Why is every d-system a monotone class? The first part to prove that a d-system $D$ is also a monotone class is to take $C_1, C_2, ... \in D, C_1 \subset C_2 ...$ and show that the countable union of these sets is in $D$. I'm struggling to show this since this is similar to one property of a d-system but only for pairwise disjoint sets. Now I need to show it for all sets that form a increasing set sequence?
 A: The definition of Dynkin system (E. Dynkin himself used the term $\lambda$-systems in Theory of Markov Processes, Pergamon Press, Oxford-London-New York-Paris, 1960,  and he attributed the notion to Sierpinski) is:

A collection of subsets $\mathcal{D}$ of a set $X$ is a $\lambda$-system if

*

*$X\in \mathcal{D}$.

*For any $A,B\in\mathcal{D}$, if $A\subset B$, then $B\setminus A\in \mathcal{D}$

*For any monotone increasing sequence of sets $B_n\in\mathcal{D}$, $\bigcup_nB_n\in\mathcal{D}$.


This definition implies in particular that



*$X\in\mathcal{D}$

*$A\in\mathcal{D}$ then $X\setminus A\in\mathcal{D}$

*For any sequence of pairwise disjoint sets $A_n\in\mathcal{D}$, $\bigcup_nA_n\in\mathcal{D}$.


Conditions (4-6) seem to be the starting point for the OP. Here is a short proof that the OP's notion of a $d$-system is equivalent to that of that of Dynkin's:
(1-3) imply (4-6):
First notice that $\emptyset\in\mathcal{D}$. Now, If $A, B\in\mathcal{D}$ and $A\cap B=\emptyset$, then $A\subset (X\setminus B)$; hence $(X\setminus B)\setminus A=(X\setminus B)\cap(X\setminus A)\in\mathcal{D}$. Taking complements yields $A\cup B\in\mathcal{D}$. Thus shows that $\mathcal{D}$ (under 1-3) is closed under finite union of pairwise disjoint sets. Then, $B_n=\bigcup^n_{j=1}A_j\in\mathcal{D}$ is monotone increasing and so, $\bigcup_nA_n=\bigcup_nB_n\in\mathcal{D}$.
(4-6) imply (1-3):
$A,B\in \mathcal{D}$ and $A\subset B$, then $(X\setminus B)$ and $A$ are in $\mathcal{D}$ and are disjoint; hence $(X\setminus B)\cup A\in\mathcal{D}$ and by taking complements, $B\setminus A\in\mathcal{D}$.
If $A_n\nearrow A$ and $A_n\in\mathcal{D}$, then $B_1=A_1$ and $B_n=A_n\setminus A_{n-1}$, $n\geq2$, are in $\mathcal{D}$ and they are pairwise disjoint; hence $A=\bigcup_nA_n=\bigcup_nB_n\in\mathcal{D}$.

Solution to OP's problem:
Recall that a collection $\mathcal{M}$ os sets is a monotone class if for any monotone sequence (decreasing or decreasing) os sets $A_n\in\mathcal{M}$, $\bigcup_nA_n$ and $\bigcap_nA_n$ belong to $\mathcal{M}$.
Under the original definition of Dynkin class (axioms 1-3) it is easy to check that $\mathcal{D}$ is a monotone class. It is enough to show that $\mathcal{D}$ is closed under countable intersections of decreasing sets. If $A_n\searrow A$ and $A_n\in\mathcal{D}$, then  $X\setminus A_n\nearrow X\setminus A$. Since $X\setminus A_n\in\mathcal{D}$,  $X\setminus A\in\mathcal{D}$ and so, $A\in\mathcal{D}$.
