I have been studying transition kernels recently from "Probability and Stochastics" by Erhan Cinlar. To avoid any ambiguity, a transition kernel in my textbook is defined as a mapping \begin{align*} K \colon E \times \mathcal{F} & \longrightarrow \mathbb{R}_+ \\ (x,B) & \longmapsto K(x,B) \end{align*}
such that $\forall \; B \in \mathcal{F} \; \; $ the mapping $x \longmapsto K(x,B)$ is an $\mathcal{E}$-measurable and positive function, and $\forall x \in E$ the mapping $B \longmapsto K(x,B)$ defines a measure on $(F,\mathcal{F})$.
My textbook defines the kernel to be bounded if the mapping $x \longmapsto K(x,F)$ is bounded, and finite if the measure $B \longmapsto K(x,B)$ is finite for every $x \in E$.
My issue
If the kernel is bounded, i.e., $\forall B \in \mathcal{F} \; $ the mapping $x \longmapsto K(x,B)$ is bounded, then $F \in \mathcal{F} \implies $ $\forall x \in E, \; K(x,F) < \infty$, i.e., $\forall x \in E \;$ the measure defined by $B \longmapsto K(x,B)$ is a finite measure.
Conversely, if the kernel is finite, i.e. $\forall x \in E \; $ the mapping $B \longmapsto K(x,B)$ is finite, this means that $K(x,F) < \infty \; \forall x \in E$, i.e., the kernel is bounded.
So, in my view, the two definition stress two different concepts, but are in fact equivalent. If my understanding is correct, there would be no issue, but an exercise in the book shows a result for bounded kernels and requires me to extend the result to finite kernels, which suggests that the two definitions are in fact not equivalent.
Any help would be much appreciated.
I am posting a counterexample of a finite kernel that is not bounded, in case it can be helpful to anyone.
Consider $\big( E, \mathcal{E} \big) = \big( \mathbb{N}, 2^{\mathbb{N}} \big)$, and $\big( F,\mathcal{F} \big)$ arbitrary.
Consider a transition kernel
\begin{align*} K \colon \mathbb{N} \times \mathcal{F} & \longrightarrow \mathbb{N} \end{align*}
such that $K(n,F) = n$.
Then, $K(n,F) = n < \infty \; \forall \, n \in \mathbb{N}$, i.e., $K$ is finite.
But the mapping $n \longmapsto K(n,F)=n$ is not bounded, i.e., $K$ is not bounded.
On the contrary, it looks to me that a finite kernel is also bounded.