In the proof of monotone convergence I have seen I got confused by the second $\leq $ inequality in:
$0\leq E[(g-g_{n})1_{B}]=E[(f-f_{n})1_{B}]\leq 0$
Let $(X_{n})_{n}$ be $L^{1}$ random variables and $(f_{n})_{n}$ be a sequence of their respective representatives such that $f_{n} \leq f_{n+1}$ and $f_{n} \to f$ a.s.
Next let $(g_{n})_{n}$ be the sequence of respective representatives of the $(E[X_{n}\vert \mathcal{G}])_{n}$
It then shows that for any $n \in \mathbb N$: $P(A_{n})=0$ where $A_{n}:=\{ g_{n} > g_{n+1}\}$
It then defines $B:=\{\sup\limits_{n \in \mathbb N}g_{n}<g\}\cap \bigcap\limits_{n \in \mathbb N} A_{n}^{c}$
and then states without explanation that: $0\leq E[(g-g_{n})1_{B}]=E[(f-f_{n})1_{B}]\leq 0$
I understand the first inequality but not the second. I might be missing something obvious.