Please bear with me; while the description of the setting looks complicated, the actual question in the yellow box below should be answerable with elementary measure-theoretic facts.
Let
- $(E,\mathcal E,\lambda),(E',\mathcal E',\lambda')$ be measure spaces
- $I$ be a finite set
- $p,q_i:E\to[0,\infty)$ be $\mathcal E$-measurable with $$\int p\:{\rm d}\lambda=\int q_i\:{\rm d}\lambda=1\tag1$$ for $i\in I$
- $\mu:=p\lambda$
- $\varphi_i:E'\to E$ be bijective and $(\mathcal E',\mathcal E)$-measurable with $$\lambda'\circ\varphi_i^{-1}=q_i\lambda\tag2$$ for $i\in I$
- $\sigma_{ij}':E'\times E'\to[0,\infty)$ be $\mathcal E'\otimes\mathcal E'$-measurable with $$\sigma_{ij}(x',y')=\sigma_{ji}(y',x')\;\;\;\text{for all }(i,x'),(j,y')\in I\times E'$$ and $$\sum_{j\in I}\int\lambda'({\rm d}y')\sigma'_{ij}(x',y')=1\;\;\;\text{for all }(i,x')\in I\times E'$$ and $$\sigma_{ij}(x,y):=\sigma'_{ij}\left(\varphi^{-1}_i(x),\varphi_j^{-1}(y')\right)\;\;\;\text{for }(i,x),(j,y)\in I\times E$$
- $w_i:E\to\mathbb R$ be $\mathcal E$-measurable with $$\left\{p\ne0\right\}\subseteq\left\{\sum_{i\in I}w_i=1\right\}$$
- $$\alpha_{ij}(x,y):=\left.\begin{cases}\displaystyle1\wedge\frac{w_j(y)p(y)q_i(x)}{w_i(x)p(x)q_j(y)}&\text{, if }w_i(x)p(x)q_j(y)>0\\1&\text{, otherwise}\end{cases}\right\}$$ and $$\tau_{ij}(x,y):=w_i(x)q_j(y)\sigma_{ij}(x,y)$$ for $(i,x),(j,y)\in I\times E$ and $k:=\sum_{i\in I}\sum_{j\in I}\alpha_{ij}\tau_{ij}$
- $$\tilde\kappa(x,B):=\int\lambda({\rm d}y)k(x,y)\;\;\;\text{for }(x,B)\in E\times\mathcal E$$
Are we able to show that $$\int\lambda({\rm d}x)\int\lambda({\rm d}y)k(x,y)k(y,x)<\infty\tag3;$$ at least under suitable integrability assumptions on $\sigma'$?
For example, if $I=\{1\}$, then (I dropped the indices for simplicity) $$k(x,y)k(y,x)\le q(x)q(y)|\sigma(x,y)|^2\;\;\;\text{for all }x,y\in E\tag4$$ and hence $$\int\lambda({\rm d}x)\int\lambda({\rm d}y)k(x,y)k(y,x)\le\int\lambda'({\rm d}x')\int\lambda'({\rm d}y')|\sigma'(x',y')|^2\tag5.$$
The general case looks quite complicated: $$k(x,y)k(y,x)=\sum_{i\in I}\sum_{i'\in I}\sum_{j\in I}\sum_{j'\in I}\alpha_{ij}(x,y)\tau_{ij}(x,y)\alpha_{i'j'}(y,x)\tau_{i'j'}(y,x)\tag6$$ for all $x,y\in E$ and \begin{equation}\begin{split}&\alpha_{ij}(x,y)\tau_{ij}(x,y)\alpha_{i'j'}(y,x)\tau_{i'j'}(y,x)\\&\;\;\;\;=\left.\begin{cases}\displaystyle w_i(x)q_j(y)\wedge\frac{w_j(y)p(y)q_i(x)}{p(x)}&\text{, if }p(x)>0\\w_i(x)q_j(y)&\text{, otherwise}\end{cases}\right\}\sigma_{ij}(x,y)\\&\;\;\;\;\;\;\;\;\left.\begin{cases}\displaystyle w_{i'}(y)q_{j'}(x)\wedge\frac{w_{j'}(x)p(x)q_{i'}(y)}{p(y)}&\text{, if }p(y)>0\\w_{i'}(y)q_{j'}(x)&\text{, otherwise}\end{cases}\right\}\sigma_{i'j'}(y,x)\\&\;\;\;\;\le w_i(x)q_j(y)\sigma_{ij}(x,y)w_{i'}(y)q_{j'}(x)\sigma_{i'j'}(y,x)\end{split}\tag7\end{equation} for all $x,y\in E$ and $i,i',j,j'\in E$.
Maybe we can estimate this and end up with an integrability assumption as in $(5)$, e.g. $$\sum_{i\in I}\sum_{j\in I}\int\lambda'({\rm d}x')\lambda'({\rm d}y')\sigma_{ij}(x',y')<\infty\tag8.$$