In the text "All of Statistics A Concise Course in Statistical Inference" by Larry Wassermann,and I'm inquiring of my solution to $\text{Exercise (1)}$ ? Due to the fact that resolving the exercise requires one to be familiar with the proof it has been provided for the readers convince.
$\text{Exercise (1)}$
Fill in the details of the proof of $\text{Theorem (1.8)}$. Also prove the monotone deceasing case.
$\text{Theorem (1.8) (Continuity of Probabilities).}$ 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
$$\mathcal{P(A_{n})} = \mathcal{P} \bigg( \bigcup_{i = 1}^{n} B_{i} \bigg) = \sum_{i = 1} \mathcal{P}(B_{i}) $$
Abusing axiom $(3)$ again,
$$\lim_{n \rightarrow \infty} \mathcal{P}(A_{n}) = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \mathcal{P}(B_{i}) = \sum_{i = 1}^{ \infty} \mathcal{P}(B_{i}) = \mathcal{P} \bigg( \bigcup_{i=1}^{\infty} B_{i} \bigg) = \mathcal{P}(A).$$
$\text{Solution}$
$\text{Proof of the Monotone Increasing Case}$
Recall from the proof of $(1.8)$ that the author defines $B_{1} = A_{1}$, and also the respective sets $B_{2}$, $B_{3}$. The author claims that $B_{1}, B_{2}, ...$ are disjoint. To verify this one can observe that $\big(A_{i} \big)_{i \in I}$ is a family of sets, it's easy to see that
$$\bigcap_{i\in I} A_{i} = \emptyset.$$
The author also claims that for each $n$.
$(1)$
$$A_{n} = \bigcup_{i=1}^{n}A_{i} = \bigcup_{i=1}^{n}B_{i}$$
and he also claims that,
$(2)$
$$B_{n}= \bigcup_{i=1}^{\infty}B_{i} = \bigcup_{i=1}^{\infty} A_{i}$$
To prove $(1)$ and $(2)$ take,
$$ \lim_{ n \rightarrow \infty} (A_{n}) = \lim_{ n \rightarrow \infty} \bigg(\bigcup_{i=1}^{n}A_{i} \bigg) = \lim_{ n \rightarrow \infty} \bigg( \bigcup_{i=1}^{n}B_{i} \bigg) \rightarrow A. $$
As well as,
$$\lim_{ n \rightarrow \infty} (B_{n}) = \lim_{ n \rightarrow \infty} \bigg(\bigcup_{i=1}^{n}B_{i} \bigg) = \lim_{ n \rightarrow \infty} \bigg( \bigcup_{i=1}^{n}A_{i} \bigg) \rightarrow B .$$
QED.
$\text{Proof of the monotone decreasing case }$
Suppose that $A_{n}$ is monotone decreasingso that $A_{1} \subset A_{2} \subset $ Then let$A = \lim_{n \rightarrow \infty} A_{n} = \bigcap_{i=1}^{\infty} A_{i}.$ 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, $A_{n} = \bigcap_{i=1}^{n}A_{i} = \bigcap_{i=1}^{n}B_{i}$ for each $n$ and $\bigcap_{i=1}^{\infty}B_{i} = \bigcap_{i=1}^{\infty} A_{i}$ $\big( \text{Exercise (1)} \big)$ From Axiom $(3)$, one can say that
$$\mathcal{P(A_{n})} = \mathcal{P} \bigg( \bigcap_{i = 1}^{n} B_{i} \bigg) = \sum_{i = 1} \mathcal{P}(B_{i}) $$
Abusing $\text{Axiom (3)}$ again,
$$\lim_{n \rightarrow \infty} \mathcal{P}(A_{n}) = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \mathcal{P}(B_{i}) = \sum_{i = 1}^{ \infty} \mathcal{P}(B_{i}) = \mathcal{P} \bigg( \bigcap_{i=1}^{\infty} B_{i} \bigg) = \mathcal{P}(A). $$
Recall from the proof of $(1.8)$ that the author defines $B_{1} = A_{1}$, and also the respective sets $B_{2}$, $B_{3}$. The author claims that $B_{1}, B_{2}, ...$ are disjoint. To verify this one can observe that $\big(A_{i} \big)_{i \in I}$ is a family of sets, it's easy to see that
$$\bigcap_{i\in I} A_{i} = \emptyset.$$
The author also claims that for each $n$.
$(1)$
$$A_{n} = \bigcap_{i=1}^{n}A_{i} = \bigcap_{i=1}^{n}B_{i}$$
and he also claims that,
$(2)$
$$B_{n}= \bigcap_{i=1}^{\infty}B_{i} = \bigcap_{i=1}^{\infty} A_{i}$$
To prove $(1)$ and $(2)$ take,
$$ \lim_{ n \rightarrow \infty} (A_{n}) = \lim_{ n \rightarrow \infty} \bigg(\bigcap_{i=1}^{n}A_{i} \bigg) = \lim_{ n \rightarrow \infty} \bigg( \bigcap_{i=1}^{n}B_{i} \bigg) \rightarrow A. $$
As well as,
$$\lim_{ n \rightarrow \infty} (B_{n}) = \lim_{ n \rightarrow \infty} \bigg(\bigcap_{i=1}^{n}B_{i} \bigg) = \lim_{ n \rightarrow \infty} \bigg( \bigcap_{i=1}^{n}A_{i} \bigg) \rightarrow B .$$
QED.