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Let $(E,\mathscr E)$ be a measurable space and denote by $\mathrm b\mathscr E$ the space of all Borel measurable bounded functions $f:E\to\mathbb R$. On this space the partial order is given by $$ f\leq g \quad \Leftrightarrow \quad f(x)\leq g(x)\text{ for all }x\in E $$ and let us use the notation $\mathrm b\mathscr E_{\geq f} = \{g\in \mathrm b\mathscr E:g\geq f\}$. Let $A$ be the monotone operator on the space$\mathrm b\mathscr E$, which is not necessarily linear, and let $\mathrm{fix}(A)$ be the set of fixpoints of the operator $A$: $$ \mathrm{fix}(A) = \{g\in \mathrm b\mathscr E:g = Ag\}. $$ Let $f_0\in \mathrm b\mathscr E$ be such that $f_0\leq Af_0$ and construct the sequence $f_{n} = A^n f_0$. This sequence converges point-wise non-decreasingly to a measurable function $f:E\to\mathbb R$. One can prove that whenever $f\in \mathrm{fix}(A)$ it holds that $f = \min\left\{\mathrm{fix}(A)\cap \mathrm b\mathscr E_{\geq f_0}\right\}$.

I guess that it is already known, and maybe a consequence of a more general result. I am looking for the classical reference to this fact.

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Related to this question of mine – Ilya Jun 1 '12 at 17:57
    
I don't know a reference. $Ag=g$ and $g\geq f_0$ implies by monotonicity $g\geq\lim A^nf_0=f$, so $f$ is a lower bound for $\{\mathrm{fix}(A)\cap \mathrm b\mathscr E_{\geq f_0}\}$ even if it isn't a member. – Jonas Meyer Jun 2 '12 at 2:34
    
@JonasMeyer yes, that's the way it is proved. The result is quite useful and I read it in the literature in our field always formulated for each particular case, so I decided to generalize it. However I expect the fact to be known, but don't know where to look for it. E.g. The Banach space structure seems to be unnecessary here, and maybe it can be just formulated for posets – Ilya Jun 2 '12 at 8:27

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