Monotone convergence theorem for stochastic integrals

I was wondering if there exists an equivalent of a monotone convergence theorem for stochastic integrals. I looked into plenty of books and papers, but I haven't found anything useful. I would expect something like that:
Assume that $L(t)$ is Levy process and $X_n(t)$ a sequence of stochastic processes such that $\lim_{n \to \infty}X_n(t)=X(t)$ in a monotonous way (limit defined in some sense, for example in distribution). Then $\lim_{n \to \infty} \int_{-\infty}^t X_n(s-)dL_s=\int_{-\infty}^t X(s-)dL_s$.
Is it true? If it is, do you have any hints how to prove it and/or where I could find a proof?

• How do you define $\int\dots dL_s$? – zhoraster Feb 3 '17 at 8:40
• In a "convenient way". So can be Ito integral, can be Stratonovich integral. I'd be happy to find a result for any of these definitions. – Paula Feb 3 '17 at 11:46
• And what do you need these results for? Anyway, the chances are minimal to get something making use of monotonicity and not included in existing general results. Maybe only when $L$ is a subordinator: the integral is defined pathwise in Lebesgue-Stieltjes sense, so you can use deterministic monotone convergence theorem. – zhoraster Feb 3 '17 at 13:09
• I'm actually still trying to solve the question I asked before: mathoverflow.net/questions/259991/…. Could you please explain the subordinator case? – Paula Feb 3 '17 at 13:37

Secondary, for the most definitions of stochastic integrals(in path, $L^1$, $L^2$ or locally etc.), the MCT could be deduced by DCT(dominated convergence theorem), if the $X_1$ and $X=\lim\limits_{n\to\infty}X_n$ are supposed integrable. Since $$\sup_{n,m\ge 1}|X_n-X_m|\le 2\max(|X_1|,|X|).$$