In measure theory, the probability measure has the property of sigma additivity. From that the text derives a series of theorems, the last of which is confusing to me.
Sigma continuity is stated as:
For Arbitrary events $A,B,A_1,A_2,... \epsilon\mathscr{F} $, If either $A_n \uparrow A $ or $A_n\downarrow A$ (ie,. the $A_n$ are either increasing with union A, or decreasing with intersection A), then as $n \longrightarrow \infty$ ,$P(A_n) \longrightarrow P(A) $
I don't understand what "increasing with union A" means.
Is It increasing with the A without a subscript in $A,B,A_1,A_2,... \epsilon\mathscr{F} $?
is $A_n$ the $n^{th}$ event of $A,B,A_1,A_2,... \epsilon\mathscr{F} $?
is $A$ defined as the union of all $A_1,A_2,..$?
[![Consequences of P following sigma additivity property.][1]][1]
[![Proof of result][2]][2]
If A is the union of all the $A_1,A_2,...$ (Im not sure if this is right), then in derivation of (e), I have trouble seeing why $P(A) = \sum_{i\geq 1} P(A_i$ \ $A_{i-1})$ ?
Suppose $A_1$ is the bottom right circle, $A_2$ is the bottom left circle and $A_3$ is the top circle and the Letters represent the P(Elements where the letters are). Then P(A) = X+Y+Z+T+M+N+L. Yet according to the formula, $P(A) = \sum_{i\geq 1} P(A_i$ \ $A_{i-1})$...
P(A)=P($A_1$)+P($A_2$ \ $A_1$) + P($A_3$ \ $A_2$)
=(N+Z+T+L) + (M+Y) + (X+T) $\neq$ X+Y+Z+T+M+N+L
We see that T has been double counted.
Sequence of events $A_1,A_2,A_3,\dots$ is an increasing sequence with $A_n\uparrow A$ if $A_n\subseteq A_{n+1}$ for $n=1,2,3,\dots$ and $A=\bigcup_{n=1}^{\infty} A_n$.
So in your example we cannot speak of an increasing sequence.
If $A_1,A_2,A_3,\dots$ is an increasing sequence with $A_n\uparrow A$ then the sets $B_1=A_1$ and $B_{n+1}=A_{n+1}\setminus A_n$ for $n=1,2,3,\dots$ are disjoint with $A_n=B_1\cup\cdots\cup B_n$ and $A=\bigcup_{n=1}^{\infty}B_n$.
Then $P(A)=\sum_{n=1}^{\infty}P(B_n)$ by $\sigma$-additivity so that: $$P(A)=\lim_{n\to\infty}\sum_{k=1}^nP(B_k)=\lim_{n\to\infty}P(A_n)$$
This with $P(A_1)\leq P(A_2)\leq\cdots$ so can also write $P(A_n)\uparrow P(A)$.