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I quote Øksendal (2003). My doubts along the below proof will be written in $\color{red}{\text{red}}$.

Let $\mathcal{V}=\mathcal{V}(S,T)$ be the class of functions $f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$ such that $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$-measurable (where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $[0,\infty)$), $f(t,\omega)$ is $\mathcal{F}_t$-adapted and $\mathbb{E}\bigg[\int_{S}^T f(t,\omega)^2 dt\bigg]<\infty$.
[...] Recall that a function $\phi\in\mathcal{V}$ is called elementary if it has the form $$\phi(t,\omega)=\sum_j e_j(\omega)\cdot\chi_{[t_j, t_{j+1}]}(t)\tag{1}$$ [...]

Statement Let $h\in\mathcal{V}$ be bounded. Then there exist bounded functions $g_n\in\mathcal{V}$ such that $g_n(\cdot,\omega)$ is continuous for all $\omega$ and $n$ and $$\mathbb{E}\left[\int_S^T\left(h-g_n\right)^2dt\right]\to 0\tag{2}$$
Proof Suppose $|h(t,\omega)|\le M$ for all $(t,\omega)$. For each $n$ let $\psi_n$ be a nonnegative continuous function on $\mathbb{R}$ such that:
(i) $\psi_n(x)=0$ for $x\le -\frac{1}{n}$ and $x\ge0$;
(ii)$\displaystyle{\int_{-\infty}^{+\infty}\psi_n(x)dx}=1$.
Define $$g_n(t,\omega)=\int_0^t\psi_n(s-t)h(s,\omega)ds\tag{3}$$ Then, $g_n(\cdot,\omega)$ is continuous for each $\omega$
$\color{red}{\text{(1. Does this follow just from the fact that for each }n, \psi_n\text{ is a continous function?)}}$
and $|g_n(t,\omega)\le M|$
$\color{red}{\text{(2. In the same spirit of question/observation 1., does this just follow from the fact that}}$
$\color{red}{|h(t,\omega)|\le M\text{ for all }(t,\omega)\text{?)}}$
Since $h\in\mathcal{V}$, we can show that $g_n(t,\cdot)$ is $\mathcal{F}_t$-measurable for all $t$.
$\color{red}{\text{(3. How is that possible to show that, since }h\in\mathcal{V}\text{, }g_n(t,\cdot)\text{ is }\mathcal{F}_t\text{-measurable for all }t\text{?)}}$
Moreover, $$\int_S^T(g_n(s,\omega)-h(s,\omega))^2 ds\to 0\hspace{0.5cm}\text{as }n\to\infty\text{ for each }\omega\tag{4}$$ since $\left\{\psi_n\right\}$ constitutes an approximate identity.
$\color{red}{\text{(4. Could you please help me understand why }\psi_n\text{ constitutes an approximate identity}}$
$\color{red}{\text{and why this implies }(4)\text{?)}}$

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  • $\begingroup$ What do you mean by $\mathcal F_t$-measurable? Is it the same as $\mathcal F$-measurable (since it is mentioned for a function on $\Omega$)? $\endgroup$ – supinf Oct 20 '20 at 12:55
  • $\begingroup$ I refer to the filtration $\left(\mathcal{F}_t\right)$ @supinf $\endgroup$ – Strictly_increasing Oct 20 '20 at 14:23
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  1. For $t>v$, $$ |g_n(t,\omega)-g_n(v,\omega)|\le M\int_v^t \psi_n(s-t)\, ds\le M'(t-v). $$ because $\psi_n$ is bounded (it is continuous on $[-1/n,0]$). Therefore, $g_n(\cdot,\omega)$ is continuous, i.e., for each $\epsilon>0$, $|g(t,\omega)-g(v,\omega)|\le \epsilon$ whenever $|t-v|\le \epsilon/M'$. (In fact, it is uniformly continuous.) Also $g_n$ is bounded because $$ |g_n(t,\omega)|\le M\int_0^t\psi_n(s-t)\, ds\le M\int_{-\infty}^{\infty}\psi_n(x)\,dx=M. $$

  2. Suppose that $h$ is elementary. Note that for all $j> J$ such that $t\in [t_J,t_{J+1})$, we have $\chi_{[t_j,t_{j+1})}(s)=0$, and so $$ g_n(t,\omega)=\sum_{j\le J}e_j(\omega)\int_0^t \psi_n(s-t)\chi_{[t_j,t_{j+1})}(s)\, ds, $$ where each $e_j$, $j\le J$, is $\mathcal{F}_t$-measurable.

  3. The sequence $\{\psi_n\}$ is an approximate identity. Look at section 2.5 in this note for the definition and relevant results.

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  • $\begingroup$ I have some questions. $$$$ Point 1.: Isn't $\psi_n$ continuous on all $\mathbb{R}$ by definition? Why did you stress "it is continuous on $[-1/n,0]$"? Secondly, why "whenever $|t-v|\le \epsilon/M'$"? Is $\epsilon/M'$ an arbitrary small value or is it a value being there for some specific reason?$$$$ Point 2. Did you suppose here that, for $\omega$ fixed, $h=\chi_{[t_j,t_{j+1})}(s)$? If not, which is the role of $h$ and what does $h$ correspond to along your reasoning? $$$$ [continue] $\endgroup$ – Strictly_increasing Oct 29 '20 at 23:14
  • $\begingroup$ [continue] $$$$ Point 3. According to the notes you adviced, why would it hold that $\lim\limits_{n\to\infty}\int|\psi_n(x)|dx=0$ (see third "property" of an approximate identity in section $2.5$)? Finally, along your notes where could I find the result ensuring that since $\left\{\psi_n\right\}$ is an approximate identity, then $$\int_S^T(g_n(s,\omega)-h(s,\omega))^2 ds\to 0\hspace{0.5cm}\text{as }n\to\infty\text{ for each }\omega$$?$$$$Thank you a lot in advance. $\endgroup$ – Strictly_increasing Oct 29 '20 at 23:14
  • $\begingroup$ (1) EVT (2) $h\in\mathcal{V}$, i.e., it can be written in that form (3) For $\delta>0$, $\psi_n(x)=0$ on $\{|x|>\delta\}$ for $n$ large enough. Also the last result follows from Theorem 2.5.3. $\endgroup$ – d.k.o. Oct 30 '20 at 8:42
  • $\begingroup$ (1) Hence, one has $\epsilon/M'\epsilon/M'$ at that point by application of EVT? Isn't it just an arbitrarily small quantity? (2) OK, but you did suppose $h$ elementary, didn't you? $\endgroup$ – Strictly_increasing Oct 30 '20 at 8:50
  • $\begingroup$ (1) It says that $M'=M\times \max \psi_n<\infty$. (2) You're right, I misread the question. Measurability is easy with elementary processes. For a general $h\in \mathcal{V}$ we approximate $h$ by a sequence of elementary processes and take the limit. However, one needs to use the completeness of $\mathcal{F}_t$ to get the result. $\endgroup$ – d.k.o. Oct 30 '20 at 9:49

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