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arjun
arjun
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Values appearing with density in an ergodic system

Let $(X,\mu)$ be a probability space with invertible, measure preserving, totally-ergodic map $T:X \to X$. ($(X,\mu,T)$ is a $\mathbb{Z}$ dynamical system). Suppose $f:X \to \mathbb{R}$ is a measurable function such that $\mu(x \colon f(x) = a) = 0$ for all $a \in \mathbb{R}$.

Let

$$ D(x) = \left\{ a \in \mathbb{R} \colon \liminf_{n \to \infty} \frac{ \{0 \leq k < n \colon f(T^{k}x) = a \}}{n} > 0 \right\} $$$$ D(x) = \left\{ a \in \mathbb{R} \colon \liminf_{n \to \infty} \frac{ \# \{0 \leq k < n \colon f(T^{k}x) = a \}}{n} > 0 \right\} $$

be the set of numbers that appear with density. Is $D(x) = \emptyset$ with probability $1$? Are there standard conditions on $(X,\mu,T)$ like mixing, positive entropy and so on that guarantee that $D(x) = \emptyset$ with probability $1$? One such is the example where $X = [0,1]^\mathbb{Z}$ $\mu = \mu_0^{\otimes \mathbb{Z}}$ where $\mu_0$ is the uniform measure on $[0,1]$, and $T$ is the shift map.

Speculation: Obviously this is true for all the rationals; i.e., $D(x) \cap \mathbb{Q} = \emptyset$ with probability 1. So maybe I need some sort of continuity for this to have a chance to be true? I'm in the fairly general setting of a shift system, and do have some flexibility in the way I formulate my problem. My only constraint is that $f$ should have the right "finite dimensional distributions" that are specified, and $X$ is a factor of a Bernoulli system ($\mu$ is a product measure). So perhaps I can construct $X$ and $f$, and put a topology on it such that it is continuous?

Values appearing with density in an ergodic system

Let $(X,\mu)$ be a probability space with invertible, measure preserving, totally-ergodic map $T:X \to X$. ($(X,\mu,T)$ is a $\mathbb{Z}$ dynamical system). Suppose $f:X \to \mathbb{R}$ is a measurable function such that $\mu(x \colon f(x) = a) = 0$ for all $a \in \mathbb{R}$.

Let

$$ D(x) = \left\{ a \in \mathbb{R} \colon \liminf_{n \to \infty} \frac{ \{0 \leq k < n \colon f(T^{k}x) = a \}}{n} > 0 \right\} $$

be the set of numbers that appear with density. Is $D(x) = \emptyset$ with probability $1$? Are there standard conditions on $(X,\mu,T)$ like mixing, positive entropy and so on that guarantee that $D(x) = \emptyset$ with probability $1$? One such is the example where $X = [0,1]^\mathbb{Z}$ $\mu = \mu_0^{\otimes \mathbb{Z}}$ where $\mu_0$ is the uniform measure on $[0,1]$, and $T$ is the shift map.

Speculation: Obviously this is true for all the rationals; i.e., $D(x) \cap \mathbb{Q} = \emptyset$. So maybe I need some sort of continuity for this to have a chance to be true? I'm in the fairly general setting of a shift system, and do have some flexibility in the way I formulate my problem. My only constraint is that $f$ should have the right "finite dimensional distributions" that are specified, and $X$ is a factor of a Bernoulli system ($\mu$ is a product measure). So perhaps I can construct $X$ and $f$, and put a topology on it such that it is continuous?

Values appearing with density in an ergodic system

Let $(X,\mu)$ be a probability space with invertible, measure preserving, totally-ergodic map $T:X \to X$. ($(X,\mu,T)$ is a $\mathbb{Z}$ dynamical system). Suppose $f:X \to \mathbb{R}$ is a measurable function such that $\mu(x \colon f(x) = a) = 0$ for all $a \in \mathbb{R}$.

Let

$$ D(x) = \left\{ a \in \mathbb{R} \colon \liminf_{n \to \infty} \frac{ \# \{0 \leq k < n \colon f(T^{k}x) = a \}}{n} > 0 \right\} $$

be the set of numbers that appear with density. Is $D(x) = \emptyset$ with probability $1$? Are there standard conditions on $(X,\mu,T)$ like mixing, positive entropy and so on that guarantee that $D(x) = \emptyset$ with probability $1$? One such is the example where $X = [0,1]^\mathbb{Z}$ $\mu = \mu_0^{\otimes \mathbb{Z}}$ where $\mu_0$ is the uniform measure on $[0,1]$, and $T$ is the shift map.

Speculation: Obviously this is true for all the rationals; i.e., $D(x) \cap \mathbb{Q} = \emptyset$ with probability 1. So maybe I need some sort of continuity for this to have a chance to be true? I'm in the fairly general setting of a shift system, and do have some flexibility in the way I formulate my problem. My only constraint is that $f$ should have the right "finite dimensional distributions" that are specified, and $X$ is a factor of a Bernoulli system ($\mu$ is a product measure). So perhaps I can construct $X$ and $f$, and put a topology on it such that it is continuous?

Source Link
arjun
arjun
  • 941
  • 4
  • 15

Values appearing with density in an ergodic system

Values appearing with density in an ergodic system

Let $(X,\mu)$ be a probability space with invertible, measure preserving, totally-ergodic map $T:X \to X$. ($(X,\mu,T)$ is a $\mathbb{Z}$ dynamical system). Suppose $f:X \to \mathbb{R}$ is a measurable function such that $\mu(x \colon f(x) = a) = 0$ for all $a \in \mathbb{R}$.

Let

$$ D(x) = \left\{ a \in \mathbb{R} \colon \liminf_{n \to \infty} \frac{ \{0 \leq k < n \colon f(T^{k}x) = a \}}{n} > 0 \right\} $$

be the set of numbers that appear with density. Is $D(x) = \emptyset$ with probability $1$? Are there standard conditions on $(X,\mu,T)$ like mixing, positive entropy and so on that guarantee that $D(x) = \emptyset$ with probability $1$? One such is the example where $X = [0,1]^\mathbb{Z}$ $\mu = \mu_0^{\otimes \mathbb{Z}}$ where $\mu_0$ is the uniform measure on $[0,1]$, and $T$ is the shift map.

Speculation: Obviously this is true for all the rationals; i.e., $D(x) \cap \mathbb{Q} = \emptyset$. So maybe I need some sort of continuity for this to have a chance to be true? I'm in the fairly general setting of a shift system, and do have some flexibility in the way I formulate my problem. My only constraint is that $f$ should have the right "finite dimensional distributions" that are specified, and $X$ is a factor of a Bernoulli system ($\mu$ is a product measure). So perhaps I can construct $X$ and $f$, and put a topology on it such that it is continuous?