The Unapologetic Mathematician

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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