Let $(X_n,P_x)$ be a time homogeneous markov chain with one step transition kernel $\pi$. Denote the corresponding generator $$(\mathscr{L}f)(x) = (\pi f)(x)-f(x)=E_x[f(X_1)-f(X_0)]$$.
Then we know that $M_n^{[f]} = f(X_n)-\sum_{i<n} (\mathscr{L}f)(X_i)$ is a martingale w.r.t. $(\mathscr{F}_n^X)$ and $P_x$ for any $x$ and bounded measurable $f$.
Now, if the inequality $\mathscr{L}f\le -c$ holds for nonnegative measurable $f,c$, then why is the process $$M_n^{[f,c]}= f(X_n)+\sum_{i<n}c(X_i)$$ a non-negative supermartingale?
$\mathscr{L}f \le -c$ is equivalent to $c \le -\mathscr{L}f$. We need to show that $$E[f(X_n)+\sum_{i<n} c(X_i)|\mathscr{F}_{n-1}]\le f(X_{n-1})+\sum_{i<n-1}c(X_i).$$
We have $E[f(X_n)+\sum_{i<n} c(X_i)|\mathscr{F}_{n-1}]\le E[f(X_n)-\sum_{i<n} \mathscr{L}f(X_i)|\mathscr{F}_{n-1}]=f(X_{n-1})-\sum_{i<n-1} \mathscr{L}f(X_i).$ But then $c \le -\mathscr{L}f$, so I can't get the RHS bound above. How do we show this? I would greatly appreciate any help.
I think you can take the following expectation and use the inequality: $$\begin{aligned}E[M_{n}^{[f,c]}-(c(X_1)+c(X_2)+...+c(X_{n-2}))|\mathscr{F}_{n-1}]&=E[f(X_n)|\mathscr{F}_{n-1}]+c(X_{n-1})\leq\\ &\leq E[f(X_n)|\mathscr{F}_{n-1}]-(\mathscr{L}f)(X_{n-1})=\\ &=f(X_{n-1})\\ \end{aligned}$$ $$\implies E[M_{n}^{[f,c]}|\mathscr{F}_{n-1}]\leq f(X_{n-1})+(c(X_1)+...+c(X_{n-2}))=M_{n-1}^{[f,c]}$$