Question about B. Host paper 'Nombres, normaux entropie, translations' 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. 
 A: 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.
