Why is this infinite pairwise disjoint union an element of the semi-algebra? Let
$$
P(\cup_{i=1}^n D_n ) = \sum_{i=1}^n P(D_n)\text{ for pairwise disjoint } D_1,D_2,\dots,D_n \in \mathcal{J} \text{ with } \cup_{i=1}^n D_n \in \mathcal{J} \tag{1}
$$
$\mathcal{J}$ is a semi-algebra. Let
$$
A_n = (\cup_{j=1}^\infty) D_j) \setminus (\cup _{i=1}^n D_i)
$$
is $A_n \in \mathcal{J}$ If so, why?
I want to say that, for $s>r$, 
$$
\cup_{j=1}^s D_j \setminus \cup _{i=1}^r D_i = (\cup _{i=1}^r d_i)^c \cap \cup_{j=1}^s D_j \in \mathcal{J}
$$
using (1). However, in the question I am asking $s=\infty$ so I don't think that (1) would apply?
Edit: The book I am looking at has an errata and, in more recent versions, the question requires that $A_n$ is a finite pairwise disjoint union of elements of $\mathcal{J}$, which is trivial because $D_i$ are defined as pairwise disjoint elements of $\mathcal{J}$.
I am leaving the question up because I am still interested in whether or not $A_n$ is an element of $\mathcal{J}$. I suspect not, but if that is the case perhaps someone can provide a proof. 
If the mods disagree with my leaving this question open feel free to close it.
 A: By definition $J \in \mathscr J \to \exists$ PWDJ $C_1, ..., C_m \in \mathscr J$ s.t. $J^C = \bigcup_{i=1}^{m} C_i$.
However, the converse is true as well.
(Btw, in either case pairwise disjointness is not necessary?)
Assume:


*

*$D_1, ..., D_n, D_{n+1}, ... \in \mathscr J$ (extends past $n$?) is PWDJ.

*Partition $\Omega = D^C \cup A_n \cup B_n$ where

*$$D := \bigcup_{i=1}^{\infty} D_i \in \mathscr J \ (also?)$$

*$$B_n := \bigcup_{i=1}^{n} D_i \in \mathscr J$$

*$$A_n := D \setminus B_n $$
Let us try to find PWDJ $C_1, ..., C_m$ s.t. $A_n^C = \bigcup_{i=1}^{m} C_i$:
Case 1: $D = \Omega$
$$A_n^C = B_n$$
Choose $m=n$ and $C_i = D_i$
Case 2: $D \subset \Omega$
$$A_n^C = B_n \cup D^C$$
Note that:


*

*$D^C \in \mathscr J \to A_n \in \mathscr J$

*$\exists$ PWDJ $E_1, ..., E_m \in \mathscr J$ s.t. $D^C = \bigcup_{i=1}^{e} E_i \to A_n \in \mathscr J$
Try to show hypothesis holds:


*

*Idk

*$D \in \mathscr J$
Choose $m = n + e$ and $C_i = D_i$ for $i = 1, ..., n$ and $C_i = E_{i-n}$ for $i = n+1, ..., n+e$
QED
