I am wondering about a proof or reference to the following claim by Getoor; Let $(E,\mathscr E)$ and $(F,\mathscr F)$ be two measurable spaces. A (positive) {kernel} from $(F,\mathscr F)$ to $(E,\mathscr E)$ is a map $$k: F\times \mathscr E \to [0,\infty],\quad (x,E)\mapsto k(x,E)$$ such that $x\mapsto K(x,A)$ is $\mathscr F$ measurable for each fixed $A\in \mathscr E$ and $A\mapsto k(x,A)$ is a measure on $(E,\mathscr E)$ for each fixed $x\in F$. If $k$ is a bounded kernel from $(F,\mathscr F)$ to $(E,\mathscr E)$, then $$K: \mathcal{B}(\mathscr E) \to \mathcal B(\mathscr F),\quad K f(x)=K(x,f)=\int_E f(y) k(x,dy)$$ defines a bounded linear map (from bounded measurable to bounded measurable functions) with $$ (f_n)_{n\in \mathbb{N}} \subseteq \mathcal{B}(\mathscr E)^+ \uparrow f \in \mathcal{B}(\mathscr E) \Rightarrow K f_n \uparrow K f.$$ Claim: Conversely every bounded linear positive map $ \mathcal{B}(\mathscr E) \to \mathcal B(\mathscr F)$ that satisfies the above condition is given by a kernel
How do you go about showing something like this? Is it some Banach algebra characterisation theorem? We can define the composition of kernels as If $k$ is a kernel from $(F,\mathscr F)$ to $(E,\mathscr E)$ and $l$ is a kernel from $(G,\mathscr G)$ to $(F,\mathscr F)$, then the composition of $k$ and $l$ is a kernel $l\star k$ from $(G,\mathscr G)$ to $(E,\mathscr E)$ defined by $$l\star k (x,A)=\int_F k(y,A) l(x,dy)$$ for $x\in G$ and $A \in \mathscr E$. On the bounded kernels with the sup norm we get a submultiplicative norm. I'm guessing this is a Banach algebra with $$L\star K f(x)=\int_F f(y) l\star k(x, dy)=\int_F f(y) \int_F k(z,dy)l(x,dz)$$ But I'm not sure this is the right way to define the composition operator
For $x\in F$ and $A\in\mathscr{E}$ let $k(x,A)=K1_A(x)$. Since $K1_A$ is measurable by assumption, it suffices to check that $A\mapsto k(x,A)$ is a measure for every $x\in F$. For that purpose let $(A_n)$ be a disjoint sequence in $\mathscr{E}$ and $f_n=1_{\bigcup_{k=1}^n A_k}=\sum_{k=1}^n 1_{A_k}$. Then $f_n\nearrow 1_{\bigcup_{k=1}^\infty A_k}$, hence $$ \sum_{k=1}^n k(x,A_k)=Kf_n(x)\nearrow K1_{\bigcup_{k=1}^\infty A_k}(x)=k\left(x,\bigcup_{k=1}^\infty A_k\right). $$ Note that the space you constructed is not a Banach algebra because you cannot multiply arbitrary elements. In any case, even if $(E,\mathscr{E})=(F,\mathscr{F})$, I do not see how this result should follow from an abstract representation theorem for Banach algebras.