Ito integral show $\int_0^t Z X_u dM_u = Z\int_0^tX_u dM_u$ Fix a continuous local martingale $M$ starting at $0$. Suppose $X \in \mathscr{P}^*(M)$, i.e. $X$ is progressively measurable and $\int_0^tX_u^2 d\langle M \rangle_u<\infty$ a.s. Then suppose $Z$ is $\mathscr{F}_0$-measurable (we are NOT assuming $\mathscr{F}_0$ is trivial, so $Z$ does not necessarily have to be constant). How do I show that
$$
\int_0^t Z X_udM_u = Z \int_0^t X_u dM_u.
$$
What I have tried so far: If $X$ simple and $M$ is a martingale, then the result is immediate by definition using sums (all we need to note is that $ZX_u$ will still be progressive since $Z$ is known at time $0$). If $Z$ is bounded then taking simple $X^{(n)} \to X$ (in the sense of $E \int_0^t|X_u^{(n)}-X_u|^2d\langle M \rangle_u \to 0$) we also have $Z X^{(n)} \to Z X$, which gives the result for $Z$ bounded and $M$ a martingale. By taking a localizing sequence we may easily extend this to $Z$ bounded and $M$ a local martingale.
Both sides of the proposed equality make sense without assuming any extra assumptions on $Z$, but if $Z$ blows up in $\omega$, a localizing sequence $\tau_n$ for $X$ may not be a localizing sequence for $ZX$, and we do not necessarily have $ZX^{(n)} \to ZX$ in the sense $L^2(P,\langle M\rangle)$ above. Intuitively this doesn't matter, but I don't know how to show it.
How can I get rid of this requirement that $Z$ be bounded? Maybe there is a more direct way to do this without resorting to simple functions?
 A: Suppose that $M$ is a martingale and $Z$ satisfies
$$\mathbb{E}\left( \int_0^t |Z \cdot X_u|^2 \, d\langle M \rangle_u \right) < \infty \tag{1}$$
Define cut-off functions
$$Z_k := (-k) \vee Z \wedge k, \quad (k \in \mathbb{N})$$
By applying the dominated convergence theorem, we see
$$\|Z_k \cdot X-Z \cdot X\|_{L^2(\langle M \rangle)}^2 = \mathbb{E} \bigg( \int_0^t \underbrace{|Z_k \cdot X_u-Z \cdot X_u|^2}_{\leq 2|X_u \cdot Z|^2} \, d\langle M \rangle_u \bigg) \stackrel{k \to \infty}{\to} 0$$
Using the well-known isometry, we thus get
$$\int_0^t Z_k \cdot X_u \, dM_u \stackrel{k \to \infty}{\to} \int_0^t Z \cdot X_u \, dM_u \tag{2}$$
in $L^2(\mathbb{P})$. Consequently, there exists a subsequence which converges almost surely. Since $Z_k$ is bounded,
$$\int_0^t Z_k \cdot X_u \, dM_u = Z_k \cdot \int_0^t X_u \, dM_u.$$
From $Z_k \to Z$ a.s. we conclude
$$Z_k \cdot \int_0^t X_u \, dM_u \stackrel{k \to \infty}{\to} Z \cdot \int_0^t X_u \, dM_u \quad \text{a.s.} \tag{3}$$
Combining $(2)$ and $(3)$ finishs the proof.
