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I found different statements of the Monotone Class Theorem. On probability Essentials (Jean Jacod and Philip Protter) the Monotone Class Theorem (Theorem 6.2, page 36) is stated as follows:

Let $\mathcal{C}$ be a class of subsets of $\Omega$ under finite intersections and containing $\Omega$. Let $\mathcal{B}$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $\mathcal{B} = \sigma ( \mathcal{C})$.

While on Wikipedia (https://en.wikipedia.org/wiki/Monotone_class_theorem) the theorem is:

Let $G$ be an algebra of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-algebra generated by $G$, i.e. $\sigma(G) = M(G)$.

Where a monotone class in a set $R$ is a collection $M$ of subsets of $R$ which contains $R$ and is closed under countable monotone unions and intersections.

It looks like the second theorem should be a special case of the first. Does the first prove the second? Is it possible to prove the first from the second? Is there a decent literature on those two theorems?

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Both results are actually equivalent. You can prove one from the other.

Regarding the first result:

Let $\mathcal{C}$ be a class of subsets of $\Omega$ under finite intersections and containing $\Omega$. Let $\mathcal{B}$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $\mathcal{B} = \sigma ( \mathcal{C})$.

Some books call it "Monotone Class Theorem", although this is not the most usual naming.

A class having $\Omega$, closed under increasing limits and by difference is called a "Dynkin $\lambda$ system". A non-empty class closed under finite intersections is called a "Dynkin $\pi$ system".

The result above can be divided in two results

1.a. A $\lambda$ system which is also a $\pi$ system is a $\sigma$-algebra. 1.b. Given a $\pi$ system, the smallest $\lambda$ system containing it is also a $\pi$ system.

Some books call result 1 (or result 1.b.) "Dynkin $\pi$-\lambda$ Theorem.

Some quick references is https://en.wikipedia.org/wiki/Dynkin_system

The second result

Let $G$ be an algebra of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-algebra generated by $G$, i.e. $\sigma(G) = M(G)$.

Where a monotone class in a set $R$ is a collection $M$ of subsets of $R$ which contains $R$ and is closed under countable monotone unions and intersections.

is usually called "Monotone Class Lemma" (or theorem) you can find it in books like Folland's Real Analysis or Halmos' Measure Theory. In fact, Halmos presents a version of this result for $\sigma$-rings.

Let $G$ be ring of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-ring generated by $G$.

Let us prove that the results are equivalent

Result 1: Let $\mathcal{C}$ be a class of subsets of $\Omega$ under finite intersections and containing $\Omega$. Let $L(\mathcal{C})$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $L(\mathcal{C}) = \sigma ( \mathcal{C})$.


Result 2: Let $G$ be an algebra of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-algebra generated by $G$, i.e. $\sigma(G) = M(G)$.

Where a monotone class in a set $R$ is a collection $M$ of subsets of $R$ which contains $R$ and is closed under countable monotone unions and intersections.

Proof:

(2 $\Rightarrow$ 1). Note that any class containing $\mathcal{C}$ which is closed under increasing limits and by difference is close by complement because $\Omega \in \mathcal{C}$, and so it is also closed by decreasing limits. So it is closed under countable monotone unions and intersections. It means: any class containing $\mathcal{C}$ which is closed under increasing limits and by difference is monotone class.

Note also that any class containing $\mathcal{C}$ which is closed under increasing limits and by difference contains $A(\mathcal{C})$ the algebra generated by $\mathcal{C}$.

Then using Result 2 we have $$ \sigma(\mathcal{C}) = \sigma(A(\mathcal{C})) = M(A(\mathcal{C})) \subseteq L(A(\mathcal{C}))=L(\mathcal{C}) $$ Since $\sigma(\mathcal{C})$ is a class containing $\mathcal{C}$ which is closed under increasing limits and by difference, we have $L(\mathcal{C}) \subseteq \sigma(\mathcal{C})$, so $L(\mathcal{C}) = \sigma(\mathcal{C})$.

(1 $\Rightarrow$ 2). First let us prove that $M(G)$ is a class containing $G$ which is closed under increasing limits and by difference. Since $M(G)$ is monotone, we have that $M(G)$ is closed under increasing limits.

Now, for each $E\in M(G)$, define

$$M_E=\{ F \in M(G) : E\setminus F , F \setminus E \in M(G) \}$$

Since $M(G)$ is a monotone class, $M_E$ is a monotone class. Moreover, if $E\in G$ then for all $F \in G$, $F\in M_E$, because $G$ is an algebra. So, if $E\in G$, $G \subset M_E$. So, if $E\in G$, $M(G) \subset M_E$. It means that for all $E\in G$, and all $F \in M(G)$, $F \in M_E$. So, for all $E\in G$, and all $F \in M(G)$, $E \in M_F$. So, for all $F \in M(G)$, $G \subset M_F$, but since $M_F$ is a monotone class, we have, for all $F \in M(G)$, $M(G)\subset M_F$. But that means that $M(G)$ is closed by differences.

So we proved that $M(G)$ is a class containing $G$ which is closed under increasing limits and by difference.

So by Result 1, $$\sigma(G)=L(G) \subseteq M(G)$$ Since $\sigma(G)$ is a monotone class, we have $$ M(G) \subseteq \sigma(G)$$ So we have $$\sigma(G)= M(G)$$

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  • $\begingroup$ Wait... here: math.stackexchange.com/questions/1193970/… They say that they are not the same thing. Could you give me some hint or reference about how to prove one from the other? In particular I don't see the $\sigma$-algebra needed in proving the first and I don't see how the closure under difference emerges when proving the second. $\endgroup$ – ThePunisher Jun 28 '16 at 7:36
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    $\begingroup$ In the question math.stackexchange.com/questions/1193970/… what the OP calls Monotone Class Theorem is NOT what is, in general, called Monotone Class theorem. Brent Kerby highlighted it in his answer (the accepted answer). In fact, in that questions, the OP is just comparing two formulations of the Denkin's $\pi$-$\lambda$ theorem. Yes, it is yet another level of complexity: some books present Denkin's $\pi$-$\lambda$ theorem in slightly different ways and some of them call "Monotone Class Theorem" you result 1, not result 2. $\endgroup$ – Ramiro Jun 28 '16 at 14:16
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    $\begingroup$ @lanzariel, see my comment above. I have also included more details in my answer and I have also included in my answer a proof that both results are equivalent. $\endgroup$ – Ramiro Jun 28 '16 at 19:26
  • $\begingroup$ Thank you very much. Now I can clearly see the situation! Well done! $\endgroup$ – ThePunisher Jun 29 '16 at 13:51

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