Monotone class theorem for pre-Dynkin system ("finitely additive Dynkin system/λ system)

Question

A pre-Dynkin system is a set system $\mathcal D \subset \wp(\Omega)$ which contains $\Omega$ and is closed under complements and finite disjoint unions. Is it true that the monotone class generated by $\mathcal D$ equals the Dynkin system generated by $\mathcal D$,

$$ m(\mathcal D) = d(\mathcal D)? $$ (Here $m(\mathcal D)$ denote the the smallest monotone class containing a collection of sets $\mathcal D \subset \wp(\Omega)$ and $d(\mathcal D)$ denotes the smallest Dynkin system containing $\mathcal D$. A Dynkin system, also called $\lambda$-system, contains $\Omega$ and is closed under complements and countable disjoint unions.)

Commentary: I seem to be able to prove that 1.) $m(\mathcal D)$ is closed under complements, and 2.) the smallest system closed under countable increasing unions that contains $\mathcal D$ is stable under disjoint unions.

Context: The question is motivated by the monotone class theorem for sets (https://en.wikipedia.org/wiki/Monotone_class_theorem). A monotone class is a collection of sets closed under taking unions of sequences of increasing sets and taking intersections of sequences of decreasing sets. The theorem says that if $\mathcal A$ is a field/algebra, then $m(\mathcal A) = \sigma(\mathcal A)$, i.e. the monotone class generated by $\mathcal A$ coincides with the $\sigma$-field generated by $\mathcal A$.

Answer 1

Part I. $m(\mathcal{D})\subset d(\mathcal{D})$

This is true regardless of the class $\mathcal{D}$ (whether pre-Dynkin or not). This is because a Dynkin class is necessarily a monotone class -- easier to check with the alternative equivalent definition of $\lambda$-system.

Part II. $m(\mathcal{D})\supset d(\mathcal{D})$

First, you can show that

$$\mathcal{M}_1\overset{\Delta}=\left\{B\in m(\mathcal{D})\,:\,\left(B^{c}\in m(\mathcal{D})\right) \wedge \left(B\cup A \in m(\mathcal{D}),\,\forall{A}\in \mathcal{D},\,A\cap B =\emptyset\right)\right\}$$

is so that $\mathcal{M}_1\supset \mathcal{D}$ and $\mathcal{M}_1$ is a monotone class (which is easy). This implies that $\mathcal{M}_1=m(\mathcal{D})$.

Further, you can show that (i.e., upgrade $\forall{A}\in \mathcal{D}$ to $\forall{A}\in m(\mathcal{D})$ in the definition of $\mathcal{M}_1$) $$\mathcal{M}_2\overset{\Delta}=\left\{B\in m(\mathcal{D})\,:\,\left(B^{c}\in m(\mathcal{D})\right) \wedge \left(B\cup A \in m(\mathcal{D}),\,\forall{A}\in m(\mathcal{D}\right),\,A\cap B =\emptyset)\right\}$$ is so that $\mathcal{M}_2\supset \mathcal{D}$ (this follows from $\mathcal{M}_1=m(\mathcal{D})$) and $\mathcal{M}_2$ is a monotone class (straightforward). In other words, $\mathcal{M}_2=m(\mathcal{D})$. Therefore, $m(\mathcal{D})$ is pre-Dynkin (since $\mathcal{M}_2$ is pre-Dynkin by construction). This further implies that $m(\mathcal{D})$ is a $\lambda$-system and as a result, $m(\mathcal{D})\supset d(\mathcal{D})$.

Answer 2

I put in some details for https://mathoverflow.net/users/138242/augusto-santos argument:

A Dynkin system is a monotone class, hence $m(\mathcal{D}) \subset d(\mathcal{D})$. It remains to show $\mathcal{M}$ is a Dynkin system, as then also $\mathcal{M} \supset d(\mathcal{D})$.

Apply the principle of good sets: Let $$\mathcal{M}_1 = \{B \in m(\mathcal{D})\colon B^c \in m(\mathcal{D}) \text{ and } ( A \in \mathcal{D}, B\cap A = \varnothing \Rightarrow B\cup A \in m(\mathcal{D}) )\}$$ be the set of "$\mathcal{D}$-good" sets. (1) $\mathcal{D} \subset \mathcal{M}_1$. (2) $\mathcal{M}_1$ is a monotone class: Let for a sequence $(B_n) \subset \mathcal{M}_1$, $B_n \uparrow B\subset \Omega$. Then $B \in m(\mathcal{D})$. By $\mathcal{M}_1 \ni B_n^c \downarrow B^c$ also $B^c \in m(\mathcal{D})$. Now choose $A \in \mathcal{D}$ with $B\cap A = \varnothing$. As $B_n$ is increasing, also $B_n\cap A = \varnothing$, and thus $\mathcal{M}_1 \ni (B_n \cup A) \uparrow (B\cup A)$ and $(B\cup A) \in \mathcal{M}$. We conclude and $B \in \mathcal{M}_1$. Also, for a decreasing sequence $(B_n) \subset \mathcal{M}_1$, $B_n \downarrow B\subset \Omega$, by $\mathcal{M}_1 \ni B_n^c \uparrow B^c \in \mathcal{M}_1$ as just shown and as $\mathcal{M}_1$ is closed under complements, $B \in \mathcal{M}_1$.

From (1) and (2) we conclude $m(\mathcal{D}) \subset \mathcal{M}_1$.

Let us now upgrade to the set of "$m(\mathcal{D})$-good" sets using the fact $m(\mathcal{D}) \subset \mathcal{M}_1$. Let $$\mathcal{M}_2 = \{B \in m(\mathcal{D})\colon B^c \in m(\mathcal{D})\text{ and } (A \in m(\mathcal{D}), B\cap A = \varnothing \Rightarrow B\cup A \in m(\mathcal{D}) )\}.$$ (3) Still $\mathcal{D} \subset \mathcal{M}_2$: Let $B \in \mathcal{D}$. $B, B^c \in m(\mathcal{D})$ and for any $A \in m(\mathcal{D}) \subset \mathcal{M}_1$, $A \cup B \in m(\mathcal{D})$.
(4) Also $\mathcal{M}_2$ is a monotone class, this is shown as in step (2).

From (3) and (4) we conclude $m(\mathcal{D}) \subset \mathcal{M}_2$. Thus $m(\mathcal{D})$ is a Dynkin system and $m(\mathcal{D}) \supset d(\mathcal{D})$.