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In Oksendal's and Xuerong Mao's books when defining the integral wrt the Brownian motion they use the following lemma:

Let $\mathscr{M}^{2}([a,b];\mathbb{R})$ be the space of all real-valued measurable {$\mathcal{F}_{t}\}$-adapted process $f=\{f(t)\}_{a\leq t\leq b}$ such that $E\int_{a}^{b}|f(t)|^{2}dt<\infty.$ Then, for any $f\in\mathscr{M}^{2}([a,b];\mathbb{R}),$ there exists a sequence $\{g_{n}\}$ of simple processes such that $$\lim_{n\rightarrow\infty}E\int_{a}^{b}|f(t)-g_{n}(t)|^{2}dt=0.$$ They prove the lemma in three steps:

Step 1: they prove that for every $f \in \mathscr{M}^{2}([a,b];\mathbb{R})$ exists a sequence of bounded processes $\{\varphi_{n}\}$ in $\mathscr{M}^{2}([a,b];\mathbb{R})$ such that $$\lim_{n\rightarrow\infty}E\int_{a}^{b}|f(t)-\ \varphi_{n}(t)|^{2}dt=0.$$ Step 2: For every bounded $\varphi \in \mathscr{M}^{2}([a,b];\mathbb{R})$ exists a sequence of bounded continuous processes $\{\phi_{n}\}$ in $\mathscr{M}^{2}([a,b];\mathbb{R})$ such that $$\lim_{n\rightarrow\infty}E\int_{a}^{b}|\varphi(t)-\ \phi_{n}(t)|^{2}dt=0.$$ Step 3: For every bounded and continuous $\phi \in \mathscr{M}^{2}([a,b];\mathbb{R})$ exists a sequence of simple processes $\{g_{n}\}$ in $\mathscr{M}^{2}([a,b];\mathbb{R})$ such that $$\lim_{n\rightarrow\infty}E\int_{a}^{b}|\phi(t)-\ g_{n}(t)|^{2}dt=0.$$

Then, they conclude that the proof is finished. I guess that to go from these three steps to the final result is obvious but I can't see it. What is the mathematical argument that allow us to conclude the proof of the lemma from the three steps?

To be more specific, given a $\epsilon>0$, I need to find a $n_{0}\in \mathbb{N}$ such that $n\geq n_{0}$ implies $$E\int_{a}^{b}|f(t)-g_{n}(t)|^{2}dt<\epsilon.$$ How can I find '$n_{0}$' using steps: 1,2 and 3?

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3 Answers 3

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Hint: $|f(t)-g_n(t)|^2 = |(f(t)-\varphi_n(t)) + (\varphi_n(t)-\phi_n(t)) + (\phi_n(t)-g_n(t))|^2 \leq |f(t)-\varphi_n(t)|^2 + |\varphi_n(t)-\phi_n(t)|^2 + |\phi_n(t)-g_n(t))|^2$.

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  • $\begingroup$ Thanks. I thought about the idea you suggest but it's not clear to me that we can do that. The n can't be the same for every sequence, isn't it? In step 2 we show that every element of the sequence {varphi_{n}} can be aproximated by a sequence {phi_{m}}. So I guess for every n we have a m_{n} but I don't know how to continue. $\endgroup$
    – UBM
    Commented Aug 21, 2019 at 11:46
  • $\begingroup$ The inequality is not correct. $\endgroup$ Commented Aug 21, 2019 at 11:47
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I will write $\|f\|$ for $\sqrt {E\int_a^{b} |f(t)|^{2}dt}$. Given $n$ choose $n_1$ such that $\|f-\varphi_{n_1}\| <\frac 1 {3n}$. Then apply Step 2 with $\varphi$ replaced by $ \varphi_{n_1}$ and choose $n_2$ such that $\|\varphi_{n_1}-\phi_{n_2}\| <\frac 1 {3n}$. Finally apply Step 3 with $\phi$ replaced by $\phi_{n_2}$ and choose $n_3$ such that $\|\phi_{n_2}-g_{n_3}\| <\frac 1 {3n}$. By triangle inequality we see that $\|f-g_{n_3}\| <\frac 1 n$. Now rename $g_{n_3}$ as $h_n$. we get $\|f-h_n\| <\frac 1 n$ and $h_n$ is simple for each $n$.

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  • $\begingroup$ I don't understand. In order to prove the lemma, by definition of limit, given an epsilon 'e', we need to find a n_0 such that n >= n_0 implies ||f-g_n||<e (where g_n is a simple function). How do we obtain this n_0 from the numbers in your comment: n1, n2 and n3? $\endgroup$
    – UBM
    Commented Aug 21, 2019 at 13:22
  • $\begingroup$ I am using $h_n$ instead of $g_n$ for the simple process. $\|f-h_n\|<\frac 1 n$ for each $n$ and each $h_n$ is simple. So given $\epsilon >0$ simply choose $n$ so large that $\frac 1 n <\epsilon$. This gives a simple process $h_n$ with $\|f=h_n\| <\epsilon$. $\endgroup$ Commented Aug 21, 2019 at 23:16
  • $\begingroup$ So in your proof you also must impose $n_{2}>n_{1}$ and $n_{3}>n_{2}$, right? And then $n \geq n_{3}$ implies the final result, is that correct? $\endgroup$
    – UBM
    Commented Aug 22, 2019 at 9:53
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Proof. Let $\epsilon > 0$. By step 1, exists $n_{1} \in \mathbb{N}$ such that $$n \geq n_{1} \Rightarrow E\int_{a}^{b}|f(t)-\varphi_{n}(t)|^{2}dt < \frac{\epsilon}{9}.$$ Next, pick $n_{2} \in \mathbb{N}$ such that both $$n_{2} \geq n_{1} \text{ and } \forall n \geq n_{2}, E\int_{a}^{b}|\varphi_{n_{1}}(t) - \phi_{n}(t)|^{2}dt< \frac{\epsilon}{9}.$$ We know that such $n_{2}$ exists because of step 2. Next, pick $n_{3} \in \mathbb{N}$ such that both $$n_{3} \geq n_{2} \text{ and } \forall n \geq n_{3}, E\int_{a}^{b}|\phi_{n_{2}}(t) - g_{n}(t)|^{2}dt< \frac{\epsilon}{9}.$$ We know that such $n_{3}$ exists because of step 3. Therefore, $ n \geq n_{3}$ implies that

$$E\int_{a}^{b}|f(t) - g_{n}(t)|^{2}dt = E\int_{a}^{b}|(f(t)-\varphi_{n_{1}}) + (\varphi_{n_{1}} - \phi_{n_{2}}) + (\phi_{n_{2}} - g_{n}(t))|^{2}dt$$ $$\leq E\int_{a}^{b}\Big(3|f - \varphi_{n_{1}}|^{2} + 3|\varphi_{n_{1}} - \phi_{n_{2}}|^{2} + 3|\phi_{n_{2}} -g_{n}|^{2}\Big)dt$$ $$ = 3E\int_{a}^{b}|f - \varphi_{n_{1}}|^{2}dt + 3E\int_{a}^{b}|\varphi_{n_{1}} - \phi_{n_{2}}|^{2}dt + 3E\int_{a}^{b}|\phi_{n_{2}} -g_{n}|^{2}dt < \epsilon.$$ In the last part of the proof, in the first inequality, I have used the fact that $(a+b+c)^{2} \leq 3(a^{2}+b^{2}+c^{2}).$

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