Let $h\in C^1(\mathbb R)$ such that $h'$ is Lipschitz continuous and $$L\varphi:=-h'\varphi'+\varphi''\;\;\;\text{for }\varphi\in C^2(\mathbb R).$$ Now, let $(X_t)_{t\ge0}$ be the unique strong solution of $${\rm d}X_t=-h'(X_t){\rm d}t+\sqrt2{\rm d}W_t\tag1,$$ where $(W_t)_{t\ge0}$ is a Brownian motion.
I've read (in this paper, below Assumption 2.4) that if $V:\mathbb R\to[0,\infty)$ with $$V(x)\xrightarrow{x\to\infty}\infty\tag2$$ and $a,d>0$ with $$LV\le-aV+d\tag3,$$ then $$\operatorname E\left[V(X_t)\mid\mathcal F_s^X\right]\le e^{-a(t-s)}V(X_s)+\frac da(1-e^{-a(t-s)})\tag4.$$ Why does $(4)$ hold?
Obviously, $(4)$ is an application of (the Itō formula and) a Gronwall-type lemma. Actually, it's precisely Theorem 6 here. However, in order to apply that theorem, we should need that the process $\left(\operatorname E\left[V(X_t)\mid\mathcal F_s^X\right]\right)_{t\ge0}$ is continuous. This shouldn't hold, unless $V$ is (continuous and) bounded (which would allow an application of Lebesgue's dominated convergence theorem). But $V$ is clearly assumed to be unbounded by $(2)$. So, what am I missing?
(Clearly, the authors of the paper are missing assumptions on $V$ anyway. In order for $(3)$ to make sense, $V$ needs to be twice differentiable (at least in some weak sense).
It appears to be implicit that $V$ is $C^2$, hence locally bounded. By Ito, $$ V(X_t) = e^{-a(t-s)}V(X_s)+{d\over a}(1-e^{-a(t-s)})+\int_s^t K_u\,du +M_t-M_s, $$ where $M$ is a martingale and $K_u=LV(X_u)=aV(X_u)-d\le 0$. Now take he conditional expectation.