The Unapologetic Mathematician

Mathematics for the interested outsider

Monotone Classes

Now we want to move from algebras of sets closer to \sigma-algebras by defining a new structure: a “monotone class”. This is a collection \mathcal{M}\subseteq P(Z) so that if we have an increasing sequence of sets

\displaystyle E_1\subseteq E_2\subseteq E_3\subseteq\dots

with E_i\in\mathcal{M}, then the “limit” of this sequence is in \mathcal{M}. Similarly, if we have a decreasing sequence of sets

\displaystyle E_1\supseteq E_2\supseteq E_3\supseteq\dots

with E_i\in\mathcal{M}, then the “limit” of this sequence is in \mathcal{M}. In each case, we can construe this limit more simply. In the case of an increasing sequence, each term E_i contains all the terms before it, and is thus the same as the union of the whole sequence up to that point. Thus we can define

\displaystyle\lim\limits_{n\to\infty}E_n=\lim\limits_{n\to\infty}\bigcup\limits_{i=1}^nE_i=\bigcup\limits_{i=1}^\infty E_i

Similarly, for a decreasing sequence we can define

\displaystyle\lim\limits_{n\to\infty}E_n=\lim\limits_{n\to\infty}\bigcap\limits_{i=1}^nE_i=\bigcap\limits_{i=1}^\infty E_i

Now, if \mathcal{A}\subseteq P(X) is an algebra of subsets of X, and \mathcal{M} is a monotone class containing \mathcal{A}, then \mathcal{M} also contains \mathcal{S}(\mathcal{A}), the smallest \sigma-algebra containing \mathcal{A}.

As an aside, what do we mean by “smallest”? Well, it’s not hard to see that the intersection of a collection of \sigma algebras is itself a \sigma-algebra, just like we did way back when we showed that we had lower bounds in the lattice of ideals. So take all the \sigma-algebras containing \mathcal{A} — there is at least one, because P(X) itself is one — and define \mathcal{S}(\mathcal{A}) to be the intersection of all of them. This is a \sigma-algebra contained in all the others, thus “smallest”.

Similarly, there is a smallest monotone class containing \mathcal{A}; we will assume that \mathcal{M} is this one without loss of generality, since if it contains \mathcal{S}(\mathcal{A}) then all the others do as well. We will show that this \mathcal{M} is actually a \sigma-algebra, and then it must contain \mathcal{S}(\mathcal{A})!

Given A\in\mathcal{A}, define \mathcal{M}_A to be the collection of B\in\mathcal{M} so that A\cup B\in\mathcal{M}. This is itself a monotone class, and it contains \mathcal{A}, and so it must contain \mathcal{M}. But it’s defined as being contained in \mathcal{M}, and so we must have \mathcal{M}_A=\mathcal{M}. Thus A\cup B\in\mathcal{M} as long as A\in\mathcal{A} and B\in\mathcal{M}.

Now take a B\in\mathcal{M} and define \mathcal{M}_B to be the collection of D\in\mathcal{M} so that D\cup B\in\mathcal{M}. What we just showed is that \mathcal{A}\subseteq\mathcal{M}_B, and \mathcal{M}_B is a monotone class. And so \mathcal{M}_B-\mathcal{M}, and D\cup B\in\mathcal{M} whenever both D and B are in \mathcal{M}.

We can repeat a similar argument to the previous two paragraphs to show that D\setminus B\in\mathcal{M} for D,B\in\mathcal{M}. Start by defining \mathcal{M}_A as the collection of B\in\mathcal{M} so that both B\setminus A and A\setminus B are in \mathcal{M}.

So \mathcal{M} is both a monotone class and an algebra. We need to show that it’s a \sigma-algebra by showing that it’s closed under countable unions. But if E_i\in\mathcal{M} then F_n=\cup_{i=1}^nE_i is a monotone increasing sequence of set in \mathcal{M}, since they’re all finite unions. The countable union is the limit of this sequence, which \mathcal{M} contains by virtue of being a monotone class.

And so \mathcal{M} is a \sigma-algebra containing \mathcal{A}, and so it contains \mathcal{S}(\mathcal{A}).

March 18, 2010 - Posted by | Analysis, Measure Theory


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