I'm trying to prove the next:
Let $f,g\in L^{1}(E,\mathcal{E},\eta).$ Then $f=g$ $\eta-$a.s. i and only if $Mf=Mg$ $P-$a.s.
Here $M$ is a random measure.
The first direction follows because $M$ is monotone, so we have $f\leq g$ $\eta-$a.s. implies $Mf\leq Mg$\space $P-$ a.s. The other inequality is analogous.
For the other direction, is enough to prove that $Mf=0$ $P-$a.s. implies $f=0$ $\eta-$a.s.
I'm stuck in this. Because $Mf=0$ implies that $f=0$ $P-$a.s., right? I don't get the result that I desire.
Any kind o help is thanked in advanced.
Ok, the way I read this question I get the following: there's a probability space $(\Omega, \mathcal F, \mathsf P)$ and another one $(E, \mathcal E, \eta)$, a measurable stochastic kernel $M:\Omega\times \mathcal E\to [0,1]$ and we are asked about equivalence of $Mf = 0$ $\mathsf P$-a.s. and $f = 0$ $\eta$-a.s.
Since there seem to be no connection between $\mathsf P$ and $\eta$, that's not true, and an example is easy to give. Let's say $\Omega = E = [0,1]$ and they both are provided with the canonical Borel $\sigma$-algebras. Let's say that $M(x,A) = 1_A(x)$, that is it sends any point to exactly the same location, with no randomness, then $Mf = f$ and our statement becomes $f = 0$ $\mathsf P$-a.s. iff $f = 0$ $\eta$-a.s. which is obviously not true unless $\mathsf P$ and $\eta$ are the same.
That answer the OP as it is, though it's very likely that the OP is an ill-stated version of the question you really need to ask.