Let $(X, \mathcal{F}, \mu)$ be a general measure space, and let $T: X \to X$ be a measure-preserving transformation on $X$. Assume that $T$ is ergodic and satisfies the property that, for any set $A \in \mathcal{F}$ with $0 < \mu(A)$, there exists a positive integer $k$ such that $\mu(A \cap T^{-k}A) > 0$.

For any $A \in \mathcal{F}$ and $t \geq 0,~n\ge 1$, define the set $\mathcal{B}(A, t)$ as: $$ \mathcal{B}(A,t,n) := \left\{ x \in X : \sum_{k=1}^n \chi_A \circ T^k(x) \geq t \right\}, $$ where $\chi_A$ denotes the characteristic function of the set $A \in \mathcal{F}$.

For $A \in \mathcal{F}$ with $0 < \mu(A) < \infty$ and $\alpha \in (0,1)$, define the sequence $\theta_n(A, \alpha)$ as: $$ \theta_n(A, \alpha) := \sup \left\{ t \geq 0 : \mu_A\left( \mathcal{B}(A,t,n) \right) \geq \alpha \right\}, $$ where $\mu_A(B) = \frac{\mu(A \cap B)}{\mu(A)}$ for $B \in \mathcal{F}$, i.e., $\mu_A$ is the conditional probability measure induced by $A$.

For $A, B \in \mathcal{F}$ with $0 < \mu(A) < \infty$, $0 < \mu(B) < \infty$, and $0 < \alpha < \beta < 1$, I want to show that: $$ \liminf_{n \to \infty} \frac{\theta_n(A, \alpha)}{\theta_n(B, \beta)} \geq \frac{\mu(A)}{\mu(B)}. $$

First, observe that for any set $A \in \mathcal{F}$ with $0 < \mu(A) < \infty$ and $\alpha \in (0,1)$, the sequence $(\theta_n(A, \alpha))_{n=1}^{\infty}$ increases without bound as $n \to \infty$. Now, to derive a contradiction, assume that there exists a subsequence $(n_k)$ such that

$$ \lim_{k \to \infty} \frac{\theta_{n_k}(A, \alpha)}{\theta_{n_k}(B, \beta)} = \gamma,\quad \text{where }~0 \leq \gamma < \frac{\mu(A)}{\mu(B)}. $$

By the Ergodic Theorem, for almost every $x \in X$, we have the following result: $$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \chi_A \circ T^k(x) = \mu(A). $$ Can this result be applied in any meaningful way? I seem to have encountered difficulty in proceeding further with this concept. Your guidance and insights would be greatly appreciated. Thank you for your time and assistance.

Let $(X, \mathcal{F}, \mu)$ be a general measure space, and let $T: X \to X$ be a measure-preserving transformation on $X$. Assume that $T$ is ergodic and satisfies the property that, for any set $A \in \mathcal{F}$ with $0 < \mu(A)$, there exists a positive integer $k$ such that $\mu(A \cap T^{-k}A) > 0$.

For any $A \in \mathcal{F}$ and $t \geq 0,~n\ge 1$, define the set $\mathcal{B}(A, t)$ as: $$ \mathcal{B}(A,t,n) := \left\{ x \in X : \sum_{k=1}^n \chi_A \circ T^k(x) \geq t \right\}, $$ where $\chi_A$ denotes the characteristic function of the set $A \in \mathcal{F}$.

For $A \in \mathcal{F}$ with $0 < \mu(A) < \infty$ and $\alpha \in (0,1)$, define the sequence $\theta_n(A, \alpha)$ as: $$ \theta_n(A, \alpha) := \sup \left\{ t \geq 0 : \mu_A\left( \mathcal{B}(A,t,n) \right) \geq \alpha \right\}, $$ where $\mu_A(B) = \frac{\mu(A \cap B)}{\mu(A)}$ for $B \in \mathcal{F}$, i.e., $\mu_A$ is the conditional probability measure induced by $A$.

For $A, B \in \mathcal{F}$ with $0 < \mu(A) < \infty$, $0 < \mu(B) < \infty$, and $0 < \alpha < \beta < 1$, I want to show that: $$ \liminf_{n \to \infty} \frac{\theta_n(A, \alpha)}{\theta_n(B, \beta)} \geq \frac{\mu(A)}{\mu(B)}. $$

First, observe that for any set $A \in \mathcal{F}$ with $0 < \mu(A) < \infty$ and $\alpha \in (0,1)$, the sequence $(\theta_n(A, \alpha))_{n=1}^{\infty}$ increases without bound as $n \to \infty$. Now, to derive a contradiction, assume that there exists a subsequence $(n_k)$ such that

$$ \lim_{k \to \infty} \frac{\theta_{n_k}(A, \alpha)}{\theta_{n_k}(B, \beta)} = \gamma,\quad \text{where }~0 \leq \gamma < \frac{\mu(A)}{\mu(B)}. $$

I seem to have encountered difficulty in proceeding further with this concept. Your guidance and insights would be greatly appreciated. Thank you for your time and assistance.

Let $(X, \mathcal{F}, \mu)$ be a general measure space, and let $T: X \to X$ be a measure-preserving transformation on $X$. Assume that $T$ is ergodic and satisfies the property that, for any set $A \in \mathcal{F}$ with $0 < \mu(A)$, there exists a positive integer $k$ such that $\mu(A \cap T^{-k}A) > 0$.

For any $A \in \mathcal{F}$ and $t \geq 0$, define the set $\mathcal{B}(A, t)$ as: $$ \mathcal{B}(A,t) := \left\{ x \in X : \sum_{k=1}^n \chi_A \circ T^k(x) \geq t \right\}, $$ where $\chi_A$ denotes the characteristic function of the set $A \in \mathcal{F}$.

For $A \in \mathcal{F}$ with $0 < \mu(A) < \infty$ and $\alpha \in (0,1)$, define the sequence $\theta_n(A, \alpha)$ as: $$ \theta_n(A, \alpha) := \sup \left\{ t \geq 0 : \mu_A\left( \mathcal{B}(A,t) \right) \geq \alpha \right\}, $$ where $\mu_A(B) = \frac{\mu(A \cap B)}{\mu(A)}$ for $B \in \mathcal{F}$, i.e., $\mu_A$ is the conditional probability measure induced by $A$.

For $A, B \in \mathcal{F}$ with $0 < \mu(A) < \infty$, $0 < \mu(B) < \infty$, and $0 < \alpha < \beta < 1$, I want to show that: $$ \liminf_{n \to \infty} \frac{\theta_n(A, \alpha)}{\theta_n(B, \beta)} \geq \frac{\mu(A)}{\mu(B)}. $$

First, observe that for any set $A \in \mathcal{F}$ with $0 < \mu(A) < \infty$ and $\alpha \in (0,1)$, the sequence $(\theta_n(A, \alpha))_{n=1}^{\infty}$ increases without bound as $n \to \infty$. Now, to derive a contradiction, assume that there exists a subsequence $(n_k)$ such that

$$ \lim_{k \to \infty} \frac{\theta_{n_k}(A, \alpha)}{\theta_{n_k}(B, \beta)} = \gamma,\quad \text{where }~0 \leq \gamma < \frac{\mu(A)}{\mu(B)}. $$

By the Ergodic Theorem, for almost every $x \in X$, we have the following result: $$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \chi_A \circ T^k(x) = \mu(A). $$ Can this result be applied in any meaningful way? I seem to have encountered difficulty in proceeding further with this concept. Your guidance and insights would be greatly appreciated. Thank you for your time and assistance.

Let $(X, \mathcal{F}, \mu)$ be a general measure space, and let $T: X \to X$ be a measure-preserving transformation on $X$. Assume that $T$ is ergodic and satisfies the property that, for any set $A \in \mathcal{F}$ with $0 < \mu(A)$, there exists a positive integer $k$ such that $\mu(A \cap T^{-k}A) > 0$.

For any $A \in \mathcal{F}$ and $t \geq 0,~n\ge 1$, define the set $\mathcal{B}(A, t)$ as: $$ \mathcal{B}(A,t,n) := \left\{ x \in X : \sum_{k=1}^n \chi_A \circ T^k(x) \geq t \right\}, $$ where $\chi_A$ denotes the characteristic function of the set $A \in \mathcal{F}$.

For $A \in \mathcal{F}$ with $0 < \mu(A) < \infty$ and $\alpha \in (0,1)$, define the sequence $\theta_n(A, \alpha)$ as: $$ \theta_n(A, \alpha) := \sup \left\{ t \geq 0 : \mu_A\left( \mathcal{B}(A,t,n) \right) \geq \alpha \right\}, $$ where $\mu_A(B) = \frac{\mu(A \cap B)}{\mu(A)}$ for $B \in \mathcal{F}$, i.e., $\mu_A$ is the conditional probability measure induced by $A$.

For $A, B \in \mathcal{F}$ with $0 < \mu(A) < \infty$, $0 < \mu(B) < \infty$, and $0 < \alpha < \beta < 1$, I want to show that: $$ \liminf_{n \to \infty} \frac{\theta_n(A, \alpha)}{\theta_n(B, \beta)} \geq \frac{\mu(A)}{\mu(B)}. $$

First, observe that for any set $A \in \mathcal{F}$ with $0 < \mu(A) < \infty$ and $\alpha \in (0,1)$, the sequence $(\theta_n(A, \alpha))_{n=1}^{\infty}$ increases without bound as $n \to \infty$. Now, to derive a contradiction, assume that there exists a subsequence $(n_k)$ such that

$$ \lim_{k \to \infty} \frac{\theta_{n_k}(A, \alpha)}{\theta_{n_k}(B, \beta)} = \gamma,\quad \text{where }~0 \leq \gamma < \frac{\mu(A)}{\mu(B)}. $$

By the Ergodic Theorem, for almost every $x \in X$, we have the following result: $$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \chi_A \circ T^k(x) = \mu(A). $$ Can this result be applied in any meaningful way? I seem to have encountered difficulty in proceeding further with this concept. Your guidance and insights would be greatly appreciated. Thank you for your time and assistance.