combined with (C2) with $p=1$, does not guarantee stability.

When (C2) holds with $p=2$ it appears one has to do a little more to engineer a counterexample; specifically, we line up all the jumps out of $(-\infty , J)$ to amass at a fixed time $M$. Fix a large integer $M$, and define a time-inhomogeneous Markov chain by setting $X_0=0$, and, for $0 \le n \le M-1$, letting $$\P(X_{n+1} = X_n-1 \, | \, X_n > 0) = 1$$ $$\P(X_{n+1} = 0 \, | \, X_n = 0 ) = 1 - (M-n)^{-2}$$ $$\P(X_{n+1} = 2(M-n) \, | \, X_n = 0 ) = (M-n)^{-2}$$ Then it is easily verified that (C1) is again satisfied with $a=1$ and $J=0$. Furthermore, (C2) is satisfied with $p=2$ and $V=4$.

On the other hand, setting $A = \exp\left( - \sum_{i=1}^\infty 1/i^2 \right) > 0$, we compute that $$\E(X_M) =
\sum_{k=0}^M (M-k+1) \, \P (X_k > 0 \ {\rm and} \
 X_j = 0 \ {\rm for} \ j < k)$$ $$\ge \ \sum_{k=0}^{M} (M-k+1) \left( A / (M-k+1)^2 \right)$$ $$= \ \sum_{k=0}^{M} A / (M-k+1)
                \ = \ \sum_{j=1}^{M+1} A / j$$ which goes to infinity (like $A \log M$) as $M \to 