I was studying a lemma from my notes on ergodic theory and encountered a difficulty. The lemma states:

Let $(X, \mathcal{B}, \mu)$ be an infinite non-atomic measure space, and let $T$ be an ergodic non-singular transformation on $X$. For $N \geq 1$ and $\lambda > 0$, there exists $E \in \mathcal{B}$ such that $\mu(E) = \lambda$ and $\{T^{-k}E\}_{k=0}^N$ are mutually disjoint.

Outline of the solution provided: Since $\mu$ is non-atomic, we select $A \in \mathcal{B}$ appropriately. Then consider $E = T^{-N}A \setminus \bigcup_{k=0}^{N-1} T^{-k}A$.

From this outline, I managed to prove that $\{T^{-k}E\}_{k=0}^N$ are indeed mutually disjoint. However, I am struggling to understand how to ensure $\mu(E) = \lambda$ using an appropriate choice of $A$. According to Wacław Sierpiński's theorem on non-atomic measures, we can select $A$ such that $\mu(A) = \lambda'$ for any $\lambda' > 0$. But how can I choose $A$ in a way that guarantees $\mu(E) = \lambda$ exactly? Could you please help me solve this? Thank you for your time and assistance.

I was studying a lemma from my notes on ergodic theory and encountered a difficulty. The lemma states:

Let $(X, \mathcal{B}, \mu)$ be an infinite non-atomic measure space, and let $T$ be an ergodic non-singular transformation on $X$. For $N \geq 1$ and $\lambda > 0$, there exists $E \in \mathcal{B}$ such that $\mu(E) = \lambda$ and $\{T^{-k}E\}_{k=0}^N$ are mutually disjoint.

Outline of the solution provided: Since $\mu$ is non-atomic, we select $A \in \mathcal{B}$ appropriately. Then consider $E = T^{-N}A \setminus \bigcup_{k=0}^{N-1} T^{-k}A$.

From this outline, I managed to prove that $\{T^{-k}E\}_{k=0}^N$ are indeed mutually disjoint. However, I am struggling to understand how to ensure $\mu(E) = \lambda$ using an appropriate choice of $A$. According to Wacław Sierpiński's theorem on non-atomic measures, we can select $A$ such that $\mu(A) = \lambda'$ for any $\lambda' > 0$. But how can I choose $A$ in a way that guarantees $\mu(E) = \lambda$ exactly?

I was studying a lemma from my notes on ergodic theory and encountered a difficulty. The lemma states:

Let $(X, \mathcal{B}, \mu)$ be an infinite non-atomic measure space, and let $T$ be an ergodic non-singular transformation on $X$. For $N \geq 1$ and $\lambda > 0$, there exists $E \in \mathcal{B}$ such that $\mu(E) = \lambda$ and $\{T^{-k}E\}_{k=0}^N$ are mutually disjoint.

Outline of the Solution provided: Since $\mu$ is non-atomic, we select $A \in \mathcal{B}$ appropriately. Then consider $E = T^{-N}A \setminus \bigcup_{k=0}^{N-1} T^{-k}A$.

From this outline, I managed to prove that $\{T^{-k}E\}_{k=0}^N$ are indeed mutually disjoint. However, I am struggling to understand how to ensure $\mu(E) = \lambda$ using an appropriate choice of $A$. According to Wacław Sierpiński's theorem on non-atomic measures, we can select $A$ such that $\mu(A) = \lambda'$ for any $\lambda' > 0$. But how can I choose $A$ in a way that guarantees $\mu(E) = \lambda$ exactly? Could you please help me solve this? Thank you for your time and assistance.

I was studying a lemma from my notes on ergodic theory and encountered a difficulty. The lemma states:

Let $(X, \mathcal{B}, \mu)$ be an infinite non-atomic measure space, and let $T$ be an ergodic non-singular transformation on $X$. For $N \geq 1$ and $\lambda > 0$, there exists $E \in \mathcal{B}$ such that $\mu(E) = \lambda$ and $\{T^{-k}E\}_{k=0}^N$ are mutually disjoint.

Outline of the solution provided: Since $\mu$ is non-atomic, we select $A \in \mathcal{B}$ appropriately. Then consider $E = T^{-N}A \setminus \bigcup_{k=0}^{N-1} T^{-k}A$.

From this outline, I managed to prove that $\{T^{-k}E\}_{k=0}^N$ are indeed mutually disjoint. However, I am struggling to understand how to ensure $\mu(E) = \lambda$ using an appropriate choice of $A$. According to Wacław Sierpiński's theorem on non-atomic measures, we can select $A$ such that $\mu(A) = \lambda'$ for any $\lambda' > 0$. But how can I choose $A$ in a way that guarantees $\mu(E) = \lambda$ exactly? Could you please help me solve this? Thank you for your time and assistance.