(This question is partially related to another one on this forum.)
In Karatzas and Shreve, II edition, Chapter 3, we see in equation (2.2) the definition of the following measure on the product space $[0,\infty)\times\Omega$
$$
\mu_M(A) = \mathbb{E}\left[\int_0^{\infty}1_{A}(s,\omega)d\left<M\right>_s(\omega)\right] \quad (1)
$$
where $M$ is a continuous square integrable martingale and $A\in\mathcal{B}([0,\infty))\otimes\mathcal{F}$ is a measurable set of the product sigma algebra. So if $t\rightarrow\left<M\right>_t(\omega)$ is absolutely continuous (for almost all $\omega\in\Omega$) with respect to the Lebesgue measure $\lambda$ we have that the measure in $(1)$ becomes
$$
\mu_M(A) = \mathbb{E}\left[\int_0^{\infty}1_{A}(s,\omega)f(s,\omega)ds\right] = \int_{\Omega}\left(\int_0^{\infty}1_{A}(s,\omega)f(s,\omega)ds\right)d\mathbb{P}\quad (2)
$$
with $f(s,\omega)\geq 0$. Later, in Lemma 2.4, it is stated and proved that if $X$ is a bounded, measurable and adapted process then there exists a sequence $\xi^{(n)}$ of simple processes such that
$$
\sup_{t>0}\lim_{n\rightarrow\infty}\mathbb{E}\left[\int_0^t(\xi^{(n)}_s-X_s)^2ds\right] = 0\quad (3)
$$
I understand that this is $L^2$ convergence in the product measure $\lambda\otimes\mathbb{P}$ defined on the product $\sigma$ algebra
$$
\left(\lambda\otimes\mathbb{P}\right)(A\times B) = \lambda(A)\mathbb{P}[B]
$$
In fact the $(3)$ is equivalent to the integral with respect to $\lambda\otimes\mathbb{P}$, i.e.
$$
\int_{\left[0,\infty\right)\times\Omega}g(s,\omega)d\left(\lambda\otimes\mathbb{P}\right) = \int_{\Omega}\left(\int_{0}^{\infty}g(s,\omega)ds\right)d\mathbb{P}
$$
for a $\lambda\otimes\mathbb{P}$-measurable+integrable function $g$. Since $L^2$ convergence implies convergence in measure, we get that for all $\varepsilon>0$ it holds
$$
\int_{\Omega}\left(\int_{0}^{\infty}1_{\{\left|\xi^{(n)}_s(\omega)-X_s(\omega)\right|>\varepsilon\}}ds\right)d\mathbb{P}\rightarrow 0.
$$
The problem is: can I conclude now that
$$
\int_{\Omega}\left(\int_{0}^{\infty}1_{\{\left|\xi^{(n)}_s(\omega)-X_s(\omega)\right|>\varepsilon\}}f(s,\omega)ds\right)d\mathbb{P}\rightarrow 0.
$$
where $f$ is defined in $(2)$ ? Should $f$ be bounded, the implication would be obvious. But can I assume that $f$ is bounded? I think I am missing something important.
Best Answer
I think I have figured out the solution. I added a new post with a more generic statement HERE.