Do we need an integrability assumption on $X$ in order to prove that $X\int_{t_0}^T\Phi\:{\rm d}B=\int_{t_0}^TX\Phi\:{\rm d}B$?

Let $B$ be a Brownian motion on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\in[0,\:T]},\operatorname P)$ with $T>0$ and $\Phi\in L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right)$ be $\mathcal F$-predictable. Now, let $t_0\in[0,T]$ and $X:\Omega\to\mathbb R$ be $\mathcal F_0$-measurable.

I want to show that $$X\int_{t_0}^T\Phi\:{\rm d}B=\int_{t_0}^TX\Phi\:{\rm d}B\;.\tag1$$

If $\Phi$ is elementary, i.e. $$\Phi=\sum_{i=1}^k1_{(s_{i-1},\:s_i]}^{[0,\:T]}\eta_i\tag2$$ for some $k\in\mathbb N$, $0\le s_0<\cdots<s_k\le T$ and $\mathcal F_{s_{i-1}}$-measurable $\eta_i:\Omega\to\mathbb R$ with $|\eta_i(\Omega)|\in\mathbb N$, then it's easy to see that $(1)$ holds.

In the general case, there is a sequence $(\Phi^n)_{n\in\mathbb N}$ of elementary processes with $$\left\|\Phi-\Phi^n\right\|_{L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right)}\xrightarrow{n\to\infty}0\tag3$$ and $$\left\|\Phi\cdot B-\Phi^n\cdot B\right\|_{M^2(\operatorname P)}\xrightarrow{n\to\infty}0\;,\tag4$$ where $\Phi\cdot B$ denotes the integral process $\left(\int_0^t\Phi\:{\rm d}B\right)_{t\in[0,\:T]}$ and $M^2(\operatorname P)$ denotes the space of square integrable continous $\mathcal F$-martingales.

Let $\tilde\Phi:=1_{(t_0,\:T]}\Phi$ and $\tilde\Phi^n:=1_{(t_0,\:T]}\Phi^n$. By the (omitted) proof of the elementary case, $\tilde\Phi^n$ and $X\tilde\Phi^n$ are elementary (and hence their Itō integral is well-defined). In order to conclude by approximation, we need that $X\tilde\Phi\in L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right)$ is $\mathcal F$-predictable (such that the Itō integral is well-defined) and $$\left\|X\tilde\Phi-X\tilde\Phi^n\right\|_{L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right)}\xrightarrow{n\to\infty}0\;.\tag5$$

However, I don't see why $X\tilde\Phi$ should even satisfy the integrability assumption, since their is no integrability assumption on $X$. So, is the statement wrong? Do we need $X\in L^\infty(\operatorname P)$? Or is there something that I'm missing?

For $n=1,2,\ldots$ define $X_n:=X1_{\{|X|\le n\}}$. Your argument applies to each $X_n$ to show that the desired result holds a.s. on each $\{|X|\le n\}$, hence a.s. on $\{|X|<\infty\}$.
• Are you sure? I think that there is a problem in your argumentation: Even when $\omega\in\left\{|X|\le n\right\}$, $\int_{t_0}^TX\Phi\:{\rm d}B(\omega)$ doesn't depend solely on $\omega$. In any case, your argument cannot work unless you're able to show that $X\tilde\Phi\in L^2\left(\operatorname P\otimes\left.\lambda^1\right|_{[0,\:T]}\right)$ (since otherwise the object $\int_{t_0}^TX\Phi\:{\rm d}B$ is not even defined). – 0xbadf00d Sep 30 '17 at 20:33
• For that object to be defined you only need $X\Phi$ to be locally in $L^2$, which it is. A localizing sequence of stopping times is given by $\tau_n:=T$ on $\{|X|\le n\}$; $\tau_n:=0$ on $\{|X|>n\}$. – John Dawkins Oct 1 '17 at 15:37