# An integral with respect to a coupling for a Markov chain

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.

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The idea is the following: we approximate $V$ by simple functions. Let $V_N:=\sum_{j=1}^Na_j\chi_{A_j}$ a simple function such that $0\leq s_N\leq V$. We have \begin{align} \int_{X^2}V_N(x')B(x,y,x',y')&=\sum_{j=1}^Na_j\int_{X^2}\chi_{A_j}(x')\chi_X(y')B(x,y,x',y')\\ &=\sum_{j=1}^Na_jB(x,y,A_j,X)\\ &=\sum_{j=1}^Na_jP(x,A_j)\\ &=\int_X V_N(x')P(x,x')\\ &\leq \int_X V(x')P(x,x')\\ &\leq aV(x)+b. \end{align} Then we can choose $V_N$ converging almost everywhere to $V$, and we use Fatou's lemma. This yields $$\int_{X^2}V(x')B(x,y,x',y')= \int_{X^2}\liminf_{N\to +\infty}V_N(x')B(x,y,x',y')\\ \leq \liminf_{N\to +\infty}\int_{X^2}V_N(x')B(x,y,x',y')\leq aV(x)+b.$$ The same job for $\int_{X^2}V_N(y')B(x,y,x',y')$ gives the result.