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


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.

  • $\begingroup$ Thanks I'll try this out. I know this it not strictrly related to the question but maybe you can help me: - Is the proof of theorem 4.5 as straightforward as it appears on the paper? I can't seem to make the measure delta appear when integrating. - When they define H as the minimal invariant sigma algebra for which the functions delta(y,n) are measurable, according to what I undestand, as functions of y, the delta(y,n) takes values in the function space of the probability density functions, so in what sense are they measurable? What is the sigma algebra of the final space of the functions? $\endgroup$
    – Jarana
    Oct 5 '14 at 3:41

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