If a process is natural increasing, does that imply that the stopped process is natural increasing? Let $A = \{A_t\}_{t\geq 0}$ be a right-continuous stochastic process on the filtration $\mathcal F = \{\mathcal{F}_t\}_{t\geq 0}$ and let $T$ be a stopping time. Suppose also that $A_t$ is natural increasing; that is, $A_t$ is increasing and $$E\int_0^t M_sdA_s = \int_0^t M_{s^-}dA_s$$ for any bounded, right-continuous martingale $\{M_s\}_{s\geq 0}$. Is it true that the stopped process $\{A_{T\wedge t}\}_{t\geq 0}$ is natural increasing?

I found this proof: $$\int_0^t M_s dA_{T\wedge s} = \int_0^t M_{T\wedge s} dA_s - M_T(A_t - A_{T\wedge t}),$$ as one can verify by checking the cases ($T > t$, $T \leq t$). Applying this identity to $\int_0^t M_{s^-} dA_{T\wedge s}$ as well yields the result.
