Consider two measurable spaces: $\big( E, \mathcal{E} \big)$ and $\big( F, \mathcal{F} \big)$.
Consider a function $g \colon E \longrightarrow \mathbb{R_+}$ that is $\mathcal{E}$-measurable.
Consider a function $f \colon E \times F \longrightarrow \mathbb{R_+}$ that is $\mathcal{E} \otimes \mathcal{F}$-measurable, where $\mathcal{E} \otimes \mathcal{F}$ denotes the product $\sigma$-algebra of $E \times F$.
Consider a transition kernel $K \colon E \times \mathcal{F} \longrightarrow \bar{\mathbb{R}}_+$
My textbook ("Probability and Stochastics" by Erhan Cinlar, for reference) shows that the function
\begin{align*} Kg \, \colon E & \longrightarrow \bar{\mathbb{R}}_+ \\ x & \longmapsto Kg(x)=\int_FK(x,dy)g(y) \end{align*}
is $\mathcal{E}$-measurable and positive whatever the kernel $K$.
It then shows that the function
\begin{align*} Kf \, \colon E & \longrightarrow \bar{\mathbb{R}}_+ \\ x & \longmapsto Kf(x)=\int_FK(x,dy)f(x,y) \end{align*}
is $\mathcal{E}$-measurable and positive provided that $K$ is $\Sigma$-finite, i.e.,
$\forall \, x \in E \;$ the measure on $\big( F, \mathcal{F} \big)$ defined by $B \longmapsto K(x,B)$ is finite, or in other words $K(x,F) < \infty \; \forall \, x \in E$.
Precisely, the proof for the latter result is first done under the assumption that $K$ is bounded by means of a monotone class argument, and then the result is extended to $\Sigma$-finite kernels upon observing that boudned kernels are also finite.
As I tried to prove the theorem myself at first, I followed the steps below. They lead me to the $\mathcal{E}$-measurability of $Kf$ without the assumption that $K$ is $\Sigma$-finite, so there must be something wrong in my approach. It is as follows:
My approach
Consider $f=\mathbb{I}_{A \times B}$. Then
\begin{equation} Kf(x)=\int_FK(x,dy)\mathbb{I}_{A \times B}(x,y)=\int_FK(x,dy)\mathbb{I}_A(x)\mathbb{I}_B(y)=\mathbb{I}_A(x)\int_FK(x,dy)\mathbb{I}_B(y)=\mathbb{I}_A(x)K(x,B) \end{equation}
which is $\mathcal{E}$-measurable and positive as the product of two $\mathcal{E}$-measurable and positive functions (recall that $K(x,B)$ is $\mathcal{E}$-measurable and positive by definition).
Now consider $f=\sum_{n=1}^{N}a_n\mathbb{I}_{A_n \times B_n}$, with $a_n \geq 0 \; \forall \, n$. It is $\mathcal{E} \otimes \mathcal{F}$-measurable, positive, and simple, and it is easy to see that
\begin{equation} \int_F K(x,dy) \sum_{n=1}^{N}a_n\mathbb{I}_{A_n \times B_n} = \int_F K(x,dy) \sum_{n=1}^{N}a_n \mathbb{I}_{A_n}(x) \mathbb{I}_{B_n}(y) = \\ \sum_{n=1}^{N} a_n \mathbb{I}_{A_n}(x) \int_F K(x,dy) \mathbb{I}_{B_n}(y) = \sum_{n=1}^{N} a_n \mathbb{I}_{A_n}(x) K(x,B_n) \end{equation}
and $\mathbb{I}_{A_n}(x) K(x,B_n)$ is $\mathcal{E}$-measurable and positive $\forall \, n=1,..,N$
Now, consider an arbitrary $\mathcal{E \otimes F}$-measurable and positive $f$. We know that $\exists \big( f_n \big)_{n=1}^{\infty}$ simple and $\mathcal{E}$-measurable and positive s.t. $f_n \nearrow f$. So, by monotone convergence theorem with respect to the measure $K$,
$Kf(x)=\int_F K(x,dy) f(x,y) = \lim_{n \rightarrow \infty}\int_F K(x,dy) f_n(x,y)$
so it is $\mathcal{E}$-measurable and postive as it is the limit of a sequence of $\mathcal{E}$-measurable and positive functions $\big( Kf_n \big)_{n=1}^{\infty}$.
My approach must be wrong: my idea of where this could be the case
My approach has to be wrong because, if it were correct, it would extend the boundaries of the theorem. I believe that my approach is correct as long as the function $f$ is simple, but applying the monotone convergence theorem only tells me that $Kf$ is well defined for any $x \in E$.
One possibility could be: Since the measure defined by $B \longmapsto K(x,B)$ varies according to $x$, if it varies "not as nicely" as it would do if K were $\Sigma$-finite, then $Kf$, although well-defined over $E$, might be non-measurable with respect to $\mathcal{E}$. I do not really expect this to be the case, otherwise $\Sigma$-finiteness of $K$ should have been required even when $f$ is a function of one variable.
Another possibility: More simply, Monotone convergence theorem cannot be applied to functions of multiple variables, at least if the integration happens with respect to only one variable.
Even if any of these is correct, I should at least prove that the resulting $Kf$ is not necessarily $\mathcal{E}$-measurable, perhaps by finding a counterexample. Any help would me more than appreciated as I have been stuck on this for long.