I'm studying the following theorem in (Schindler, 2014: Set Theory Exploring Independence and Truth), p. 178-180:
Theorem 9.16 (Mokobodzki) No rapid filter F $\subset$ ${}^\omega 2$ is Lebesgue measurable.
Proof (sketch):
Let $\epsilon_n = \frac{1}{2n+1}$ for n < $\omega$. By Lemma 9.14 (A non-trivial filter on $\omega$ which extends the Fréchet filter and is Lebesgue measurable is small.) one can write
$F \subset (I, J)$, where for every $n < \omega$,
$$μ(\lbrace {}^\omega 2: x \restriction_{I_n} \in J_n \rbrace) < \epsilon_n.$$
For $n<\omega$, let $$J_n^{\star} = \lbrace s\in J_n: \forall t \in {}^{I_n}2 (\forall k\in I_n s(k) \leq t(k) \rightarrow t \in J_n)\rbrace.$$
One can show that $F\subset (I, J^{\star})$.
Then one defines $$\blacklozenge(n) = min \lbrace \overline{\overline{\lbrace k\in I_n: s(k) = 1 \rbrace}} : s \in J_n^{\star} \rbrace$$ and
$$J_n^{\star,min} = min \lbrace s\in J_n^{\star}:\overline{\overline{\lbrace k\in I_n: s(k) = 1 \rbrace}} = \blacklozenge(n) \rbrace.$$
We have $$\blacklozenge(n) \geq n+1.$$
Then one defines $f: \omega \rightarrow \omega$, $$f(n) = max\lbrace\lbrace max(k): s(k) = 1\rbrace: s \in J_n^{\star, min}\rbrace$$ for $n < \omega$.
As $F$ is rapid, there is $b\in F$ such that
$\forall n< \omega$ $$\overline{\overline{\lbrace k: b(k) = 1 \rbrace \cap f(n)}} \leq n.$$
$b$ must be element of $(I, J^{\star})$ but this leads to a contradiction.
Hence one can deduce that $F$ is not rapid.
Questions:
From the definition of $J_n^{\star}$ one can deduce: if $s \in J_n^{\star}$ and $s(k) \leq s'(k) \forall k \in I_n$ then $s' \in J_n^{\star}$. Is this intended for the rest of the argumentation? Don't we need $m := |I_n| − \blacklozenge(n)$ for the rest of the argumentation?
Why do we define the function f in this way and how do we find the final contradiction that $b$ is not in $(I, J^{\star})$?
I can't follow Schindler's argumentation concerning my last question: assume $b \in (I, J^{\star})$ and that $b \restriction_{I_n} \in J_n^{\star}$ for $n < \omega$. As $\blacklozenge(n) \geq n+1$ we get $\lbrace k \leq max(I_n): b(k) = 1 \rbrace \subset I_n$ and has maximum $f(n)$. Consequently, there is at most one such $n < \omega$.
Thank you very much in advance for any help regarding these questions. Please excuse that I post my question here although it's not directly a research question. I spent a lot of time with it and it's not so easy to find other resources.
