I'm trying to understand the proof given by D. Rudolph in his paper "x2 and x3 invariant measures and entropy". I'm particularly trying to undestand the proof of lema 4.4.


Let's consider a secuence of probability measures $\delta(\hat{y},n)$ which concentrates on $\{\frac{0}{p^n}, \dots, \frac{p^n-1}{p^n}\}$ for each $n \in \mathbb{N}$. One shows that each of the measures $\delta(\hat{y},n)$ is invariant under a shift (mod 1) by a number $a\_n$ which is a fraction, in least terms, with denominator $\geq 2^{n-i\_0+1}$ where $i\_0 \geq 0$ is a fixed index. Then the paper states:


" Thus the group of shifts (mod 1) preserving $\delta(\hat{y},n)$ is of order at least $2^{n - i\_0+1}$, and its minimal element $d\_n \leq \frac{1}{2^{n-i\_0+1}}$. For any continuous function $f$ on $[0,1), f(0) = 0, f(1) = 1$, this forces
$$
\lim\_{n \to \infty} \int f \ d\delta(\hat{y},n) = \int f dm
$$
and hence $\delta(\hat{y},n)$ converges weakly to $m$ the Lebesgue measure."


I'm having trouble understanding the fragment exposed before. Particularly that does it mean that the group of shifts preserveving the measure is of certain **order**? What is a **minimal element**? How does this force the stated convergence?


Any help is aprecciated



Weak Convergence to Lebesgue Measure
I think Dan is saying for a given $y$, you have a sequence of measures $\delta(\hat y,n)$ supported on the unit interval. Each one of these is invariant (he claims) is under some group of translations: that is a subgroup of [0,1) considered as a group (with mod 1 addition). The finite subgroups of [0,1) are just the multiples of $1/q$ for some $q$. The number $1/q$ is then the minimal element of the subgroup (i.e. the one closest to 0) and the order is $q$. 
He proves that $q$ is at least $2^{n-i\_0}$ (where $i\_0$ does not depend on $n$). The claim is that any limit of measures like this must be Lebesgue measure. 


Why? Recall that Lebesgue measure is the unique measure $m$ such that for each continuous function $f$, $\int f(x)\,dm(x)=\int f(x+t\bmod 1)\,dm(x)$ for each $t$. It then suffices to show that any limit point of the sequence $(\delta(\hat y,n))\_n$ has this invariance property. Suppose that $\delta(\hat y,n\_i)$ converges weakly to $m$. Let $f$ be continuous and let $t$ be arbitrary. Let $\epsilon>0$. There exists (by uniform continuity of $f$) an $\eta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\eta$. Now if $1/2^{n-i\_0}<\eta$, let $1/q\_n$ be the minimal element. There exists a $p$ such that $|p/q\_n-t|<\eta$. Now we have
$$
\left|\int f(x)\,dm-\int f(x+t)\,dm\right|=
\left|\int f(x)\,dm-\int f(x+t-p/q\_n)\,dm\right|<\epsilon.
$$
Since $\epsilon$ is arbitrary, we deduce $\int f(x)\,dm=\int f(x+t)\,dm$ for each continuous
$f$ and $t$. Hence $m$ is Lebesgue measure. 


