Continuity of Probability Measure and monotonicity

Question

In every textbook or online paper I read, the proof of continuity of probability measure starts by assuming a monotone sequence of sets $(A_n)$. Or it assumes the $\liminf A_n = \limsup A_n$

But what about the following proof. It seems we don't need this property (monotonic).

If $\{A_i, i ≥ 1\}$ are events (not necessarily disjoint nor monotonic), then

$$P [\cup_{i=1}^∞ A_i] = \lim_{m\to\infty} P [\cup_{i=1}^m A_i]$$

This result is known as continuity of probability measures.

Proof:- Define a new family of sets $$B_1 = A_1, \ B_2 = A_2 - A_1,\ ..., B_n = A_n-\bigcup_{i=1}^{n-1} A_i,.... $$ Then, the following claims are placed:

Claim 1:- $B_i ∩ B_j = ∅, ∀i \neq j$.

Claim 2:- $\bigcup_{i=1}^∞ A_i = \bigcup_{i=1}^∞ B_i$

Since $\{B_i, i ≥ 1\}$ is a disjoint sequence of events, and using the above claims, we get

$$P (\bigcup_{i=1}^∞ A_i) = P(\bigcup_{i=1}^∞ B_i) = \sum_{i=1}^∞ P(B_i)$$

Therefore,

$$P (\bigcup_{i=1}^∞ A_i) = \sum_{i=1}^∞ P(B_i)$$ (a)

$$= \lim_{m\to\infty} \sum_{i=1}^m P(B_i)$$ (b)

$$= \lim_{m\to\infty} P(\bigcup_{i=1}^m B_i)$$ (c)

$$= \lim_{m\to\infty} P(\bigcup_{i=1}^m A_i)$$

Here, (a) follows from the definition of an infinite series, (b) follows from Claim 1 in conjunction with Countable Additivity axiom of probability measure and (c) follows from the intermediate result required to prove Claim 2. Hence proved.

So my original $A_n$'s were NOT a monotonic sequence of sets, so why do we require them to be?

Answer 1

Yes, you can first prove $P[\cup_{i=1}^nA_i]\rightarrow P[\cup_{i=1}^{\infty} A_i]$ and, as a corollary, we get that if $C_n\nearrow C$ then: $$ P[C_n] = P[\cup_{i=1}^n C_i] \rightarrow P[\cup_{i=1}^{\infty} C_i] = P[C]. $$

Or, you can first prove the fact $C_n\nearrow C \implies P[C_n]\nearrow P[C]$ and, as a corollary, we get by defining $C_n = \cup_{i=1}^n A_i$ (as in my comment) and noting that $C_n\nearrow \cup_{i=1}^{\infty} A_i$: $$ P[\cup_{i=1}^n A_i] =P[C_n]\nearrow P[\cup_{i=1}^{\infty} A_i]. $$

Answer 2

I think it is not precise.

Firstly, the theorem needs to state that $\bigcup_{n=1}^{\infty}{A_n}$ exists. If $\bigcup_{n=1}^{\infty}{A_n}$ doesn't exist, then we cannot figure out what $\mathbb{P} \left( \bigcup_{n=1}^{\infty}{A_n} \right)$ really is.

Answer 3

I think the following is a complete proof of this theorem.

Divide the proof into 3 steps.

Step 1: If $\left\{ A_n \right\} _{n=1}^{\infty}$ is increasing, we have $\lim_{n\rightarrow \infty} \mathbb{P} \left( A_n \right) =\mathbb{P} \left( A \right) $;

Step 2 ：If $\left\{ A_n \right\} _{n=1}^{\infty}$ is decreasing, we have $\lim_{n\rightarrow \infty} \mathbb{P} \left( A_n \right) =\mathbb{P} \left( A \right) $;

Step 3: Since $\bigcap_{k=n}^{\infty}{A_k}\subset A_n\subset \bigcup_{k=n}^{\infty}{A_k}$ Then $$\mathbb{P} \left( \bigcap_{k=n}^{\infty}{A_k} \right) \leqslant \mathbb{P} \left( A_n \right) \leqslant \mathbb{P} \left( \bigcup_{k=n}^{\infty}{A_k} \right) $$ And $\bigcap_{k=n}^{\infty}{A_k}$ is increasing, $\bigcup_{k=n}^{\infty}{A_k}$ is decreasing, so \begin{align*} \mathbb{P} \left( \mathop {\lim\mathrm{inf}} \limits_{n\rightarrow \infty}A_n \right) &=\mathbb{P} \left( \bigcup_{n=1}^{\infty}{\bigcap_{k=n}^{\infty}{A_k}} \right) =\lim_{n\rightarrow \infty} \mathbb{P} \left( \bigcap_{k=n}^{\infty}{A_k} \right) \leqslant \lim_{n\rightarrow \infty} \mathbb{P} \left( A_n \right) \\&\leqslant \lim_{n\rightarrow \infty} \mathbb{P} \left( \bigcup_{k=n}^{\infty}{A_k} \right) =\mathbb{P} \left( \lim_{n\rightarrow \infty} \bigcup_{k=n}^{\infty}{A_k} \right) =\mathbb{P} \left( \bigcap_{n=1}^{\infty}{\bigcup_{k=n}^{\infty}{A_k}} \right) =\mathbb{P} \left( \mathop {\lim\mathrm{sup}} \limits_{n\rightarrow \infty}A_n \right) \end{align*} Note that $\lim_{n\rightarrow \infty} A_n=A$ exists, and thus, $\mathop {\lim\mathrm{inf}} \limits_{n\rightarrow \infty}A_n=\mathop {\lim\mathrm{sup}} \limits_{n\rightarrow \infty}A_n=A$, then $$\mathbb{P} \left( A \right) \leqslant \lim_{n\rightarrow \infty} \mathbb{P} \left( A_n \right) \leqslant \mathbb{P} \left( A \right) $$ which leads to $\lim_{n\rightarrow \infty} \mathbb{P} \left( A_n \right) =\mathbb{P} \left( A \right) $.