Continuity of Probability Measure and monotonicity 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?

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) $.
