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 and , we have defined to be the smallest σ-algebra that contains the measurable rectangles. Show that if both measures are σ-finite, then is the only measure on that assigns the value to each measurable rectangle .

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 consist of countable unions of finite intersections of measurable rectangles , 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 be some kind of *universal set* with power set . A subset is called a *-system of subsets of * provided that (i) and (ii) . Also, a subset is called a *-algebra* provided is nonempty and closed with respect to countable unions and complements. Finally, a subset is called a *Dynkin System* (or a -system, or a -system) *of subsets of * provided (i) , (ii) , and (iii) if is a pairwise disjoint collection of elements of , then .

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 is both a – and a -system, then is a -algebra.

*Proof.* This is immediate given the conditions satisfied by each of these objects.

**Proposition 2.** Given a subset , there exists a unique smallest -system containing .

*Proof.* Let be the collection of all Dynkin systems containing and define to be

.

Because each such is a -system, each such 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 , thereby proving that *is* a Dynkin system. As is always the case, the intersection is also the smallest such system, hence the result.

**Theorem (Dynkin’s – Theorem).** Let and be -, respectively -, systems of subsets of for which . Then where is the -algebra generated by , that is, the smallest -algebra containing .

*Proof (Sketch).* Let . By definition of , it follows that because is the unique smallest -system containing . Moreover, because it’s easily verified that is also a -system, is therefore a -algebra containing . By definition, however, is the *smallest* -algebra containing , and so it follows that

,

and so in particular, . Hence, the result.

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 – theorem) is useful when it comes to determining uniqueness for -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 and consider where is the -algebra of all Borel measurable subsets of and where is standard Lebesgue measure. Suppose that is another measure on for which . The claim here is that the collection of sets for which is actually , whereby it follows that (among other things) Lebesgue measure is unique. Here’s how the proof of that goes:

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

be the collection of all intervals which are sub-intervals of (not necessarily proper). Note that is then contained in , that by definition, and that is closed under finite intersections, i.e. that is a -system. Thus, the -system is contained in the -system and so, by the above theorem, . But is clearly contained in by definition of , and so . Hence, on all of , whereby it follows that the Lebesgue measure is unique.

**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.