Is the Itō integral $\int_0^TΦ_t\:{\rm d}B_t$ the mean-square limit of $\sum_{i=1}^nΦ_{t_{i-1}}(B_{t_i}-B_{t_{i-1}})$ as $\max_i(t_i-t_{i-1})\to 0$?

Let

• $T>0$
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\in[0,\:T]}$ be a right-continuous filtration on $(\Omega,\mathcal A)$
• $B$ be a Brownian motion on $(\Omega,\mathcal A,\mathcal F,\operatorname P)$

Note that $$\mathcal E_0\left\{\Phi\in\mathcal L^2(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}):\Phi\text{ is }\mathcal F\text{-progressively measurable}\right\}$$ is a closed subspace of $\mathcal L^2(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]})$, $$\mathcal E:=\left\{\Phi\in\mathcal E_0:\Phi=\sum_{i=1}^n\zeta_{i-1}1_{(t_{i-1},\:t_i]}\text{ for some }n\in\mathbb N\text{, }0\le t_0<\cdots<t_n\le T\text{ and }\zeta_0,\ldots,\zeta_{n-1}\in\mathcal L^0(\operatorname P)\right\}$$ is a dense closed subspace of $\mathcal E_0$, $$\mathcal M^2:=\left\{X\subseteq\mathcal L^2(\operatorname P):X\text{ is an almost surely right-continuous }\mathcal F\text{-martingale with }\operatorname P\left[X_0=0\right]=1\right\}$$ equipped with $$\left\|X\right\|_{\mathcal M^2}^2:=\sup_{t\in[0,\:T]}\left\|X_t\right\|_{\mathcal L^2(\operatorname P)}^2=\left\|X_T\right\|_{\mathcal L^2(\operatorname P)}^2$$ is a complete semi-normed space and $$\mathcal M^2_c:=\left\{X\in\mathcal M^2:X\text{ is almost surely continuous}\right\}$$ is a closed subspace of $\mathcal M^2$. Let $$(\Phi\cdot B)_t:=\sum_{i=1}^n\zeta_{i-1}\left(B_{t_i\:\wedge\:t}-B_{t_{i-1}\:\wedge\:t}\right)\;\;\;\text{for }t\in[0,\:T]$$ for $\Phi\in\mathcal E$ of the form as in the definition of $\mathcal E$. Then $\Phi\cdot B\in\mathcal M^2_c$ and $$\mathcal E\to\mathcal M^2_c\;,\;\;\;\Phi\mapsto\Phi\cdot B\tag 1$$ is a linear isometry. Thus, $(1)$ can be uniquely extended to a linear isometry $\mathcal E_0\to\mathcal M^2_c$.

Question:$\;\;\;$Let $\Phi\in\overline{\mathcal E}$ and $$\Phi^\varsigma:=\sum_{i=1}^n\Phi_{t_{i-1}}1_{(t_{i-1},\:t_i]}$$ for some $n\in\mathbb N$, $0=t_0<\cdots<t_n=T$ and $\varsigma:=\left\{t_0,\ldots,t_n\right\}$. Can we show that $$\left\|\Phi^\varsigma-\Phi\right\|_{\mathcal E}\xrightarrow{\left|\varsigma\right|\to 0+}0\tag 2$$ where $$\left|\varsigma\right|:=\max_{1\le i\le n}\left(t_i-t_{i-1}\right)$$ such that we could conclude that $$\left\|\left(\Phi^\varsigma\cdot B\right)_T-\left(\Phi\cdot B\right)_T\right\|_{\mathcal L^2(\operatorname P)}=\left\|\Phi^\varsigma\cdot B-\Phi\cdot B\right\|_{\mathcal M^2}\xrightarrow{\left|\varsigma\right|\to 0+}0\tag 3$$ by definition of $\Phi\cdot B$?

It's easy to show that $(2)$ holds, if $\Phi$ is continuous in mean-square, i.e. $$\Phi_t\in\mathcal L^2(\operatorname P)\;\;\;\text{for all }t\in[0,T]\tag 4$$ and $$\left(t\mapsto\Phi_t\right)\in C^0\left([0,T],\mathcal L^2(\operatorname P)\right)\;,\tag 5$$ but I don't think that $(2)$ holds in general.

However, in the book Numerical Solution of Stochastic Differential Equations, the authors state (they use a different notation, but that's exactly what they state) that $(3)$ holds in general (i.e. for all $\Phi\in\overline{\mathcal E}$).

That's weird cause they've defined $\Phi\cdot B$ (again, using a different notation) in the same way as I did. Moreover, they've stated that $(2)$ holds, if $\Phi$ is continuous in mean-square, in their proof of density of $\mathcal E$ in $\overline{\mathcal E}$. If $\Phi$ is not continuous in mean-square, they show that it can be approximated by processes which are continuous in mean-square. However, that doesn't mean that the $\Phi^\varsigma$ form an approximating sequence, does it? So, what am I missing?

• Since the Ito martingale is an isometry, this is equivalent to asking whether $\|\Phi^\varsigma-\Phi\| \to 0$ as $|\varsigma| \to 0$, for mean-square-discontinuous $\Phi \in \overline{\mathcal E}$. It seems that the answer should be no, by letting $t$ be a point of mean-square-discontinuity of $\Phi$, and a sequence of partitions $\varsigma$ in which $t$ is always one of the points picked. – Elena Yudovina Oct 21 '16 at 23:09