Deriving Dynkin’s theorem from the monotone class theorem

One of the first major results of any measure theory course is that the Lebesgue measure is the unique measure that assigns to each interval its length. A related result is the extension theorem¹

Every probability measure on a field $\mcF$ can be extended to a unique probability measure on the generated $\sigma$-field.

Here, a probability measure $P$ on a field means it has $\sigma$-additivity: if $A_1,A_2,\ldots\in\mcF$ and $\bigcup_n A_n\in\mcF$, then $P(\bigcup_n A_n) = \sum_n P(A_n)$. We can prove the preceding result as follows: the intervals of $[0, 1]$ generate a field of disjoint unions of intervals. We can define a natural measure on this interval that is finitely additive, which can be shown to be, in fact, countably additive, so it has a unique extension to the generated $\sigma$-field of Borel sets $\mcB$. This unique extension is the Lebesgue measure on $[0,1]$.

We can construct this extension by defining the outer measure, designating subsets of $\Omega$ as measurable based on Carathéodory’s criterion, and then showing that all sets in the generated $\sigma$-field are measurable.

Cool! What about uniqueness?

$\pi$ and $\lambda$-systems

Let $\Omega$ be a set. A $\pi$-system is a class of subsets of $\Omega$ closed under finite intersections. A $\lambda$-system is a class that contains $\Omega$, is closed under complement and under disjoint countable unions. These two definitions constitute the essence of a $\sigma$-field: a class is a $\sigma$-field if and only if it is both a $\pi$-system and a $\lambda$-system. Of special interest here is Dynkin’s $\pi\text{-}\lambda$ theorem:

If a $\pi$-system $\msP$ is a subset of a $\lambda$-system $\msL$, then $\sigma(\msP) \subset\msL$.

How does this help us? Suppose $P_1$ and $P_2$ are probability measures on $\sigma(\mcF)$ that agree on $\mcF$. Let $\msL$ be the maximal subset of $\sigma(\mcF)$ on which these two measures agree; $\msL$ can be shown to be a $\lambda$-system. Since the field $\mcF$ is certainly a $\pi$-system and $\mcF\subset\msL$, Dynkin’s theorem gives us $\sigma(\msF)\subset\msL$, so the two measures agree on the generated $\sigma$-field.

Monotone classes

A class $\msM$ of subsets of $\Omega$ is monotone if it is closed under monotone unions and intersections:

• $A_1, A_2, \ldots\in\msM$ and $A_n\uparrow A$ imply $A\in\msM;$
• $A_1, A_2, \ldots\in\msM$ and $A_n\downarrow A$ imply $A\in\msM$.

Similar to $\pi$-systems and $\lambda$-systems, the requirements of a field and a monotone class capture the essence of a $\sigma$-field: a class is a $\sigma$-field if and only if it is both a field and a monotone class. Note that the $\lambda$-systems are a subset of the monotone classes; here is a helpful Venn diagram:

Perhaps propitiously, we have an analogous version of Dynkin’s theorem called Halmos’s monotone class theorem:

If a field $\mcF$ is a subset of a monotone class $\msM$, then $\sigma(\mcF)\subset\msM$.

Can this be used to prove the uniqueness in the extension theorem? Yes, since all $\lambda$-systems are monotone. The class $\msL$ above is a monotone class containing a field $\mcF$, so it contains $\sigma(\mcF)$ too.

But how do we prove either of these theorems? Given their similarity, are they equivalent to each other?

They are indeed equivalent, and it is a fun exercise to prove this. First, we need some groundworks on the representation of a generated field.

Generated field

For a non-empty class $\msA$ of subsets of $\Omega$, the field generated by $\msA$, or $f(\msA)$, can be described explicitly: its members have the form $\bigcup_{i=1}^m \bigcap_{j=1}^{n_i} A_{ij}$, where $A_{ij}\in\msA$ or $A_{ij}^c\in\msA$ for each $i$ and $j$, and where the $m$ sets $\bigcap_{j=1}^{n_i}$ are disjoint.

Proof. If a set $A\subset\Omega$ satisfies $A\in\msA$ or $A^c\in\msA$, we call it $\msA$-compatible (just a made-up name). Let $F(\msA)$ be the class of sets of the given form, i.e., a finite union of finite intersections of $\msA$-compatible sets. We can see that $\msA$ is a subset of $F(\msA)$. Suppose $S = \bigcup_{i=1}^m\bigcap_{j=1}^{n_i} A_{ij}$ is a member of $F(\msA)$. Its complement is²

where $M=\{\bm{j}=(j_1,\ldots,j_m)\colon j_i\in [1, n_i]\}$. Note that $A_{ij_i}^c$ is the complement of some $A_{ij}$, so it is $\msA$-compatible. Hence, $F(\msA)$ contains $S^c$ and is closed under complementation.

Now, consider the intersection of two members of $F(\msA)$:

Each $C_i\cap D_k$ is a finite intersection of $\msA$-compatible sets, and if $(i,k)\ne (i',k')$, then $C_i\cap D_k$ is disjoint from $C_{i'}\cap D_{k'}$ (since at least one index must be different, say, $i\ne i'$, then $C_i$ and $C_{i'}$ are disjoint). Thus, the above set is also a member of $F(\msA)$, which means $F(\msA)$ is closed under finite intersection and is a field. Since it also contains $\msA$, we have $f(\msA)\subset F(\msA)$.

For any field $\msF$ containing $\msA$, every $\msA$-compatible set is a member of $\msF$, and so the finite union of finite intersections of $\msA$-compatible sets is also a member of $\msF$. In other words, $F(\msA)$ is a subset of any field containing $\msA$, so it is a subset of $f(\msA)$. Thus, $f(\msA)=F(\msA)$. $\blacksquare$

Equivalence

We now come to the main task: deriving Dynkin’s $\pi$-$\lambda$ theorem from the monotone class theorem. Suppose the $\lambda$-system $\msL$ contains the $\pi$-system $\msP$. We want to show that $\msL$ contains the $\sigma$-field generated by $\msP$, but the monotone class theorem only gives us that if $\msL$ already contains a field. Hence, we shall fill in this gap by proving that $\msL$ contains the field generated by $\msP$.

If the $\lambda$-system $\msL$ contains the $\pi$-system $\msP$, then $\msL$ contains $f(\msP)$.

Proof. By the previous section, $f(\msP)$ consists of sets of the form $\bigcup_{i=1}^m\bigcap_{j=1}^{n_i} A_{ij}$, where each $A_{ij}$ is $\msP$-compatible and the $m$ sets $\bigcap_{j=1}^{n_i}A_{ij}$ are disjoint. Since $\msL$ is closed under countable disjoint union, it suffices to show that $\bigcap_{i=1}^n A_i\in\msL$, where each $A_i$ is $\msP$-compatible.

Without loss of generality, assume that $A_1^c,\ldots, A_m^c\in\msP$ and $A_{m+1},\ldots,A_n\in\msP$. Let $B_i=A_i^c$ and $C=\bigcap_{i=m+1}^n A_i\in\msL$ (not necessarily in $\msP$: if $m=n$ then $C=\Omega$). We have

Note that each $C\cap B_i$ is in $\msP$, even when $C=\Omega$. In light of this last expression, we shall shift our goal to showing that finite union of $\msP$-sets is in $\msL$. For the case of two sets, if $D_1$ and $D_2$ are $\msP$-sets, we have $D_1\cup D_2 = D_1\cup (D_2\setminus (D_1\cap D_2))$, a disjoint union of sets in $\msL$. Suppose the union of any $m-1$ sets from $\msP$ lies in $\msL$. Given $D_1,\ldots,D_m\in\msP$, we have

By the induction hypothesis, $\bigcup_{i=2}^m D_1\cap D_i\in\msL$, so the above is a disjoint union of sets in $\msL$. By induction, the result holds for all finite $m$.

To recap, by showing that the finite union of $\msP$-sets is in $\msL$, we deduce the finite intersection of $\msP$-compatible sets is in $\msL$, it follows that $\msL$ contains the field $f(\msP)$. But $\msL$ is a monotone class as well, so by the monotone class theorem, $\msL$ contains $\sigma(f(\msP))=\sigma(\msP)$, which gives us Dynkin’s theorem.