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Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely:

Question: (Furstenberg) Let $\mu$ be a continuous probability measure on the circle such that $$\int_{S^1} f(z) d\mu = \int_{S^1} f(z^2) d\mu = \int_{S^1} f(z^3) d\mu, \quad \forall f \in C(S^1).$$ Is $\mu$ the Lebesgue measure?

The assumption implies that the Fourier coefficients $\hat\mu(n)$ satisfy $$\hat\mu(n) = \hat \mu(2^k3^ln), \quad \forall k,l \in \mathbb N, n \in \mathbb Z.$$ Furstenberg's question is known to have an affirmative answer if one makes additional assumptions on the entropy of the measure.

The basic strategy is usueally to show that a non-vanishing (non-trivial) Fourier coefficient implies the existence of an atom. The standard tool to construct atoms in a measure on the circle is Wiener's Lemma, which says that as soon as there exists $\delta>0$ such that the set $\lbrace n \in \mathbb Z \mid |\hat\mu(n)| \geq \delta \rbrace$ has positive density in $\mathbb Z$, $\mu$ has an atom. More precisely, the following identity holds: $$\sum_{x \in S^1} \mu(\lbrace x \rbrace)^2 = \lim_{n \to \infty} \frac1{2n+1} \sum_{k=-n}^n |\hat \mu(k)|^2.$$

Clearly, $$|\lbrace 2^k3^l \mid k,l \in \mathbb N \rbrace \cap [-n,n] | \sim (\log n)^2 $$ so that Wiener's Lemma can not be applied directly. My question is basically, whether this problem can be overcome for any other subsemigroup of $\mathbb N$.

Question: Is there any subsemigroup $S \subset \mathbb N$ of zero density known, such that every $S$-invariant continuous probability measure on $S^1$ is the Lebesgue measure? What about the subsemigroup generated by $2,3$ and $5$?

Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely:

Question: (Furstenberg) Let $\mu$ be a continuous probability measure on the circle such that $$\int_{S^1} f(z) d\mu = \int_{S^1} f(z^2) d\mu = \int_{S^1} f(z^3) d\mu, \quad \forall f \in C(S^1).$$ Is $\mu$ the Lebesgue measure?

The assumption implies that the Fourier coefficients $\hat\mu(n)$ satisfy $$\hat\mu(n) = \hat \mu(2^k3^ln), \quad \forall k,l \in \mathbb N, n \in \mathbb Z.$$ Furstenberg's question is known to have an affirmative answer if one makes additional assumptions on the entropy of the measure.

The basic strategy is usually to show that a non-vanishing (non-trivial) Fourier coefficient implies the existence of an atom. The standard tool to construct atoms in a measure on the circle is Wiener's Lemma, which says that as soon as there exists $\delta>0$ such that the set $\lbrace n \in \mathbb Z \mid |\hat\mu(n)| \geq \delta \rbrace$ has positive density in $\mathbb Z$, $\mu$ has an atom. More precisely, the following identity holds: $$\sum_{x \in S^1} \mu(\lbrace x \rbrace)^2 = \lim_{n \to \infty} \frac1{2n+1} \sum_{k=-n}^n |\hat \mu(k)|^2.$$

Clearly, $$|\lbrace 2^k3^l \mid k,l \in \mathbb N \rbrace \cap [-n,n] | \sim (\log n)^2 $$ so that Wiener's Lemma can not be applied directly. My question is basically, whether this problem can be overcome for any other subsemigroup of $\mathbb N$.

Question: Is there any subsemigroup $S \subset \mathbb N$ of zero density known, such that every $S$-invariant continuous probability measure on $S^1$ is the Lebesgue measure? What about the subsemigroup generated by $2,3$ and $5$?

Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely:

Question: (Furstenberg) Let $\mu$ be a continuous probability measure on the circle such that $$\int_{S^1} f(z) d\mu = \int_{S^1} f(z^2) d\mu = \int_{S^1} f(z^3) d\mu, \quad \forall f \in C(S^1).$$ Is $\mu$ the Lebesgue measure?

The assumption implies that the Fourier coefficients $\hat\mu(n)$ satisfy $$\hat\mu(n) = \hat \mu(2^k3^ln), \quad \forall k,l \in \mathbb N, n \in \mathbb Z.$$ Furstenberg's question is known to have an affirmative answer if one makes additional assumptions on the entropy of the measure.

The standard tool to construct atoms in a measure on the circle is Wiener's Lemma, which says that as soon as there exists $\delta>0$ such that the set $\lbrace n \in \mathbb Z \mid \hat\mu(n) \geq \delta \rbrace$ has positive density in $\mathbb Z$, $\mu$ has an atom. More precisely, the following identity holds: $$\sum_{x \in S^1} \mu(\lbrace x \rbrace)^2 = \lim_{n \to \infty} \frac1{2n+1} \sum_{k=-n}^n |\hat \mu(k)|^2.$$

Clearly, $$|\lbrace 2^k3^l \mid k,l \in \mathbb N \rbrace \cap [-n,n] | \sim (\log n)^2 $$ so that Wiener's Lemma can not be applied directly. My question is basically, whether this problem can be overcome for any other subsemigroup of $\mathbb N$.

Question: Is there any subsemigroup $S \subset \mathbb N$ of zero density known, such that every $S$-invariant continuous probability measure on $S^1$ is the Lebesgue measure? What about the subsemigroup generated by $2,3$ and $5$?

Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely:

Question: (Furstenberg) Let $\mu$ be a continuous probability measure on the circle such that $$\int_{S^1} f(z) d\mu = \int_{S^1} f(z^2) d\mu = \int_{S^1} f(z^3) d\mu, \quad \forall f \in C(S^1).$$ Is $\mu$ the Lebesgue measure?

The assumption implies that the Fourier coefficients $\hat\mu(n)$ satisfy $$\hat\mu(n) = \hat \mu(2^k3^ln), \quad \forall k,l \in \mathbb N, n \in \mathbb Z.$$ Furstenberg's question is known to have an affirmative answer if one makes additional assumptions on the entropy of the measure.

The basic strategy is usueally to show that a non-vanishing (non-trivial) Fourier coefficient implies the existence of an atom. The standard tool to construct atoms in a measure on the circle is Wiener's Lemma, which says that as soon as there exists $\delta>0$ such that the set $\lbrace n \in \mathbb Z \mid |\hat\mu(n)| \geq \delta \rbrace$ has positive density in $\mathbb Z$, $\mu$ has an atom. More precisely, the following identity holds: $$\sum_{x \in S^1} \mu(\lbrace x \rbrace)^2 = \lim_{n \to \infty} \frac1{2n+1} \sum_{k=-n}^n |\hat \mu(k)|^2.$$

Clearly, $$|\lbrace 2^k3^l \mid k,l \in \mathbb N \rbrace \cap [-n,n] | \sim (\log n)^2 $$ so that Wiener's Lemma can not be applied directly. My question is basically, whether this problem can be overcome for any other subsemigroup of $\mathbb N$.

Question: Is there any subsemigroup $S \subset \mathbb N$ of zero density known, such that every $S$-invariant continuous probability measure on $S^1$ is the Lebesgue measure? What about the subsemigroup generated by $2,3$ and $5$?

Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely:

Question: (Furstenberg) Let $\mu$ be a continuous probability measure on the circle such that $$\int_{S^1} f(z) d\mu = \int_{S^1} f(z^2) d\mu = \int_{S^1} f(z^3) d\mu, \quad \forall f \in C(S^1).$$ Is $\mu$ the Lebesgue measure?

The assumption implies that the Fourier coefficients $\hat\mu(n)$ satisfy $$\hat\mu(n) = \hat \mu(2^k3^ln), \quad \forall k,l \in \mathbb N, n \in \mathbb Z.$$ Furstenberg's question is known to have an affirmative answer if one makes additional assumptions on the entropy of the measure.

The standard tool to construct atoms in a measure on the circle is Wiener's Lemma, which says that as soon as there exists $\delta>0$ such that the set $\lbrace n \in \mathbb Z \mid \hat\mu(n) \geq \delta \rbrace$ has positive density in $\mathbb Z$, $\mu$ has an atom. More precisely, the following identity holds: $$\sum_{x \in S^1} \mu(\lbrace x \rbrace)^2 = \lim_{n \to \infty} \frac1{2n+1} \sum_{k=-n}^n |\hat \mu(k)|^2.$$

Clearly, $$|\lbrace 2^k3^l \mid k,l \in \mathbb N \rbrace \cap [-n,n] | \sim (\log n)^2 $$ so that Wiener's Lemma can not be applied directly. My question is basically, whether this problem can be overcome for any other subsemigroup of $\mathbb N$.

Question: Is there any subsemigroup $S \subset \mathbb N$ of zero-density known, such that every $S$-invariant continuous probability measure on $S^1$ is the Lebesgue measure? What about the subsemigroup generated by $2,3$ and $5$?

Hillel Furstenberg conjectured that the only $2$-$3$-invariant probability measure on the circle without atoms is the Lebesgue measure. More precisely:

Question: (Furstenberg) Let $\mu$ be a continuous probability measure on the circle such that $$\int_{S^1} f(z) d\mu = \int_{S^1} f(z^2) d\mu = \int_{S^1} f(z^3) d\mu, \quad \forall f \in C(S^1).$$ Is $\mu$ the Lebesgue measure?

The assumption implies that the Fourier coefficients $\hat\mu(n)$ satisfy $$\hat\mu(n) = \hat \mu(2^k3^ln), \quad \forall k,l \in \mathbb N, n \in \mathbb Z.$$ Furstenberg's question is known to have an affirmative answer if one makes additional assumptions on the entropy of the measure.

The standard tool to construct atoms in a measure on the circle is Wiener's Lemma, which says that as soon as there exists $\delta>0$ such that the set $\lbrace n \in \mathbb Z \mid \hat\mu(n) \geq \delta \rbrace$ has positive density in $\mathbb Z$, $\mu$ has an atom. More precisely, the following identity holds: $$\sum_{x \in S^1} \mu(\lbrace x \rbrace)^2 = \lim_{n \to \infty} \frac1{2n+1} \sum_{k=-n}^n |\hat \mu(k)|^2.$$

Clearly, $$|\lbrace 2^k3^l \mid k,l \in \mathbb N \rbrace \cap [-n,n] | \sim (\log n)^2 $$ so that Wiener's Lemma can not be applied directly. My question is basically, whether this problem can be overcome for any other subsemigroup of $\mathbb N$.

Question: Is there any subsemigroup $S \subset \mathbb N$ of zero density known, such that every $S$-invariant continuous probability measure on $S^1$ is the Lebesgue measure? What about the subsemigroup generated by $2,3$ and $5$?