# Dominated convergence theorem for stochastic processes integrated with respect to Lebesgue measure

I consider a filtered probability space $$(\Omega, \mathcal{F}, P)$$ with filtration $$(\mathcal{F}_t)$$ on which $$(X_t)$$ is an $$(\mathcal{F}_t)$$-adapted stochastic process. In addition to this, $$(X^n_t)$$ is a sequence of stochastic processes where for each $$n$$, $$(X_t^n)$$ is $$(\mathcal{F}_t)$$-adapted. I want to show that for every fixed $$t \geq 0$$ $$\lim_{n \to \infty} \int_0^t \vert X_s^n -X_s \vert^2 ds = 0, \quad \text{P-a.s}$$ My questions is, does their exist a dominated convergence result which can be used in this context? More precisely, is it enough if I can show that for every fixed $$t \geq 0$$ $$\lim_{n \to \infty} \vert X_t^n - X_t \vert^2 = 0, \quad \text{P-a.s}$$ and for every $$t \geq 0$$ there exist a stochastic variable $$Y_t$$ with $$E[\vert Y_t \vert ]< \infty$$ such that for every $$n$$ $$\vert X_t^n - X_t \vert^2 \leq Y_t \quad \text{P-a.s}$$ I'm looking for a reference.

unfortunately I don't know a reference, but I can give you a simple proof

I think we need an additional hypothesis: $$X^n,X$$ need to have a.s. càdlàg (or càglàd) paths for all $$n$$. For simlicity we assume it to hold for all $$\omega\in\Omega$$ and not only a.s. (otherwise intersect all the following sets with an appropriate set). Another additional hypothesis would be to hold your two equation/inequation not only for all $$t\geq0$$ a.s., but a.s. for all $$t\geq0$$ (that's a significant difference!). The proof is very similar of the given one below.

By assumption for every $$t\geq0$$ there is a set $$A_t\subset\Omega$$, $$P(A_t^\complement)=0$$ s.t. $$\forall \omega\in A_t: \lim_{n\to\infty}|X^n_t(\omega)-X_t(\omega)|^2=0\tag{1}$$ and a set $$B_t\subset\Omega$$, $$P(B_t^\complement)=0$$ s.t. $$\forall \omega\in B_t\forall n\in\Bbb N: |X^n_t(\omega)-X_t(\omega)|^2\leq Y_t(\omega).\tag{2}$$ Now consider the set $$M:=\bigcap_{t\in\Bbb Q_{\geq0}}(A_t\cap B_t).$$ Note: as $$\left(A\cap B\right)^{\complement}=A^{\complement}\cup B^{\complement}$$ and we have a countable intersection, $$P(M^\complement)=0$$ holds.

The trick is now to consider the integral path-wise. Let $$\omega\in M$$ be fixed. As the processes are càdlàg (or càglàd) there is only a countable set $$S\subset\Bbb R$$ on which $$X^n(\omega),X(\omega)$$ $$(n\in\Bbb N)$$ have jumps. So consider function $$f_n(t):=|X^n_t(\omega)-X_t(\omega)|^2\cdot1_{S^\complement}(t)$$ instead of $$|X^n_t(\omega)-X_t(\omega)|^2$$. By continuity on $$S^\complement$$ the properties $$(1)$$ and $$(2)$$ hold for $$f_n$$ for all $$t$$ and not only on $$\Bbb Q_{\geq0}$$. For $$f_n$$ you can use the 'traditional' dominated convergence theorem for $$\lim_{n\to\infty}\int_0^tf_n(s)\Bbb ds=0$$ and as $$S$$ is a Lebesgue null-set $$\int_0^tf_n(s)\Bbb ds=\int_0^t \vert X_s^n(\omega) -X_s(\omega) \vert^2\Bbb ds$$ holds.

Alltogether we found a set $$M$$, $$P(M^\complement)=0$$ s.t. for all $$\omega\in M$$ $$\lim_{n\to\infty}\int_0^t\vert X_s^n(\omega) -X_s(\omega) \vert^2\Bbb ds=0$$ holds.

• Thank you very much. I really appreciate your answer. Mar 26, 2021 at 13:24
• I have tried to prove that the continuity on $S^C$ implies that (1) and (2) hold for all t; If $a<b$ are two consecutive numbers from $S$ then $f_n$ and $f$ are continuous for all $n$ on the interval $(a,b)$. Then if $t^*$ is a irrational number from $(a,b)$ and the sequence $(f_n)$ turns out to be triangular functions like on the photo en.wikipedia.org/wiki/Triangular_function#/media/…, with a common fixed top at $t^*$, then it seems to me that we cannot be sure that (1), (2) are satisfied on the entire interval $(a,b)$. Apr 21, 2021 at 7:58
• Probably you are right. I think we need the addational assumption that the convergence $X^n_t\to X_t$ is uniformly in $t$ for each path. But I'm not sure if this is already implies the mentioned alternative hypothesis that the convergence holds for all $t\geq0$ a.s., which is definitely sufficient to show the statement.
– mag
Apr 21, 2021 at 11:16