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I was exploring this question: "Prove that the only Borel probability rotation-invariant measure on unit circle is normalized Lebesgue measure".

I have a question about this remark. I proved that if $\lambda$ is the Lebesgue measure and $n\in\Bbb N_0$ then: $$\int\limits_{\mathbb T}z^n d\lambda=\begin{cases}0, \ \ \text{if} \ n\not=0 \newline 1 \ \ \text{otherwise} \end{cases}$$

But I have two questions I didn't find answers to:

  1. How to prove the converse, that if $\lambda$ is a measure on $\Bbb T$ satisfying the above integral identities then $\lambda$ must be the Lebesgue measure? (I thought about denseness of polynomials in $L^2$, but it seems useless)
  2. How this fact linked to the question about uniqueness of invariant measure?

I know that a measure $\mu$ is invariant under $T: X\rightarrow X$ iff $$\int \limits_X f \ d\mu = \int\limits_X f\circ T \ d\mu\ \ \ \ \ \ \ \forall f\in L^1_\mu(X)$$ But I don't see how to apply this fact in our case.

Any hint is appreciated! Thanks in advance!

Mittens
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Big Coconut
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  • What is "the converse", exactly? – FShrike Aug 06 '23 at 13:15
  • @FShrike, it was stated here that if integrals of $z^n$ behave the way they do if measure is Lebesgue, then the measure is indeed Lebesgue – Big Coconut Aug 06 '23 at 13:45
  • The conclusion holds for any regular Borel complex valued measure with bounded variation. If such measure is rotation invariant with respect to an angle $\alpha$ such that $\alpha/\pi$ is irrational, then it is a constant multiple of the Lebesgue measure. The Riesz representation theorem can be applied and the linear density of monomials $z^n$, $n\in\mathbb{Z},$ in $C(\mathbb{T}).$ – Ryszard Szwarc Aug 16 '23 at 17:18

1 Answers1

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With regards to question (1) in the OP, suppose $\mu$ is a (nonnegative) Borel measure on $\mathbb{S}^1$ such that $\int_{\mathbb{S}^1}z^n\,\mu(dz)=\int_{\mathbb{S}^1}z^n\,\lambda(dz)$for all $n\in\mathbb{Z}_+=\{0\}\cup\mathbb{N}$, where $\lambda$ for a is the Lebesgue measure on $\mathbb{S}^1$. Then, by complex conjugation, $$\begin{align}\int_{\mathbb{S}^1}p(z)\,\mu(dz)=\int_{\mathbb{S}^1}p(z)\,\lambda(dz)\tag{0}\label{zero}\end{align}$$ for any trigonometric polynomial function $p$ (i.e., for any $p\in\operatorname{span}(z^n:n\in\mathbb{Z})$ Clearly $\mu$ is finite: $\mu(\mathbb{S}^1)=1=\lambda(\mathbb{S}^1)$. Since trigonometric polynomials are dense in $(C(\mathbb{S}^1;\mathbb{C}),\|\;\|_u)$, \eqref{zero} extends to all $f\in C(\mathbb{S}^1;\mathbb{C})$. The extension to all bounded Borel functions follows by monotone class theorem for functions:

Theorem Suppose $\mathcal{V}\subset\mathbb{R}^\Omega$ (resp. $\mathcal{V}\subset\mathcal{B}_b(\Omega;\mathbb{R})$) is a real vector space containing the constant functions, and a monotone (resp. a bounded monotone) class. If $\mathcal{M}\subset\mathcal{V}\cap\mathcal{B}_b(\Omega;\mathbb{R})$ is a multiplicative class, then $\mathcal{V}$ contains all real valued $\sigma(\mathcal{M})$--measurable functions.

For the purposes of the OP, $\mathcal{V}$ is the space of all bounded Borel measurable functions $f$ that satisfy $\int_{\mathbb{S}^1}g\,d\mu=\int_{\mathbb{S}^1}g\,d\lambda$, $\mathcal{M}$ is the collections of all real continuous functions on $\mathbb{S}^1$. Hence $\lambda=\mu$.

As for question 2, $\mathcal{S}^1$ with the complex product and the Euclidean topology is a topological compact Abelian group. The Lebesgue measure $\lambda$ on $\mathbb{S}^1$ is the unique Haar measure (rotation invariant Borel measure) with $\lambda(\mathbb{S}^1)=1$. Any other Haar measure is a constant multiple of $\lambda$.

Mittens
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