Ensuring the measure condition $\mu(E) = \lambda$ in a lemma: need some clarification regarding the selection of $A$

Question

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?

Answer 1

If $\mu(E)\geq\lambda$ you can choose a subset $E'$ with $\mu(E')>\lambda$. The following exhaustion argument ensures you can have an $E$ with $\mu(E)>\lambda$.

Do as you said and get an $E$. If $\mu(E)<\lambda$, using the fact that $\uplus_{j=0}^{n-1}T^{-j}E$ has finite measure and $\mu(X)>0$ you can find a $B$ of positive measure such that $\cup_{j=0}^{N-1}T^{-j}B \cap\left(\uplus_{j=0}^{n-1}T^{-j}E\right)=\emptyset.$ Now let $E'=B\setminus \cup_{j=0}^{N-1}T^-jB$, and setting $E_2=E\cup E'$ you get a new tower with a base of strictly larger measure. Continue this way until you found a tower with a base set whose measure is greater than $\lambda$.