# The Intersection of the Collection of Dreams and the Collection of Maths, or Dynkin π-λ Systems

I’ve posted before about how easily my sleep can be dominated by math stuff after a hard day or thirty of being cooped up in an office, grinding away at theorems and postulates and proofs with hardly a break in the mix.

Over the summer, the same thing happens after only a medium-hard day or three.

I woke up twice this morning, about two hours apart, and both times I was thinking about a random piece of mathematics not related to anything I’ve been actively studying recently. When I finally awoke a third time – this time, for good – I of course couldn’t remember it at all.

Then, finally, I sat in silence and forced my synapses to make connections they didn’t want to make and eventually, after a solid twenty minutes of mental strain, it all came flooding back in.

This is an exposition about so-called Dynkin (π-λ) Systems and the corresponding Dynkin π-λ Theorem. Feel free to stick around.

Until last semester, I didn’t even know either of those topics existed. If I’m honest, I didn’t even know that Dynkin was a surname at all. At some point during the semester, my professor assigned one of his usual homework assignments – lengthy and challenging as were all of the assignments he gave in that, the second class in the sequence of Measure and Integration Theory – and my level of stumped led (indirectly) to my introduction thereto. The question, from Chapter 20 of the 4th edition of Royden and Fitzpatrick’s Real Analysis:

Exercise 12.(i). For two measure spaces $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$, we have defined $\mathcal{A}\times\mathcal{B}$ to be the smallest σ-algebra that contains the measurable rectangles. Show that if both measures are σ-finite, then $\mu\times\nu$ is the only measure on $\mathcal{A}\times\mathcal{B}$ that assigns the value $\mu(A)\cdot\nu(B)$ to each measurable rectangle $A\times B$.

Immediately, I tried to prove this using brute force: Suppose there is another measure that assigns the same values, etc. etc., try to form a contradiction. From what I see now, using contradiction may not necessarily be that difficult: Because elements of the σ-algebra $\mathcal{A}\times\mathcal{B}$ consist of countable unions of finite intersections of measurable rectangles $A\times B$, one can, in theory, argue directly that the measure assigned to such measurable rectangles therefore determine the measure of everything else in the algebra.

At the time, though, I had no idea whatsoever how to tackle the logistics of this problem, and because my professor at the time was one of the few people who had ever encouraged us to seek out new ideas online, I took to Google to try to find something to help. From what I can recollect, Google led me to a series of similar solutions, one of which was on the Math Stack Exchange, and there was my introduction to the topic at hand.

First, a little background:

Throughout, let $\Omega$ be some kind of universal set with power set $\mathcal{P}(\Omega)=2^\Omega$. A subset $P\subset 2^\Omega$ is called a $\pi$-system of subsets of $\Omega$ provided that (i) $P\neq\varnothing$ and (ii) $A,B\in P\implies A\cap B\in P$. Also, a subset $\Sigma\subset 2^\Omega$ is called a $\sigma$-algebra provided $\Sigma$ is nonempty and closed with respect to countable unions and complements. Finally, a subset $D\subset 2^\Omega$ is called a Dynkin System (or a $D$-system, or a $\lambda$-system) of subsets of $\Omega$ provided (i) $\Omega\subset D$, (ii) $A\in D\implies A^c\in D$, and (iii) if $A_1,A_2,\ldots$ is a pairwise disjoint collection of elements of $D$, then $\cup_\alpha A_\alpha\in D$.

It’s obvious based on sheer intuition that all these notions must be somewhat interconnected. Here are some results that make this intuition a bit more concrete.

Proposition 1. If $X\subset 2^\Omega$ is both a $\pi$– and a $\lambda$-system, then $X$ is a $\sigma$-algebra.

Proof. This is immediate given the conditions satisfied by each of these objects.   $\square$

Proposition 2. Given a subset $X\subset 2^\Omega$, there exists a unique smallest $\lambda$-system $D(X)$ containing $X$.

Proof. Let $\mathcal{D}$ be the collection of all Dynkin systems containing $X$ and define $D(X)$ to be

$\displaystyle D(X)\overset{\text{def}}{=}\bigcap_{\Lambda\in\mathcal{D}}\Lambda$.

Because each such $\Lambda$ is a $\lambda$-system, each such $\Lambda$ satisfies the three properties in the definition above. However, because each of these three properties is a containment property, they therefore carry over to the intersection of all such $\Lambda$, thereby proving that $D(X)$ is a Dynkin system. As is always the case, the intersection is also the smallest such system, hence the result.  $\square$

Theorem (Dynkin’s $\pi$$\lambda$ Theorem). Let $P$ and $L$ be $\pi$-, respectively $\lambda$-, systems of subsets of $\Omega$ for which $P\subset L$. Then $\sigma(P)\subset L$ where $\sigma(P)$ is the $\sigma$-algebra generated by $P$, that is, the smallest $\sigma$-algebra containing $P$.

Proof (Sketch). Let $P\subset L\subset 2^\Omega$. By definition of $D(P)$, it follows that $D(P)\subset L$ because $D(P)$ is the unique smallest $\lambda$-system containing $P$. Moreover, because it’s easily verified that $D(P)$ is also a $\pi$-system, $D(P)$ is therefore a $\sigma$-algebra containing $P$. By definition, however, $\sigma(P)$ is the smallest $\sigma$-algebra containing $P$, and so it follows that

$P\subset \sigma(P)\subset D(P)\subset L\subset 2^\Omega$,

and so in particular, $\sigma(P)\subset L$. Hence, the result.   $\square$

Well okay then, that’s all well and good, but there’s a lot left here to be desired. In particular: Who cares? Why would I type a ridiculously-long entry about some problem on which I floundered and then follow said floundering up with a bunch of abstract nonsense that doesn’t seem to address anything at all? So far, all I’ve done is give a bunch set theory, apply a bunch of conditions to sets of sets of sets, and show that these conditions play off each other in what seems to be a pre-scripted manner.

Why would anybody care about that at all?

As it turns out, Dynkin systems (and more precisely, the $\pi$$\lambda$ theorem) is useful when it comes to determining uniqueness for $\sigma$-algebras, among which are the ones that arise naturally in the study of measure theory / real analysis. I’ll conclude this with an example that’s pretty standard and can be found on the internet almost everywhere.

Example 1. Let $\Omega=[0,1]$ and consider $(\Omega,\mathcal{B},m)$ where $\mathcal{B}$ is the $\sigma$-algebra of all Borel measurable subsets of $\Omega$ and where $m$ is standard Lebesgue measure. Suppose that $\mu$ is another measure on $[0,1]$ for which $\mu([a,b])=b-a$. The claim here is that the collection $D$ of sets $S$ for which $\mu[S]=m[s]$ is actually $D=\mathcal{B}$, whereby it follows that (among other things) Lebesgue measure is unique. Here’s how the proof of that goes:

First, note that $D$ is easily shown to be a Dynkin system. Next, let

$I=\{(a,b),[a,b),(a,b],[a,b]\,:\,0\leq a\leq b\leq 1\}$

be the collection of all intervals which are sub-intervals of $[0,1]$ (not necessarily proper). Note that $I$ is then contained in $D$, that $\sigma(I)=\mathcal{B}$ by definition, and that $I$ is closed under finite intersections, i.e. that $I$ is a $\pi$-system. Thus, the $\pi$-system $I$ is contained in the $\lambda$-system $D$ and so, by the above theorem, $\mathcal{B}=\sigma(I)\subset D$. But $D$ is clearly contained in $\mathcal{B}$ by definition of $m$, and so $\mathcal{B}=D$. Hence, $\mu=m$ on all of $\mathcal{B}$, whereby it follows that the Lebesgue measure $m$ is unique.  $\square$

Resources.
For more information on the items used here, you can check Wikipedia (here, here, or here), as well as the course notes from LSU and the presentation from Probability.net.