If a measure takes infinitely many values then there is a disjoint sequence of sets of positive measure I want to prove that if a measure takes infinitely many values then there is a disjoint sequence of sets of positive measure. The notes I'm reading says there is a sequence $$0<\mu(Y_1)<\mu(Y_2)<\dots$$ and that we can construct the sequence $(A_n)$ from $(Y_n)$ recursively. The first idea was to define $$A_n=Y_n\setminus(Y_1\cup\dots\cup Y_{n-1})$$ but by considering a particular example I saw that it didn't work.
Edit : From the comments I realized that I cannot extract such a sequence either.
 A: Let $(S,\Sigma,\mu)$ be a measure space.
Let us start with a definition and some remarks.

We say $A$ is an finite atom if $0<\mu(A)<+\infty$ and, for every $B\subseteq A$, $\mu(B)=0$ or $\mu(B)=\mu(A)$

Note that if $A_1$ and $A_2$ are finite atoms, then either $\mu(A_1 \Delta A_2)=0$ or $\mu(A_1 \cap A_2)=0$. (It can be easily proved by considering the  $\mu(A_1 \cap A_2)=0$ or $\mu(A_1 \cap A_2)=\mu(A_i)$, $i=1,2$).
If $A_1$ and $A_2$ are finite atoms and $\mu(A_1 \Delta A_2)=0$  we say that $A_1$ and $A_2$ are equal $a.e.$ (or almost equal). We write $A_1=A_2$ $a.e.$ (the $a.e.$ comes from the fact that the $\chi_{A_1} = \chi_{A_2}$ $a.e.$).
If $A_1$ and $A_2$ are finite atoms and $\mu(A_1 \cap A_2)=0$  we say that $A_1$ and $A_2$ are disjoint $a.e.$ (or almost disjoint).
Now let us prove that

If $\mu$ takes infinitely many values, then there is a disjoint sequence of sets of positive measure.

Let us prove by cases.
Case 1.  There are infinitely many almost disjoint finite atoms. In this case there is a sequence $\{A_i\}_i$ of almost disjoint finite atoms. In this case it is clear that $\mu$ takes infinitely many values (just consider $\mu(A_1)$, $\mu(A_1 \cup A_2)$ , ...).
Let us prove that there is a disjoint sequence of sets of positive measure.
For each $i$, take
$$E_i= A_i \setminus \left ( \bigcup_{j>i} (A_j\cap A_i) \right )$$
It is easy to see $\{E_i\}_i$ is a sequence of disjoint finite atoms. So $\{E_i\}_i$ is a disjoint sequence of sets of positive measure.
Case 2. There are finitely many almost disjoint finite atoms $A_1, \dots A_n$ and, for any $B\in \Sigma$, if $B \subset S\setminus  \bigcup_{i=1}^n A_i$, then $\mu(B)=0$ or $\mu(B)=\infty$. In this case $\mu$ has only finitely many values. The result is vacuously true.
Case 3. There are finitely many almost disjoint finite atoms $A_1, \dots A_n$ and, there is $B\in \Sigma$, $B \subset S\setminus  \bigcup_{i=1}^n A_i$, such that $0<\mu(B)<\infty$. In this case $B$ is not a finite atom and it does not contain any finite atom. Take $B_1=B$. There is $B_2 \subset B_1$  such that $0<\mu(B_2)<\mu(B_1)$. Since $B_2$ is not a finite atom, there is $B_3 \subset B_2$, such that $0<\mu(B_3)<\mu(B_2)<\mu(B_1)$.
So, by induction, we have a sequence $\{B_i\}_i$ such that
$$ B_1 \supset B_2 \supset B_3 \supset \cdots$$
and
$$\infty > \mu(B_1) > \mu(B_2) > \mu(B_3) > \cdots $$
So it is clear that, in this case, $\mu$ takes infinitely many values.
Let us prove that there is a disjoint sequence of sets of positive measure.
Define, for all $i$, $E_i=B_i\setminus B_{i+1}$.  It is clear that $\{E_i\}_i$ is a sequence of disjoint sets. And, for all $i$, we have
$$\mu(E_i) = \mu(B_i\setminus B_{i+1})= \mu(B_i)-\mu(B_{i+1})>0$$
So $\{E_i\}_i$ is a disjoint sequence of sets of positive measure.
This completes the proof.
