# Doubts on procedure of approximation of bounded functions by means of bounded, continuous functions

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.
$$\color{red}{\text{(4. Could you please help me understand why }\psi_n\text{ constitutes an approximate identity}}$$
$$\color{red}{\text{and why this implies }(4)\text{?)}}$$

• 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 $\Omega$)? – supinf Oct 20 '20 at 12:55
• I refer to the filtration $\left(\mathcal{F}_t\right)$ @supinf – Strictly_increasing Oct 20 '20 at 14:23

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.
• 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] – Strictly_increasing Oct 29 '20 at 23:14
• [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. – Strictly_increasing Oct 29 '20 at 23:14
• (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. – d.k.o. Oct 30 '20 at 8:42
• (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? – Strictly_increasing Oct 30 '20 at 8:50
• (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. – d.k.o. Oct 30 '20 at 9:49