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===Statement===
Let G be an [[field of sets|algebra of sets]] and define M(G) to be the smallest monotone class containing G. Then M(G) is precisely the [[Sigma-algebra|σ-algebra]] generated by G, i.e. σ(G) = M(G)
===Proof===
The following was taken from Probability Essentials, by Jean Jacod and Philip Protter.<ref name="Jacod">{{cite book|last=Jacod|first=Jean|author2=Protter, Phillip |year=2004|title=Probability Essentials|publisher=Springer|page=36|isbn=978-3-540-438717}}</ref> The idea is as follows: we know that the sigma-algebra generated by an algebra of sets G contains the smallest monotone class generated by G. So, we seek to show that the monotone class generated by G is in fact a sigma-algebra, which would then show the two are equal.
To do this, we first construct monotone classes that correspond to elements of G, and show that each equals the M(G), the monotone class generated by G. Using this, we show that the monotone classes corresponding to the other elements of M(G) are also equal to M(G). Finally, we show this result implies M(G) is indeed a sigma-algebra.
Let <math>\mathcal{B} = M(G)</math>, i.e. <math>\mathcal{B}</math> is the smallest monotone class containing <math>G</math>. For each set <math>B</math>, denote <math>\mathcal{B}_B</math> to be the collection of sets <math>A \in \mathcal{B}</math> such that <math>A \cap B \in \mathcal{B}</math>. It is plain to see that <math>\mathcal{B}_B</math> is closed under increasing limits and differences.
Consider <math>B \in G</math>. For each <math>C \in G</math>, <math> B \cap C \in G \subset \mathcal{B}</math>, hence <math>C \in \mathcal{B}_B </math> so <math>G \subset \mathcal{B}_B </math>. This yields <math>\mathcal{B}_B = \mathcal{B}</math> when <math> B \in G</math>, since <math>\mathcal{B}_B</math> is a monotone class containing <math>G</math>, <math>\mathcal{B}_B \subset \mathcal{B}</math> and <math>\mathcal{B}</math> is the smallest monotone class containing <math>G</math>
Now, more generally, suppose <math>B \in \mathcal{B}</math>. For each <math>C \in G</math>, we have <math> B \in \mathcal{B}_C</math> and by the last result, <math>B \cap C \in \mathcal{B}</math>. Hence, <math>C \in \mathcal{B}_B</math> so <math>G \subset \mathcal{B}_B</math>, and so <math>\mathcal{B}_B = \mathcal{B}</math> for all <math>B \in \mathcal{B}</math> by the argument in the paragraph directly above.
Since <math>\mathcal{B}_B = \mathcal{B}</math> for all <math>B \in \mathcal{B}</math>, it must be that <math>\mathcal{B}</math> is closed under finite intersections. Furthermore, <math>\mathcal{B}</math> is closed by differences, so it is also closed under complements. Since <math>\mathcal{B}</math> is closed under increasing limits as well, it is a sigma-algebra. Since every sigma-algebra is a monotone class, <math>\mathcal{B} = \sigma\,(G)</math>, i.e. <math>\mathcal{B}</math> is the smallest sigma-algebra containing G
==Monotone class theorem for functions==
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