Let $\{X_t: t\geq 0\}$ be a stochastic process on the probability space $(\Omega,\mathcal{A},\mathbb{P})$ with values in $\mathbb{R}$ and $T:(\Omega,\mathcal{A},\mathbb{P})\rightarrow\mathbb{R}^{+}_0$ a random variable such that T is stochastically independent from the process $\{X_t: t\geq 0\}$. Furthermore there should hold that $\frac{\mathrm{d}\mathbb{P}_{X_t}}{\mathrm{d}\mu}=f_t(x)$ and $\frac{\mathrm{d}\mathbb{P}_{T}}{\mathrm{d}\nu}=g(x)$.
I want to show that $\mathbb{E}(X_T)=\int_{\mathbb{R}}\int_{\mathbb{R}} xf_t(x)\mathrm{d}\mu(x)\,g(t)\mathrm{d}\nu(t)$. To do this I used a transformation to write
$$\mathbb{E}(X_T)=\int_\Omega X_{T(\omega)}(\omega)\mathbb{P}(\omega)=\int_{\mathbb{R}\times\Omega}X_t(\omega)\mathrm{d}\mathbb{P}_{(T,\mathrm{id})}(t,\omega)$$
I now want to use the independence of $T$ and $X_t$ and then Fubini's theorem but unfortunately there doesn't hold $\mathbb{P}_{(T,\mathrm{id})}=\mathbb{P}_T\otimes \mathbb{P}$ in which case I could then conclude that
\begin{align} \int_{\mathbb{R}\times\Omega}X_t(\omega)\mathrm{d}\mathbb{P}_{(T,\mathrm{id})}(t,\omega)&=\int_{\mathbb{R}\times\Omega}X_t(\omega)\mathrm{d}(\mathbb{P}_T\otimes \mathbb{P})(t,\omega)=\int_{\mathbb{R}}\int_\Omega X_t(\omega)\mathrm{d}\mathbb{P}(\omega)\mathrm{d}\mathbb{P}_T(t)\\ &=\int_{\mathbb{R}}\int_\Omega x\mathrm{d}\mathbb{P}_{X_t}(x)\,g(t)\,\mathrm{d}\nu (t)=\int_{\mathbb{R}}\int_{\mathbb{R}} xf_t(x)\mathrm{d}\mu(x)\,g(t)\mathrm{d}\nu(t) \end{align} How can I show that in this case I can write $\mathbb{P}_{(T,\mathrm{id})}=\mathbb{P}_T\otimes \mathbb{P}$?