In the book "Brownian Motion and Stochastic Calculus" by Karatzas & Shreve we have a proposition on page 194, that I am having trouble with:
Proposition 3.5.4.
Fix $0 \leq T < \infty$ and assume that $Z(X)$ is a martingale. If $M \in \mathcal{M}_T^{c, \: loc}$, then the process \begin{equation} \widetilde M_t := \left\{ M_t - \sum_{i=1}^d \int_0^t X_s^{(i)} \: d \langle{M, W^{(i)}}\rangle_s, \mathcal{F}_t \right\}, \qquad 0 \leq t \leq T \end{equation} is in $\widetilde{\mathcal{M}}_T^{c, \: loc}$. If $N \in {\mathcal{M}}_T^{c, \: loc}$ and \begin{equation*} \widetilde N_t := N_t - \sum_{i=1}^d \int_0^t X_s^{(i)} \: d \langle{N, W^{(i)}}\rangle_s, \qquad 0 \leq t \leq T, \end{equation*} then \begin{equation*} \langle{\widetilde M, \widetilde N}\rangle_t = \langle{M, N}\rangle_t, \qquad 0 \leq t \leq T, \qquad \text{$P$- and $\widetilde P_T$-a.s.} \end{equation*} where the cross-variations are computed under the appropriate measures.
In the above \begin{equation*} Z_t(X) = 1 + \sum_{i=1}^d \int_0^t Z_s(X)X_s^{(i)} \: d W_s^{(i)}, \end{equation*} where $X = \{ X_t = ( X_t^{(1)}, \dots, X_t^{(d)} ), \mathcal{F}_t \}_{0 \leq t < \infty}$ is a vector of measurable, adapted processes satisfying \begin{equation*} P \left[ \int_0^T (X_t^{(i)})^2 \: d t < \infty \right] = 1, \qquad 1 \leq i \leq d, \qquad 0 \leq T < \infty \end{equation*} and $W = \{ W_t = ( W_t^{(1)}, \dots, W_t^{(d)} ), \mathcal{F}_t \}_{0 \leq t < \infty}$ is a $d$-dimensional Brownian motion defined on the probability space $(\Omega, \mathcal F, P)$.
I believe this is all the required information for my problem.
In the proof of the proposition, there are three calculations I am having trouble with:
\begin{equation} Z_t(X) \widetilde M_t = \int_0^t Z_u(X) \: d M_u + \sum_{i=1}^d \int_0^t \widetilde M_u X_u^{(i)} Z_u(X) \: d W_u^{(i)}. \end{equation}
Here are my calculations for the first statement:
\begin{align} Z_t(X) \widetilde M_t = Z_0(X) \widetilde M_0 + \int_0^t Z_u(X) \: d \widetilde M_u + \int_0^t \widetilde M_u \: d Z_u(X) + \langle Z(X), \widetilde M \rangle_t \\ = \int_0^t Z_u(X) \: d \widetilde M_u + \sum_{i=1}^d \int_0^t \widetilde M_u X_u^{(i)} Z_u(X) \: d W_u^{(i)} + \langle Z(X), \widetilde M \rangle_t, \end{align}
since $Z_0(X) = 1$ and $\widetilde M_0 = 0$. As you can see the first integral is with respect to $\widetilde M$ and not $M$ and we also have a covariation process I am not sure what to do with. I am sure they are connected in some way, but cannot figure it out.
\begin{equation*} \begin{split} \widetilde M_t \widetilde N_t - \langle{M,N}\rangle_t &= \int_0^t \widetilde M_u \: d N_u + \int_0^t \widetilde N_u \: d M_u \\ &- \sum_{i=1}^d \left( \int_0^t \widetilde M_u \: d \langle{N, W^{(i)}}\rangle_u + \int_0^t \widetilde N_u \: d \langle{M, W^{(i)}}\rangle_u \right) \end{split} \end{equation*}
Here are my calculations for the second statement: \begin{align} \widetilde M_t \widetilde N_t &= \widetilde M_0 \widetilde N_0 + \int_0^t \widetilde M_u \: d \widetilde N_u + \int_0^t \widetilde N_u \: d \widetilde M_u + \langle \widetilde M, \widetilde N \rangle_t \\ % &= \int_0^t \widetilde M_u \: d N_u + \int_0^t \widetilde N_u \: d M_u \\ &- \sum_{i=1}^d \left( \int_0^t \widetilde M_u \: d \langle{N, W^{(i)}}\rangle_u + \int_0^t \widetilde N_u \: d \langle{M, W^{(i)}}\rangle_u \right) + \langle \widetilde M, \widetilde N \rangle_t, \end{align}
as you can see, I am very close to the desired statement, but I get $\langle \widetilde M_t, \widetilde N_t \rangle$ instead of $\langle M_t, N_t \rangle$, which is very strange. The statement of the proposition is to show these are almost surely equal to each other.
\begin{equation*} \begin{split} &Z_t(X) \left( \widetilde M_t \widetilde N_t - \langle{M,N}\rangle_t \right) \\ &= \int_0^t Z_u(X) \widetilde M_u \: d N_u + \int_0^t Z_u(X) \widetilde N_u \: d M_u \\ &+ \sum_{i=1}^d \int_0^t \left( \widetilde M_u \widetilde N_u - \langle{M,N}\rangle_u \right) X_u^{(i)} Z_u(X) \: d W_u^{(i)}. \end{split} \end{equation*}
Here are my calculations for the second statement:
\begin{align} Z_t&(X) \left( \widetilde M_t \widetilde N_t - \langle {M,N} \rangle_t \right) \\ % &= Z_t(X) \widetilde M_t \widetilde N_t - Z_t(X) \langle {M,N}\rangle_t \\ % &= Z_0(X) \widetilde M_0 \widetilde N_0 + \int_0^t Z_u(X) \: d (\widetilde M_u \widetilde N_u) + \int_0^t \widetilde M_u \widetilde N_u \: d Z_u(X) + \langle Z(X), \widetilde M \widetilde N \rangle_t \\ &- \left( Z_0(X) \langle M,N\rangle_0 + \int_0^t Z_u(X) \: d \langle {M,N} \rangle_u + \int_0^t \langle {M,N}\rangle_u \: d Z_u(X) + \langle Z(X), \langle {M,N}\rangle \rangle_t \right), \end{align}
since $Z_0(X) = 1$ and $\widetilde M_0 = \widetilde N_0 = \langle M,N \rangle_0=0$, the product of these terms disappear. Any help on how to proceed with these calculations would be greatly appreciated. I am sure things get easier here at the end, if the previous calculations are accounted for, I seem to missing a fundamental clue, which is used in all calculations.