Let $(X,\mathcal{B}(X))$ be a measurable space with Borel algebra and $P\colon X\times \mathcal{B}(X) \rightarrow \left[0,1 \right]$ a stochastic kernel. We assume that $X$ is a separable metric space which ensures that $\mathcal{B}(X^2)=\mathcal{B}(X)\otimes \mathcal{B}(X)$.
A stochastic kernel $B\colon X^2 \times \mathcal{B}(X^2) \to \left[0,1 \right]$ is said to be a coupling for $P$ if $B(x,y, A\times X)= P(x,A)\;\;\mbox{and}\;\; B(x,y, X \times A)= P(y,A), $ for $x,y \in X$ and $A\in \mathcal{B}(X)$.
Let $V:X \rightarrow \mathbb{R}^+$ be a measurable function. We know that $\int_{X} V(x')\, P(x,dx')\leq aV(x)+b.$
How to show that for every coupling $B$ for $P$ we have $\int_{X^2}(V(x')+ V(y'))\, \tag{1} B(x,y,dx' \times dy')\leq a(V(x)+V(y))+2b\;?$
It is clear that if $B(x,y,\cdot)=P(x,\cdot) \otimes P(y,\cdot)$ and $B$ is measurable with respect to $(x,y)$ it is a coupling for $P$ and $(1)$ holds. However I don not see why it holds for an arbitrary $B$. I would be greatful for any hints.