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Moritz Schauer
Moritz Schauer
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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 d(\mathcal{D}) \subset \mathcal{M}_1$$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}_1$$m(\mathcal{D}) \subset \mathcal{M}_2$. Thus $m(\mathcal{D})$ is a Dynkin system and $m(\mathcal{D}) \supset d(\mathcal{D})$.

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 $A \in d(\mathcal{D}) \subset \mathcal{M}_1$, $A \cup B \in m(\mathcal{D})$.
(4) Also $\mathcal{M}_2$ is a monotone class, this is shown as in step (2).

From (3) and (4) we conclude $m(\mathcal{D}) \subset \mathcal{M}_1$. Thus $m(\mathcal{D})$ is a Dynkin system and $m(\mathcal{D}) \supset d(\mathcal{D})$.

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

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Moritz Schauer
Moritz Schauer
  • 430
  • 2
  • 9

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 $A \in d(\mathcal{D}) \subset \mathcal{M}_1$, $A \cup B \in m(\mathcal{D})$.
(4) Also $\mathcal{M}_2$ is a monotone class, this is shown as in step (2).

From (3) and (4) we conclude $m(\mathcal{D}) \subset \mathcal{M}_1$. Thus $m(\mathcal{D})$ is a Dynkin system and $m(\mathcal{D}) \supset d(\mathcal{D})$.