Deriving Dynkin’s theorem from the monotone class theorem
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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 theorem^{1}
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. 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 nonempty 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 madeup 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 is^{2}
$S^c = \left(\bigcup_{i=1}^m\bigcap_{j=1}^{n_i}A_{ij}\right)^c = \bigcap_{i=1}^m\bigcup_{j=1}^{n_i}A_{ij}^c = \bigcup_{\bm{j}\in M}\bigcap_{i=1}^m A_{i\bm{j}_i}^c,$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 $F(\msA)$:
$\left(\bigcup_{i=1}^m\bigcap_{j=1}^{n_i} A_{ij}\right) \cap \left(\bigcup_{k=1}^p\bigcap_{l=1}^{q_k} B_{kl}\right) = \bigcup_{\substack{1\le i\le m\\ 1\le k\le p}} \left(\bigcap_{j=1}^{n_i}A_{ij}\cap \bigcap_{l=1}^{q_k}B_{kl}\right) = \bigcup_{\substack{1\le i\le m\\ 1\le k\le p}} C_i\cap D_k.$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
$\bigcap_{i=1}^n A_i = C\cap\bigcap_{i=1}^m A_i = C\cap \left(\bigcup_{i=1}^m A_i^c\right)^c = C\setminus \left(C\cap \bigcup_{i=1}^m B_i \right) = C\setminus\bigcup_{i=1}^m C\cap B_i.$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 $m1$ sets from $\msP$ lies in $\msL$. Given $D_1,\ldots,D_m\in\msP$, we have
$\bigcup_{i=1}^m D_i = \left(D_1\setminus \bigcup_{i=2}^m D_i\right)\cup \bigcup_{i=2}^m D_i = \left(D_1\setminus \bigcup_{i=2}^m D_1\cap D_i\right)\cup \bigcup_{i=2}^m D_i.$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.
Footnotes

This is theorem 3.1 in “Probability and Measure” by Billingsley and is a version of the Carathéodory’s extension theorem: a premeasure on a given ring $\mcR$ of subsets of $\Omega$ can be extended to a measure on the $\sigma$ring generated by $\mcR$, and this extension is unique if the measure is $\sigma$finite. The stated result follows from this general version, as a probability measure is both a premeasure and a $\sigma$finite measure.

Intuitively, the tuples in $M$ gives us the indices for the terms of $\prod_{i=1}^m\sum_{j=1}^{n_i} a_{ij} = (a_{11}+\cdots+a_{1n_1})\cdots (a_{m1}+\cdots+a_{mn_m}) = \sum_{\textit{\textbf{j}}\in M}\prod_{i=1}^m a_{i\textit{\textbf{j}}_i}$. Morphing sums into unions and products into intersections, we get the stated identity (a rigorous proof can be given by induction on $m$).