Existence of a set $A ∈M$ with each $m(A ∩ E_n) > 0$ and each $m(E_n \setminus A) > 0$?

Question

The following is an exercise from Bruckner's Real Analysis:

Let ${E_n}$ be a sequence of measurable sets in a measure space $(X, M, m)$ with each $0<m(E_n) < ∞$. When is it generally possible to select a set $A ∈M$ with each $m(A ∩ E_n) > 0$ and each $m(E_n \setminus A) > 0$?

I have no idea how to approach the problem at all . A useful hint also would be great, thanks!

Answer 1

You can find the definition of atoms here

So, given $(X,M,m)$ a measure space and $A\in M$, then $A$ is not an atom if $m(A)=0$ or if there is $B \in M$ such that $B \subseteq A$ and $0<m(B)<m(A)$. Note that in this case it is also immediate that $m(A \setminus B) > 0$

Now let us consider:

Let ${E_n}$ be a sequence of measurable sets in a measure space $(X, M, m)$ with each $0<m(E_n) <\infty$. When is it generally possible to select a set $A \in M$ with each $m(A \cap E_n) > 0$ and each $m(E_n \setminus A) > 0$?

Answer: when $E_n$ are disjoint sets and each $E_n$ is not an atom.

*Proof: Since each $E_n$ is not an atom and $m(E_n) > 0$, we have that , for each $n$, there is $A_n\in M$ such that $A_n\subset E_n$ and $$0<m(A_n)<m(E_n)$$ Let $A = \bigcup_n A_n$.

Since $E_n$ are disjoint, and $A_n\subset E_n$, we have $A\cap E_n =A_n\cap E_n = A_n$ and $E_n \setminus A = E_n \setminus A_n $.
So,

1. $m(A \cap E_n) = m(A_n) > 0$ and

2. $m(E_n \setminus A) = m(E_n \setminus A_n)>0$.