Roth's theorem is stated in the book by Einsiedler and Ward, theorem 7.14 page 191 as:

Let $(X,\mathscr{B},\mu,T)$ be a measure-preserving probability system. Then, for any functions $f_1,f_2 \in L^{\infty}(X,\mathscr{B},\mu)$, $$\frac{1}{N}\sum\limits_{n=1}^{N}U_T^{n}f_1U_T^{2n}f_2$$ converges in $L^2(X,\mathscr{B},\mu)$ (Here $U_T g:=g\circ T$). Moreover, for any $A\in \mathscr{B}$ with $\mu(A)>0$ we have $$\lim_{N\to \infty}\frac{1}{N}\sum\limits_{n=1}^{N} \mu(A\cap T^{-n}A \cap T^{-2n}A)>0.$$

Question: Can I deduce this theorem from the special case where $(X,\mathscr{B},\mu,T)$ is taken to be an invertible, ergodic, Borel probability system?

The reason I'm asking is that Einsiedler-Ward only seem to prove this for the special case. I'm not sure if i'm misreading their proof or if the reduction to the general case is easy.

My issue is primarily with the $L^2$ convergence claim. A first attempt at the reduction could be to apply the ergodic decomposition theorem. However this isn't a valid approach since our space isn't assumed to be a Borel space.

  • $\begingroup$ Just a small remark that doesn't answer the question: you still get Roth's theorem (i.e. large subsets of Z have a 3AP) from the special case. $\endgroup$ Nov 13, 2018 at 2:03
  • $\begingroup$ Yes. Prior to this theorem, the book reduces the proof of Furstenberg's 'multiple recurrence theorem' to the invertible, ergodic, Borel case. But the $L^2$ convergence statement in this theorem seems to be much stronger. I was wondering if a similar reduction could be made. $\endgroup$
    – Latimer
    Nov 13, 2018 at 11:24
  • $\begingroup$ what's the issue if you try to go through the reductions that the book makes for Furstenberg's multiple recurrence theorem? $\endgroup$ Nov 13, 2018 at 12:35
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    $\begingroup$ one of the steps is to reduce the system to a borel system. For any measurable $A$, you consider a factor of the form $\{0,1\}^{\mathbb{N}}$ with an appropriate measure and factor map. Multiple recurrence follows for the original system if you prove it for all these borel factors. Not sure what the analogous step here would be. $\endgroup$
    – Latimer
    Nov 13, 2018 at 15:17
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    $\begingroup$ Yessir professor. $\endgroup$
    – Latimer
    May 10, 2020 at 22:59


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