Showing that If $A_{n} \rightarrow A$ than $\lim_{n \rightarrow \infty} \big( \mathcal{P}(A_{n}) \big) \rightarrow \mathcal{P}(A)$?

Question

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.

Answer 1

it's easy to see that $$\bigcap_{i\in I} A_{i} = \emptyset.$$

This is false. Since $A_i$ is increasing, $\bigcap_{i=1}^\infty A_i = A_1$. Unless $A_1 = \emptyset$, your statement is not true. Fortunately, it was also pointless, as you do not use it anywhere.

he also claims that,

(2) $$B_{n}= \bigcup_{i=1}^{\infty}B_{i} = \bigcup_{i=1}^{\infty} A_{i}$$

This is not equal to $B_n$ for any $n$. It is also not equal to "$B$" (as you use later), as you have never defined a set $B$. Nor do you have any need to.

To prove (1) and (2)

take, $$\lim_{ n \to \infty} (A_{n}) = \lim_{ n \to\infty} \bigg(\bigcup_{i=1}^{n}A_{i} \bigg) = \lim_{ n \to\infty} \bigg( \bigcup_{i=1}^{n}B_{i} \bigg) \to A.$$

You are also massively mixing up notations here. Note:

• $\{A_n\}_{n=1}^{\infty}$ is a sequence (of sets in this case).
• If the sequence converges, $\lim_{n\to\infty} A_n$ is the value that the sequence converges to. It is NOT the sequence. It is a single value. Therefore, saying $\lim_{n\to\infty} A_n \to A$ makes as much sense as saying $2 \to 2$. Sequences converge to something. Single values are just that one value. The following two statements are equivalent: $$\lim_{n\to\infty} A_n = A$$ and $$A_n \to A$$ But $\lim_{n\to\infty} A_n \to A$ is nonsensical.
• Despite the presence of $\infty$ in the notation, $$\bigcup_{i=1}^\infty A_i := \{x \mid \exists i, x \in A_i\}$$ is not defined as a limit. Part of what you need to prove is that $\lim_{n\to\infty} \bigcup_{i=1}^n A_i = \bigcup_{i=1}^\infty A_i$
• You are making use of (1) here in an attempt to prove (2), but you haven't offered anything at all towards a proof of (1).

As well as, $$\lim_{ n \rightarrow \infty} (B_{n}) = \lim_{ n\to \infty} \bigg(\bigcup_{i=1}^{n}B_{i} \bigg) = \lim_{ n \to \infty} \bigg( \bigcup_{i=1}^{n}A_{i} \bigg) \to B.$$

$B_n \ne \bigcup_{i=1}^n B_i$ and I've already mentioned that there is no set "$B$". This adds nothing to your proof.

Proof of the monotone decreasing case

Suppose that $A_n$ is monotone decreasing so that $A_1\subset A_2\subset$

$A_1\subset A_2\subset\ ...$ is the definition of monotone increasing. Decreasing would be $A_1\supset A_2\supset\ ...$

The rest of this is just a copy-paste of the first part with unions converted to intersections. It makes no sense. Since $A_1$ is the superset of all later sets, all of your $B_i$ in this case are empty, except $B_1$.

Finally, even if your proofs were correct, you have only examined the cases where $A_n$ is monotone. What if $A_n$ is neither increasing nor decreasing?