I'm dealing with the random forcing $\mathbb{P}=\mathbb{M}\mathbb{B}(X,\mu)$ defined in Kunen's 2013 book. Recall that the conditions of that forcing are the equivalence classes of positive measureble sets modulo null sets $[S]$ endowed with the order $[S]\leq [T]$ iff $S\setminus T\in \text{null}$.
I would like to show that if $A$ is a maximal antichain below a contidition $p$ then $\mu(p)=\sum_{q\in A}\mu(q)$. To prove this fact I've divided the proof in various steps:
- $\mu(\bigcup A)=\sum_{q\in A}\mu(q)$: Indeed, taking $B_n=\bigcup_{i\leq n} q_i$ we have a increasing sequence of measurable sets such that $\bigcup_n B_n=\bigcup A$, so $\mu(\bigcup A)=\lim \mu(B_n)$. Since $A$ is an antichain, its conditions are almost disjoint (i.e. modulo a null set) so $\mu(B_n)=\sum_{i=1}^n \mu(q_i)$. Finally we have $\mu(\bigcup A)=\sum_{q\in A}\mu(q)$.
- $\mu(p\setminus\bigcup A)=0$: Is this fact true? And, how can I use the notion of maximal antichain to prove it? I know that every condition below $p$ must be comparable with some element of $A$ but I'm not willing to take advantage of that information.
With these two facts in hand we can finally conclude that $\mu(p)=\mu(\bigcup A)=\sum_{q\in A}\mu(q)$. So, is anybody willing to give me some advice regarding to the second item?
Thanks in advance.
Suppose $\mu(p\setminus \bigcup A)>0$. Then we want to say that $[p\setminus\bigcup A]$ is a condition less than $p$. If so, we'd be done, since it is clearly incompatible with each element of $A$ (why?), so this would contradict the claim that $A$ was a maximal antichain.
So we just have to prove that $p\setminus\bigcup A$ is measurable. Good news: $p$, and each of the $q$s, are measurable! Bad news: an arbitrary union of measurable sets need not be measurable. Good news: what can we say about the cardinality of $A$? (HINT: Remember your previous question.)