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I put this question on mathstack but it seems more suitable to put it here:

I was reading this paper and I got stuck in a detail left for the reader that I couldn't figure out:

Let $X = \mathbb{R}/\mathbb{Z}$, $p > 1$ a integer, $D_n = \{kp^{-n}\colon 0 \leq k < p^n \}$ and $\mu$ a measure on $X$ that is $T \colon X \ni x \to (px\mod 1)$ invariant. We define the measure $$\omega_n = \sum_{\alpha \in D_n} \mu * \delta_\alpha = \sum_{\alpha \in D_n} \mu(\cdot + \alpha ) $$ Notice that $\mu = \omega_0 \ll \cdots \ll \omega_n \ll \omega_{n+1}$ then the Radon Nikodym derivative $\phi_n = \frac{d\mu}{d\omega_n}$ exists and the paper said that is easy to see that $$\phi_n = \prod_{k=0}^{n-1}\phi_1 \circ T^k$$ but I could not do it, although I reduced it to the following:

If we proove this $$\frac{d\omega_{n}}{d\omega_{n+1}} = \frac{d\omega_{n-1}}{d\omega_n} \circ T$$ we can use the chain rule for Radon Nikodym derivatives to conclude.

Any help will be appreciated.

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This relatively straight forward.
The main observation is that, in the Host Meiri terminology, the $p^{N}$-cells are exactly the inverse image of the $p^{N-1}$-cells, and the $\times p$ map is $p$ to $1$ map on the one-torus.

Explicit calculation shows that $w_{n}(T^{-1}A) = p\cdot\omega_{n-1}(A)$ by $T$-invariance.

Now for the general calculation, by the chain rule for Radon-Nykodim derivative we have - $$ d\mu/d\omega_n =\frac{d\mu}{d\omega_1}\frac{d\omega_1}{d\omega_2}\cdots\frac{d\omega_{n-1}}{d\omega_{n}}, $$ and now we calculate -

Just for technicalities,I assume $T$ is invertible, this is clearly a non-issue as nearly all the papers dealing with the Rudolph-Johnson theorem actually start by embedding the problem in the solenoid (the two-sided extension of the system, $\mathbb{R}\times\mathbb{Q_{p}}/\mathbb{Z}[1/p]$).

$$ \frac{d\omega_{n-1}}{d\omega_{n}}(A)=\frac{p\cdot d\omega_{n-2}}{p\cdot d\omega_{n-1}}(TA)=\frac{d\omega_{n-2}}{d\omega_{n-1}}(TA), $$ for any Borelian set $A$. Now repeating the argument inductively gives the required result.

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  • $\begingroup$ Thanks for your help. I proved that $\omega_n(T^{-1}(A)) = p \omega_{n-1}(A)$ but I don't understand your last calculation which leed to the conclusion. $\endgroup$
    – user90803
    Commented Sep 21, 2014 at 20:19
  • $\begingroup$ you should simply substitute $\omega_n , \omega_{n-1}$ with their respective forms, the $p$-factor is reduced at it appears both in the nominator and in the denominator, and you're left with the derivative evaluated at $TA$ (the inveratibility is not a real issue here). $\endgroup$
    – Asaf
    Commented Sep 21, 2014 at 20:41
  • $\begingroup$ But you can't simply substitute on the radon-nykodim derivative. Actually you only have that $T\omega_n = p \omega_{n-1}$, we don't really have a direct relation between $\omega_n$ and $\omega_{n-1}$. I also don't understand your notation $\frac{d\omega_{n-1}}{d\omega_n}(A)$, you are evaluating a function in a borel set? $\endgroup$
    – user90803
    Commented Oct 8, 2014 at 6:38
  • $\begingroup$ You can substitute, and about the notation, this is a common notation, which should be understood as follows - $\mu(f) = \int D_{n}fd\mu_{n}$, by the RN thm, where $D_{n}$ is the derivative. Take $f$ to be char. func. of a nice set (you can also take a nice smooth approx. if you like), we should show that $D_{n}$ equals to the above mentioned formula. For $\omega_{0},\omega_{1}$, the proof is trivial. For $\omega_{2}$ you use the fact that $\omega_{1} \ll \omega_{2}$ and the RN derivative for this relation, to compute $d\mu/d\omega_{2}$. $\endgroup$
    – Asaf
    Commented Oct 8, 2014 at 6:56
  • $\begingroup$ Now in order to compute $d\omega{1}/d\omega{2}$, you just take a nice set $A$ and its char. func. and integrate, then you use the substitution $x=Ty$ and the relation between $\omega_{1},\omega_{2}$ and the uniqueness of the RN derivative (and probably induction on $n$) to conclude the result. $\endgroup$
    – Asaf
    Commented Oct 8, 2014 at 6:56

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