This is the sixth and final post in a series about Szemerédi’s theorem. In this post I complete the proof of the Multiple Recurrence Theorem, which I showed in a previous post of this series to be equivalent to Szemerédi’s theorem. The last step missing (cf. third point of Theorem 7 in my previous post) is the following:
— 1. Some Lemmas —
First we will need a few lemmas:
Lemma 2 If an extension of factors of a measure preserving system is not weak mixing, then the relatively independent self joining is not ergodic.
The converse of this lemma is also true; in other words, it provides an alternative description of weakly mixing extensions. We omit the proof of the converse because we do not need it. One good place to look up the proof of the converse direction is Einsiedler and Ward’s book mentioned above.
Proof: I will use some facts and definitions from this previous post. In particular recall that the measure is defined by
Now assume that is an ergodic system and take such that . We need to show (using the notion of uniform Cesàro averages defined in a previous post) that
By definition, we have that . Using the triangular inequality in , (1) will follow from
We can rewrite the norm of the conditional expectation in terms of the relatively independent self joining as
Taking uniform Cesàro limits and applying the ergodic theorem in the relatively independent self joining we conclude
which finishes the proof.
Although I managed to avoid doing it so far, it seems that the most convenient way to prove Proposition 1 is to appeal to the disintegration of measures relative to a factor (see also this post for a description of additional properties of disintegration of measures). The only alternative I know of is presented in this post of Tao.
Definition 3 (Disintegration of a measure) Using the same notation and conventions as in the previous posts in this series, given a factor we define, for almost every , the measure by .
Observe that if , then the norm of with respect to each of the measures is (almost everywhere) uniformly bounded and, in particular, belongs to almost every space .
for any .
— 2. Finding one conditionally compact function —
In this section we show that there exists at least one function which is conditionally compact but is not in (observe that any function in is trivially conditionally compact).
Since we are assuming that the extension is not weak mixing, it follows from Lemma 2 that the relatively independent self joining is not ergodic. Let be a -invariant function which is not a constant ( almost everywhere).
Next, it follows from (3) that for almost every and hence the operator given by
We can glue the together to obtain an operator defined (almost everywhere) by . Since is invariant under , it follows that .
Proof: One can write as a (possibly infinite) linear combination of functions of the form , so we will simply assume that . We will first show that . Indeed, if , then we claim that in . To see this, observe that also and hence
Recall that is an extension of through the projection in either coordinate. Therefore, the function is invariant under and belongs to . Invoking the ergodicity of we deduce that would have to be constant, which is a contradiction. We conclude that .
Since is not almost everywhere, the set has positive measure. For any , let such that . Then also , so . We conclude that is -invariant, so again by ergodicity it follows that has full measure. Finally, choose . We deduce that
As a last step, one can truncate to ensure that is bounded.
Next, we will show that if is the function given by Lemma 4, then is conditionally compact. Indeed , which implies that or, equivalently, that for almost every . Moreover, , and we deduce that the set is a bounded subset of , for almost every . Since each is compact, we deduce that, for almost every , the orbit
is pre-compact. However, the number of functions necessary to -cover the orbit of in may depend on .
Define, for every , the function such that is dense in the whole orbit with respect to the norm. In symbols, we have
It is easy to check that is measurable with respect to , and it follows from the compactness of each that is almost everywhere finite. Let be such that the set defined by has positive measure. We will show that the orbit of in can be -covered by functions, for almost every . Indeed, for each , define the function
Observe that, by ergodicity, is defined almost everywhere. For every and every where the are defined we have
where is the first hitting time of , more precisely, . We just proved that is indeed conditionally compact.
— 3. Finishing the proof —
In the previous section we showed that the subspace
is not contained in . We claim that is in fact closed under products. Indeed, let , let and choose such that for every we have
Observe that truncating each by does not alter the above inequality, so we will simply assume that . Therefore for each , choosing and appropriately, we have
We just showed that is an algebra. Next let
Since is an algebra and is closed, the set is a -algebra and is dense in .
The last step is to prove that . Let and let for some . By Weierstrass’ approximation theorem, the function (where is the pushforward measure of through ) can be approximated by a polynomial (even though is not continuous, it has only one point of discontinuity, so it can be altered in a set of arbitrarily small measure to become continuous). This means that can be approximated by for the same polynomial . Since is an algebra, and so and hence
Finally, note that is invariant under , thus is invariant under and therefore it is a compact extension of . This finishes the proof of Proposition 1, and hence of the Multiple recurrence theorem of Furstenberg, and hence of Szemerédi’s theorem.