What does continuity of probability even mean intuitively? I'm reading "All of Statistics A Concise Course in Statistical Inference" by Larry Wassermann, and I came across a proof for continuity of probability like this :
$\text{Theorem (1.8)(Continuity of Probabilities)}$ If $A_{n} \rightarrow A$ then
$$ \lim_{n \rightarrow \infty} \big( \mathcal{P}(A_{n}) \big) \rightarrow \mathcal{P}(A)$$
Suppose that $A_{n}$ is monotone increasing so that $A_{1} \subset A_{2} \subset $ Then let$A = \lim_{n \rightarrow \infty} A_{n} = \bigcup_{i=1}^{\infty} A_{i}.$
Define  $B_{1} = A_{1}$ ,and also define the set $B_{2} = \big\{ \omega \in \Omega: \omega \in A_{2} , \omega \notin A_{1}\big\} $ $B_{3} = \big\{ \omega \in \Omega: \, \omega \in A_{3}, \omega \notin A_{2}, \omega \notin A_{1} \big\}, ... $ It can be shown that $B_{1}$, $B_{2}$, … are disjoint, $A_{n} = \bigcup_{i=1}^{n}A_{i} = \bigcup_{i=1}^{n}B_{i}$ for each $n$ and $\bigcup_{i=1}^{\infty}B_{i} = \bigcup_{i=1}^{\infty} A_{i}$ $\big( \text{Exercise (1)} \big)$ From Axiom $(3)$, one can say that
$$\mathcal{P(A_{n})} = \mathcal{P} \bigg( \bigcup_{i = 1}^{n} B_{i} \bigg) = \sum_{i = 1} \mathcal{P}(B_{i}) $$
Using axiom $(3)$ again,
$$\lim_{n \rightarrow \infty} \mathcal{P}(A_{n}) = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \mathcal{P}(B_{i}) = \sum_{i = 1}^{ \infty} \mathcal{P}(B_{i}) = \mathcal{P} \bigg( \bigcup_{i=1}^{\infty} B_{i} \bigg) = \mathcal{P}(A).$$
So here I understand how the proof goes, as in various steps in the proof, but I don't get why and what are we trying to prove.
Since I don't have a background in pure mathematics and measure theory and I'm having a hard time wrapping my head around this.
I don't get what exactly are we trying to prove here :
let's say $A_{n}$ is an event so are we trying to prove that as the event grows to infinity(since A is monotonically increasing) the probability of that event approaches the probability of sample space, which is essentially equal to $1$?
What does this intuitively mean, or what kind of pre-requisites do I need for understand this text?
EDIT : I would also appreciate if someone can point out some other text books which would help me understand these things better as someone with little knowledge in pure math.
 A: The idea is not necessarily that the probability approaches $1$ as $n \to \infty.$
We have an event $A$ which could be the entire sample space,
or it could just be some subset of the sample space with probability less than $1.$
And each of the events $A_1,A_2,A_3,\ldots$ is a subset of $A,$
each one possibly including more of $A$ than the previous subset.
Here's a concrete example. Suppose there is a game for $k$ players that works as follows. The players are identified as Player $1$ through Player $k$. On turn $i$ the game master randomly generates the number $X_i,$ selected from the integers from $1$ to $10k$ with uniform distribution. (You could say we roll a fair die with $10k$ sides). If $X_i \leq k$ and $X_i = j$ then Player $j$ wins the game.
The game continues indefinitely until someone wins, and then it stops.
Now suppose you are Player $1$ and $A$ is the event that you win.
The event $A_i$ is the event that you win on or before turn $i.$
Clearly, if you win on or before turn $i$ then you win on or before turn $i+1,$ so $$A_1 \subset A_2 \subset A_3 \subset \cdots \subset A_i \subset A_{i+1} \subset \cdots .$$
But in order to win you must win on some turn $i,$ so
$$A = \bigcup_{i=1}^{\infty} A_i.$$
Specifically, as $i \to \infty$ the sets $A_i$ get larger and larger and they come asymptotically closer to containing the entire event $A,$ even though no particular $A_i$ actually contains all of $A.$
On the other hand, someone else might win (and in fact this happens with probability $\frac{k-1}{k}$).
When someone else wins, of course, none of the events $A_1, A_2, \ldots, A_i, \ldots$ ever occurs. For example, if player $2$ wins on turn $7,$ then $A_7$ does not occur because you did not win then, and $A_8$ does not occur because there was not even a turn $8$ for you to win.
Intuitively, each player has an equal chance of winning, so
$\newcommand{P}{\mathcal P}\P(A) = \frac 1k.$
If you win, it will likely take you many turns to do so;
$\P(A_1)$ is only $\frac1{10k},$
while $\P(A_2) = \frac{19}{100k},$ $\P(A_3) = \frac{271}{10000k},$
and so forth.
But the sequence of probabilities
$\P(A_1), \P(A_2), \P(A_3), \ldots, \P(A_i), \ldots,$
representing your cumulative probabilities of winning by each turn,
gradually but surely approaches the limit $\frac1k.$
A: It should say if $A_n \nearrow A$, then $P(A_n) \nearrow P(A)$. More explicitly, it says that if $A = \bigcup_{n = 1}^{\infty}A_n$, then $P(A) = \lim_{n \to \infty}P(A_n)$. There is also an analagous result that if $A_n \searrow A$, then $P(A_n) \searrow P(A)$.
This theorem (and especially it's corollary, the monotone convergence theorem) is useful. Suppose you have a coin that lands heads with probability $p > 0$. Flip it until it lands heads and let $X$ be the total number of flips (so $X$ is a geometric random variable). What is $P(X < \infty)$, i.e. what is the probability that the coin eventually lands heads? We can see that with $A_n = \{X < n\}$ and $A = \{X < \infty\}$ we have $A_n \nearrow A$. We have $P(A_n) = P(X < n) = 1 - P(X \geq n) = 1 - (1 - p)^n$. By the theorem, $P(A) = \lim_{n \to \infty}P(A_n) = \lim_{n \to \infty}(1 - (1 - p)^n) = 1$.
