a question about a non-atomic probability space I am reading "Mathematical Foundations of the Calculus of Probability" by J. Neveu (the English translation). There is the following exercise with hints which I don't quite understand: Say we have a probability space ($\Omega$, $\mathbb{A}$, P) without atoms, then for every $a \in [0,1]$, there exists at least one set $A \in \mathbb{A}$ of probability $P(A)=a$.
Here is the beginning of the hint which I don't quite understand: Let $\tilde{B}$ be a maximal element of the subclass $\mathbb{B}$ of $\mathbb{A}|P$ consisting of those $\tilde{B}$ such that $P(\tilde{B}) \le a$, which subclass is inductive under inclusion.
The hint continues but I don't quite understand the statement here. Here $\mathbb{A}|P$ is the quotient space. In essence, the ensemble of equivalent classes of events such that $P(A \triangle B)=0$.
Because of that, aren't all classes in $\mathbb{B}$ disjoint? How does it then make sense to define an order under inclusion? I'm pretty sure I'm missing something here..
 A: After consulting Neveu's book, I don't think it would be fair to say that you're "missing something", since the book doesn't seem to provide a sufficiently clear exposition of how the Boolean operations on $\ \mathbb{A}\ $ get transferred to the quotient $\sigma$-algebra $\ \mathbb{A}/P\ $.  When Neveu says "inductive under inclusion", the word "inclusion" isn't referring to set-theoretical inclusion, but to an ordering $\ \preceq\ $ on the elements of $\ \mathbb{A}/P\ $ defined by
$$
X\preceq Y\ \ \text{ iff }\ P\big(A\setminus B\big)=0\ \text{ for all }\ A\in X, B\in Y,
$$
or, equivalently,
$$
X\preceq Y\ \ \text{ iff }\ A\subseteq B\ \text{ for some }\ A\in X, B\in Y\  ,
$$
or, if you like to define your relations as sets of ordered pairs,
$$
\preceq\ \triangleq\big\{(X,Y)\,\big|\,\forall A\in X,B\in Y:P\big(A\setminus B\big)=0\ \big\}\ ,
$$
where $\ X\preceq Y\ $ is then synonymous with $\ (X,Y)\in\ \preceq\ \ $.   Neveu defines the term "inductive" in relation to a partially ordered set on page viii.  Here, it means that $\ \mathbb{B}\ne\emptyset\ $, and every subset of $\ \mathbb{B}\ $ which is totally ordered with respect to $\ \preceq\ $ has an upper bound, so by Zorn's lemma $\ \mathbb{B}\ $ must contain a maximal element.
