$h\in H$ and for every $\omega\in \Omega$, $$\begin{aligned}
\left(\int_0^T Z_s\cdot dW_s\circ \tau_h\right)(\omega)&= \left(\sum_{j=1}^d \sum_{i=0}^{n_j} \lambda_{i}^j(W_{t_{i+1}^j}^j-W_{t_i^j}^j)\right) {\circ \tau_{h}} (\omega)\\
&= \sum_{j=1}^d \sum_{i=0}^{n_j} \lambda_{i}^j(\omega+h)(W_{t_{i+1}^j}^j-W_{t_i^j}^j)(\omega+h)\\
& =\sum_{j=1}^d \sum_{i=0}^{n_j} \lambda_i^j {\circ \tau_{h}} (\omega) \left(\omega^j(t_{i+1}^j)-\omega(t_i^j)+h^j(t_{i+1}^j)-h^j(t_i^j)\right)\\
&=\int_0^T Z_s\circ \tau_h \cdot dW_s(\omega)+\int_0^T Z_s{\circ \tau_{h}}(\omega)\cdot dh_s,\end{aligned}$$ which gives the desired result since $h$ is absolutely continuous. We extend this result to processes $Z$ in $\H_{ d}^2$. Let $Z\in \H_{ d}^2$, then there exists a sequence $(Z^n)_{n\in \N}$ in $\mathfrak{S}$ which converges to $Z$ in $\H_{ d}^2$. Hence, $$\begin{aligned}
&\E\left[ {\left|\int_0^T Z_s \cdot dW_s \circ \tau_h-\int_0^T Z_s\circ \tau_h \cdot dW_s-\int_0^T Z_s{\circ \tau_{h}}\cdot dh_s\right|}\right]\\
&\leq \E\left[ {\left|\int_0^T Z_s \cdot dW_s \circ \tau_h-\int_0^T Z^n_s \cdot dW_s \circ \tau_h\right|}\right]+\E\left[ {\left|\int_0^T Z^n_s \circ \tau_h \cdot dW_s -\int_0^T Z_s \circ \tau_h \cdot dW_s\right|}\right]\\
&\hspace{0.9em}+\E\left[ {\left|\int_0^T Z^n_s \circ \tau_h\cdot dh_s -\int_0^T Z_s \circ \tau_h \cdot dh_s\right|}\right]\\
&\leq  \underbrace{\E\left[ {\left|\int_0^T (Z_s - Z^n_s) \cdot dW_s  \right|}\circ \tau_h\right]}_{=:A^n}+\underbrace{\E\left[ {\left|\int_0^T (Z^n_s - Z_s)\circ \tau_h \cdot dW_s\right|}\right]}_{=:B^n}\\
&\hspace{0.9em}+\underbrace{\E\left[ {\left|\int_0^T (Z^n_s - Z_s)\circ \tau_h \cdot dh_s\right|}\right]}_{=:C^n}.\end{aligned}$$

Let us estimate these three terms. First, using Proposition \[prop.cam\], Cauchy-Schwarz Inequality, then Burkholder-Davis-Gundy Inequality, we have $$\begin{aligned}
A^n= \E\left[ {\left|\int_0^T(Z_s-Z_s^n) \cdot dW_s\right|} e^{\int_0^T\dot{h}(s) \cdot dW_s-\frac12 \int_0^T \|\dot{h}(s)\|^2ds}\right]\leq &\ \E\left[\int_0^T \|Z_s-Z_s^n\|^2 ds\right]^{1/2}\\
&\times\E\left[\mathcal E\left(\int_0^\cdot\dot{h}(s) \cdot dW_s\right)_T^2\right]^{1/2}.\end{aligned}$$ By , this clearly goes to $0$ as 