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.