Suppose I have a family of martingales continuous martingales $(X^\alpha)_{\alpha \in (0,\infty)}$, that is $X^\alpha = (X^\alpha_t)_{t \geq 0}$ is a continuous martingale, for each $\alpha$.
Are there any general theorems that state when the limiting case $\alpha \to 0$ is also a continuous martingale, assuming sufficient continuity etc.?
In my example I want to show that we may take the limit $\lambda \to 0$ in the family of martingales $$ \int_0^t\exp(-\lambda s)(\lambda - A)f(X_s)ds + \exp(-\lambda t)f(X_t) $$ where $A$ is a markov generator, and $f$ continuous on some compact state space $E$.
Thanks
The condition to be a martingale is $E[X_t \mid \mathcal{F}_s] = X_s$. Since conditional expectation is continuous with respect to $L^1$ convergence, a sufficient condition for $X^0$ to be a martingale would be to have $X^a_t \to X_t^0$ in $L^1$ for every $t$.
If you also want for $X^0$ to be continuous, then you want something like $X^a_t \to X^0_t$, uniformly in $t$ on compact sets, almost surely. If you combine this with some sort of uniform integrability assumption, you may get the $L^1$ convergence as well.