)$.
For this it is sufficient (not necessary) that $\liminf A_n=\limsup A_n$.
Then this interval is the singleton:$$\{P(\lim A_n)\}$$ and $P(\lim A_n)$ serves as limit of sequence $(P(A_n))_n$.