Stochastic kernel as linear operator

Question

Let $K$ be a stochastic kernel for a set $S$ equipped with a countably generated $\sigma$-Algebra $B(S)$, i.e.

$K:S\times B(S)\rightarrow [0,1]$

such that $K(\cdot,A)$ is a measurable function for all $A\in B(S)$ and $K(x,\cdot)$ is a probability measure for all $x\in S$.

Now $K$ operates linearly on $M_{1}(S)$, the space of probability measures on the measurable space $(S,B(S))$, by setting for all probability measures $\mu $

$K\mu := \int_S K(y,\cdot)\mu(dy)$.

Question: Is the converse also true, that each linear operator on $M_{1}(S)$, the space of probability measures on $(S,B(S))$, gives us a stochastic kernel again?

Each such linear operator O can be understood as mapping

$O:S \times B(S)\rightarrow[0,1]$

$\hphantom{O:}(x,A)\mapsto (O\delta_x)(A)$,

where $\delta_x$ is the Dirac-measure. So it basically boils down to the question if this last mapping is measurable in x, or if there are linear Operators on $M_{1}(S)$, so that the above mapping is not measurable in x.

I try it like this: Every probability measure $\mu$ on $(S,B(S))$ can be written as:

$\mu = \int_S \delta_y \mu(dy)$ (I hope this is right, but I cannot remember having seen such an argument before), then we have

$O\mu = O\int_S \delta_y \mu(dy)=\int_S O\delta_y \mu(dy)$ (last equality holds for finite sums because of the linearity of O, but I don't know why that is applicable to integrals).

This cannot be defined, if $O\delta_y$ is not measurable as a function of $y$, it then follows $O\mu$ is no probability measure. Contradiction.

Clarification: By linear operator $O$ I mean that for all $\mu,\nu\in M_{1}(S)$ and all $0\leq \alpha \leq 1$ we have $O(\alpha \mu + (1-\alpha)\nu)=\alpha O\mu + (1-\alpha)O\nu$ (thanks @ByronSchmuland for pointing this out).

Answer 1

The result is false in this generality.

Fix some $z\in S$ and define $O\mu=\mu_c+\mu_d(S)\delta_z$, where $\mu_d$ and $\mu_c=\mu-\mu_d$ are the discrete and continuous parts of $\mu$, respectively. Then $O$ satisfies your conditions, but cannot possibly be given by a kernel, if there is a non-discrete probability measure $\nu$ on $S$.

Why not?

If $O$ is given by the kernel $K$ then $O\mu=\int \mu(dx) K(x,\cdot)$. Setting $\mu=\delta_x$ shows $K(x,\cdot)=O\delta_x$.

If $\nu$ has no discrete part, then $O\nu=\nu$ but $\int\nu(dx) K(x,\cdot)=\delta_z\neq \nu.$

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In order for $O$ to be represented by a kernel you need two things:

1. $x\mapsto O\delta_x$ should be measurable from $S$ to $M_1(S)$.
2. $O\mu=\int\mu(dx) O\delta_x$ for every $\mu\in M_1(S)$.

My counterexample above satisfies 1, but not 2.