I have two Markov processes $X$ and $Y$ such that $X$ is in fact deterministic and bounded and $Y$ has values on a rather complicated space but for the purposes of this question let's say that the space is $\mathbb{N}_0$ since the state space is countable. I have a bounded continuous function $f$ which I want to show is a subduality function of $X$ and $Y$, i.e. that $$\mathbb{E}f(X_0,Y_t) \ge \mathbb{E}f(X_t,Y_0)$$ for any deterministic $X_0$ and $Y_0$ and $t \ge 0$.
I know that $$G^Yf(x,y)\ge G^Xf(x,y)$$ for all $x$ and $y$, where $G^X$ and $G^Y$ are the generators of $X$ and $Y$, respectively (in the equation above, $G^X$ is assumed to always act on the first variable and $G^Y$ on the second). Using the Theorem 4.11. in Chap. 4 of Ethier and Kurtz's Markov Processes, one can get that
$$\mathbb{E}[f(X_t,y)-f(x,Y_t)]=\mathbb{E}[\int_0^t(G^Xf(X_s,Y_{t-s})-G^Yf(X_s,Y_{t-s}))ds]$$
which would be enough for my conclusion except that to be able to use that Theorem I need to show that $$\mathbb{E}[\int_0^t G^Yf(X_s,Y_{t-s}))ds] \in \mathbb{R}$$ and $$\mathbb{E}[\int_0^t G^Xf(X_s,Y_{t-s}))ds] \in \mathbb{R}$$ for the proof to hold.
I managed to show that the first integral is real using the Martingale problem for $G^Y$ and the particular properties of $X$ (most importantly, that it is bounded and deterministic). Since $G^Yf(x,y)\ge G^Xf(x,y)$ for any $x$ and $y$, I also know that the second integral is not infinite. I might be able to show that it is not equal to $-\infty$ using again the exact way $Y$ looks, but it all seems rather clumsy to me.
Isn't there a simpler (and more general) way to show this? I know that $f(X_t,y)-f(X_0,y)-\int_0^tG^Xf(X_s,y)ds$ is a martingale for any $y$ and similarly with $Y$ and $G^Y$. Shouldn't that somehow be enough to show that the two integrals/expected values are real even without any particular knowledge of $X$ and $Y$, using perhaps just the fact that $G^Yf(x,y)\ge G^Xf(x,y)$? I feel like there should be a simple way to show that and that I am missing something (almost) obvious, hence my question.