Let $\mathscr X$ be a complete separable metric space and $\mathbb B$ be the Banach space of all real-valued bounded measurable functions on $\mathscr X$. The partial order on this space is introduced by $ f\leq g \text{ iff }f(x)\leq g(x)\text{ for all }x\in \mathscr X. $ The operator $\mathscr A:\mathbb B\to\mathbb B$ is called monotone if $f\leq g$ implies $\mathscr Af\leq \mathscr Ag$, such operator is not necessary linear. Let us consider the function $f_0\in \mathbb B$ such that $\mathscr Af_0\geq f_0$ and construct the sequence $f_{n+1} = \mathscr A f_n$. Clearly, for any fixed $x\in \mathscr X$ the limit $\lim\limits_{n}f_n(x)$ exists (though it may be infinite) and the convergence is monotone.
Let us assume that for any $x\in\mathscr X$ the limit is finite and denote it by $f(x)$. Is it true that $ f = \mathscr Af\quad? $