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?
Best Answer
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 $C_n\nearrow \cup_{i=1}^{\infty} A_i$: $$ P[\cup_{i=1}^n A_i] =P[C_n]\nearrow P[\cup_{i=1}^{\infty} A_i]. $$