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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.

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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.

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    $\begingroup$ Thank you very much. I really appreciate your answer. $\endgroup$
    – J. Goles
    Mar 26, 2021 at 13:24
  • $\begingroup$ 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 $(a,b)$. $\endgroup$
    – J. Goles
    Apr 21, 2021 at 7:58
  • $\begingroup$ 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. $\endgroup$
    – mag
    Apr 21, 2021 at 11:16

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