Let
- $(\Omega,\mathcal A,\mu)$ be a measure space
- $(E,\mathcal E)$ be a measurable space and $$\mathcal E_b:=\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\}$$ be equipped with the supremum norm
$f:\Omega\to\mathcal E_b$ is called pointwise $\mu$-integrable if $$\Omega\to\mathbb R\;,\;\;\;\omega\mapsto(f(\omega))(x)\tag1$$ is $\mu$-integrable for all $x\in E$ and $$\int(f(\omega))(x)\:\mu({\rm d}\omega)=g(x)$$ for some $g\in\mathcal E_b$. In that case, $$\int f\:{\rm d}\mu:=g.$$
Let $f:\Omega\to\mathcal E_b$ be pointwise $\mu$-integrable and $L\in\mathfrak L(\mathcal E_b)$. Are we able to show that $Lf$ is pointwise $\mu$-integrable and $$\int Lf\:{\rm d}\mu=L\int f\:{\rm d}\mu?\tag2$$
This property is well-known for the usual Bochner integral of an $\mathcal E_b$-valued function (where $\mathcal E_b$ is equipped with the supremum norm). But here I even fail to argue why $$\Omega\to\mathbb R\;,\;\;\;\omega\mapsto(Lf(\omega))(x)\tag3$$ is $\mathcal A$-measurable for all $x\in E$ ...
EDIT: Maybe we cannot show the claim for arbitrary $L$. Since I'm particilar interested in this case, please feel free to assume that $L$ is a Markov kernel on $(E,\mathcal E)$ acting on $\mathcal E_b$ via $$(Lg)(x)=\int L(x,{\rm d}y)g(y)\;\;\;\text{for }x\in E\text{ and }g\in\mathcal E_b.$$ Now, I guess we need that $$\Omega\times E\to\mathbb R\;,\;\;\;(\omega,y)\mapsto(f(\omega))(y)\tag4$$ is $\mathcal A\otimes\mathcal E$-measurable and $$\int L(x,{\rm d}y)\int\mu({\rm d}\omega)|(f(\omega))(y)|<\infty\tag5$$ (do we explicitly need to assume this or are we even able to show that it generally holds?) to obtain $\mathcal A$-measurability of $$\Omega\to\mathbb R\;,\;\;\;\omega\mapsto (Lf(\omega))(x)\tag6$$ (by Fubini's theorem) for all $x\in E$. Now, \begin{equation}\begin{split}\left(L\int f\:{\rm d}\mu\right)(x)&=\int L(x,{\rm d}y)\left(\int f\:{\rm d}\mu\right)(y)\\&=\int L(x,{\rm d}y)\int\mu({\rm d}\omega)(f(\omega))(y)\\&=\int\mu({\rm d}\omega)\int L(x,{\rm d}y)(f(\omega))(y)\\&=\int\mu({\rm d}\omega)(Lf(\omega))(x)\\&=\left(\int Lf\:{\rm d}\mu\right)(x)\end{split}\tag7\end{equation} for all $x\in E$.