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 $\mathcal A$.



Monotone class theorem for pre-Dynkin system ("finitely additive Dynkin system/λ system)
**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](https://en.wikipedia.org/wiki/Dynkin_system) 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, $m(\mathcal{D})\supset d(\mathcal{D})$.


