Let $(\Omega,\mathfrak A,P)$ be a probability space, $\Theta:\Omega\to\Omega$ be $(\mathfrak A,\mathfrak A)$-measurable with $P=P\circ\Theta^{-1}$ and $$A_n:=\frac1n\sum_{i=0}^{n-1}F\circ\Theta^i\;\;\;\text{for }F\in\mathcal L^1(P).$$
Let $F\in\mathcal L^1(P)$. I'm trying to understand the following proof of the Maximal ergodic theorem, $$\operatorname E\left[F;\max_{1\le i\le n}A_iF\ge0\right]\ge0\tag1$$ for all $n\in\mathbb N$:
First of all, shouldn't $$M_{n-1}\circ\Theta=M_n-F\tag2$$ hold everywhere (not only on $B$)? And what's the point of taking the positive part $x^+:=\max(x,0)$? It should clearly hold $$X=M_n-M_{n-1}\circ\Theta\ge M_n-M_n\circ\Theta\tag3,$$ since $M_n$ is (pointwisely) a maximum over a greater set than the set over which the maximum in $M_{n-1}$ is taken ... What am I missing?
It's not the case that $M_{n-1}\circ\Theta = M_n-F$ holds everywhere. Let's take $n=2$. Then $M_2 = \max(F,F+F\circ\Theta)$ and $M_1\circ\Theta = F\circ\Theta$. Then $M_2 = F+M_1\circ\Theta$ if and only if $F\circ\Theta \ge 0$. This is the point of taking positive part $x^+ = \max(x,0)$.