2
$\begingroup$

Let $W$ be a Brownian motion on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\in[0,\:T]},\operatorname P)$ with $T>0$. Please consider the following construction of the Itō integral:

$\Phi:\Omega\times[0,T]\to\mathbb R$ is called elementary $\mathcal F$-predictable $:\Leftrightarrow$ $$\Phi=\sum_{i=1}^k1_{(t_{i-1},\:t_i]}^{[0,\:T]}\eta_i\tag1$$ for some $k\in\mathbb N$, $0\le t_0<\cdots<t_k\le T$ and $\mathcal F_{t_{i-1}}$-measurable $\eta_i:\Omega\to\mathbb R$ with $|\eta_i(\Omega)|\in\mathbb N$ for $i\in\left\{1,\ldots,k\right\}$. Let $$\mathcal E:=\left\{\Phi:\Omega\times[0,T]\to\mathbb R\mid\Phi\text{ is elementary }\mathcal F\text{-predictable}\right\}$$ be equipped with the norm inherited from $L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right)$. If $\Phi\in\mathcal E$, then $$\int\Phi\:{\rm d}W:=\sum_{i=1}^k\eta_i\left(W_{t_i}-W_{t_{i-1}}\right)$$ and $$\int_a^b\Phi\:{\rm d}W:=\int 1_{(a,\:b]}^{[0,\:T]}\Phi\:{\rm d}W\tag2$$ for $0\le a\le b\le T$. Moreover, $$(\Phi\cdot W)_t:=\int_0^t\Phi\:{\rm d}W\;\;\;\text{for }t\in[0,T]\;.$$ Now, $$\mathcal E\ni\Phi\mapsto\Phi\cdot W\tag3$$ is a linear isometry into the space of square-integrable continuous $\mathcal F$-martingales $M_c^2(\mathcal F,\operatorname P)$ equipped with the usual norm. $\mathcal E$ is a dense subset of $$\mathcal I^2:=\left\{\Phi\in\mathcal L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right):\Phi\text{ is }\mathcal F\text{-predictable}\right\}$$ and hence there is a unique isometric and linear extension $\Xi$ of $(3)$ to $\mathcal I^2$. If $\Phi\in\mathcal I^2$, then $$\Phi\cdot W:=\Xi(\Phi)$$ and $$\int\Phi\:{\rm d}W:=(\Phi\cdot W)_T\;.$$ Moreover, $$\int_a^b\Phi\:{\rm d}W:=(\Phi\cdot W)_b-(\Phi\cdot W)_a$$ for $0\le a\le b\le T$.

From this construction, it's not clear that the relation $(2)$ is still valid for $\Phi\in\mathcal I^2$. So, I want to prove that $$\int_a^b\Phi\:{\rm d}W=\int 1_{(a,\:b]}^{[0,\:T]}\Phi\:{\rm d}W\;\;\;\text{for all }0\le a\le b\le T\text{ almost surely}\;.\tag4$$

By construction, there is a sequence $(\Phi^n)_{n\in\mathbb N}\subseteq\mathcal E$ with $$\left\|\Phi-\Phi^n\right\|_{\mathcal I^2}\xrightarrow{n\to\infty}0\tag5$$ and hence $$\left\|\Phi\cdot W-\Phi^n\cdot W\right\|_{M^2(\operatorname P)}\xrightarrow{n\to\infty}0\tag6\;.$$ Now, $$\left\|1_{(a,\:b]}^{[0,\:T]}\Phi\cdot W-1_{(a,\:b]}^{[0,\:T]}\Phi^n\cdot W\right\|_{M^2(\operatorname P)}=\left\|1_{(a,\:b]}^{[0,\:T]}\left(\Phi-\Phi^n\right)\right\|_{\mathcal I^2}\xrightarrow{n\to\infty}0\tag7$$ and $$\left\|\int_a^b\Phi\:{\rm d}W-\int1_{(a,\:b]}^{[0,\:T]}\Phi^n\:{\rm d}W\right\|_{L^2(\operatorname P)}\le 2\left\|\Phi\cdot W-\Phi^n\cdot W\right\|_{M^2(\operatorname P)}\xrightarrow{n\to\infty}0\tag8$$ for all $0\le a\le b\le T$. $(7)$ and $(8)$ together yield $$\int_a^b\Phi\:{\rm d}W=\int 1_{(a,\:b]}^{[0,\:T]}\Phi\:{\rm d}W\;\;\;\text{almost surely for all }0\le a\le b\le T\;.\tag9$$

How can we show that there is a common null set?

$\endgroup$

0

You must log in to answer this question.