Lyapunov-type function in a non locally-compact space and boundedness of the average Set-up and question.
Let $\mathcal{X}$ be a complete separable metric space which is not locally-compact. Let $V: \mathcal{X} \to [0; +\infty]$ be a function and $(X_t)_{t\geq 0}$ a Markov process in $\mathcal{X}$ such that $P\{ V(X _t) < \infty \text{ for all } t \geq 0 \} = 1$. Let $g: \mathcal{X} \to \mathbb{R}$ be such that
$$
V(X _t) - \int ^t _0 g(X _s) ds \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \quad
\quad \quad \quad \quad \quad \quad \quad 
 (1)
$$
is a martingale, where
$$
g(x) \leq c -  V(x), \quad \text{ for all } x \in \{z \in \mathcal{X}: V(z) < \infty \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
$$
for some $c >0$.
Is it possible to deduce that 
$$
\limsup\limits _{t \to \infty} \mathbb{E} V(X _t) < \infty? 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
 \ \ \  \   \  \  \ \ \ \ \ \ \ \ \ \ \ \ \  \  \   \  \   \   \    \    \  \ \ \ \ \ \  \ (3)
$$
My thoughts. If $V$ was 'norm-like' in the sense  that $\{x \in \mathcal {X} : V (x) \leq r \}$ was compact for $r >0$ and $\mathcal{X}$ was locally-compact, we would get ergodicity from (1) and (2) and thus (3) would hold. I think that without $V$ being norm-like and local compactness of $\mathcal{X}$ ergodicity cannot follow from (1) and (2), however a weaker property like (3) should hold. I have an idea how to prove (3) if additionally $(X_t)_{t\geq 0}$ is assumed to be continuous a.s. (finding an ergodic birth-death process which would stochastically dominate $(V(X_t))_{t\geq 0}$), however without continuity assumption I am currently out of ideas. I did not find any similar result in the literature, but any related reference would be kindly appreciated.
Similar posts. The set-up is similar to this, however the assumptions and the questions are different.
Edit Edited the condition on $g$. The initial version was not well formulated and resulted in a triviality as was pointed out by Iosif Pinelis.
 A: The conditions $g\ge0$, (2), and $0\le V(X_t)<\infty$ almost surely (a.s.) imply $V(X_t)\le c$ a.s. (Indeed, on the event $\{V(X_t)<\infty\ \forall t\ge0\}$ -- which is of probability $1$, we have $0\le g(X_t)\le c-V(X_t)$ and hence $0\le c-V(X_t)$ and $V(X_t)\le c$.) So, 
$$\limsup_{t\to\infty}EV(X_t)\le c<\infty$$
if $c<\infty$. 
Added in response to the edit of the OP: Even when the condition $g\ge0$ is removed, the conclusion (3) holds. Indeed, letting $h(t):=Eg(X_t)$ and $v(t):=EV(X_t)$, and using the martingale condition, we have 
\begin{equation}
 v(t)-\int_0^t h(s)\,ds=E\Big(V(X_t) - \int^t_0 g(X_s)\, ds\Big)=EV(X_0),  
\end{equation}
whence $v'=h$. 
Also, by (2), 
\begin{equation}
 h(t)=Eg(X_t)\le c-EV(X_t)=c-v(t),
\end{equation}
whence $v'(t)+v(t)\le c$, or, for $u(t):=v(t)e^t$,
\begin{equation}
 u'(t)\le ce^t
\end{equation}
and hence $u(t)\le u(0)+ce^t$, that is, 
\begin{equation}
 EV(X_t)=v(t)\le u(0)e^{-t}+c,
\end{equation}
so that 
\begin{equation}
 \limsup_{t\to\infty}EV(X_t)\le c<\infty,
\end{equation} 
as desired. 
