Show that concatenation of two processes is measurable Let

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*$(E_n,\mathcal E_n)$ be a measurable space;

*$(\Omega_n,\mathcal A_n,\operatorname P_n)$ be a probability space;

*$(\mathcal F^n_t)_{t\ge0}$ be a filtration on $(\Omega_n,\mathcal A_n)$;

*$\tau_n$ be an $(\mathcal F^n_t)_{t\ge0}$-stopping time on $(\Omega_n,\mathcal A_n)$

*$(X^n_t)_{t\ge0}$ be an $(E_n,\mathcal E_n)$-valued $(\mathcal F^n_t)_{t\ge0}$-adapted (maybe we even need to assume progressive) process on $(\Omega_n,\mathcal A_n,\operatorname P_n)$
for $n=1,2$ with $E_1\cap E_2=\emptyset$, \begin{align}E&:=E_1\cup E_2;\\\mathcal E&:=\sigma(\mathcal E_1\cup\mathcal E_2);\\\Omega&:=\Omega_1\times\Omega_2\end{align} and $$X_t:\Omega\to E\;,\;\;\;(\omega_1,\omega_2)\mapsto\begin{cases}X^1_t(\omega)&\text{, if }t<\tau_1(\omega_1);\\ X^2_{t-\tau_1(\omega_1)}(\omega_2)&\text{, otherwise}\end{cases}$$ and $$\mathcal F_t:=\sigma(\mathcal F^1_t\cup\mathcal F^2_t)$$ for $t\ge0$.

How can we show that $(X_t)_{t\ge0}$ is $(\mathcal F_t)_{t\ge0}$-adapted?

Let $t\ge0$. We've clearly got $$F:=\{t<\tau_1\}\cap\Omega_2\in\mathcal F_t$$ and it is enough to show that $\left.X_t\right|_F$ and $\left.X_t\right|_{F^c}$ are $(\left.\mathcal F_t\right|_F,\mathcal E)$ and $(\left.\mathcal F_t\right|_{F^c},\mathcal E)$-measurable, respectively.
$X^1_t$ is clearly $(\mathcal F^1_t,\mathcal E)$-measurable. So, the only tricky part is the measurability on $F^c$.
I know that if $(Y_t)_{t\ge0}$ is an $(\mathcal G_t)_{t\ge0}$-progressive process and $\zeta$ is an $(\mathcal G_t)_{t\ge0}$-stopping time, then $Y_\zeta$ is $\mathcal G_\zeta$-measurable.
 A: Let $$\tilde\tau_1(\omega):=\tau_1(\omega_1)$$ and $$\tilde X^n(\omega):=X^n(\omega_n)\;\;\;\text{for }n=1,2$$ for $\omega\in\Omega$. Since $\tau_1$ is an $(\mathcal F^1_t)_{t\ge0}$-stopping time, $$\{\tilde\tau_1\le t\}=\{\tau_1\le t\}\times\Omega_2\in\mathcal F_t\tag1$$ and \begin{align}\{\tilde X^1_t\in B\}&=\{X^1_t\in B\}\times\Omega_2\in\mathcal F_t;\tag2\\\{\tilde X^2_t\in B\}&=\Omega_1\times\{X^2_t\in B\}\in\mathcal F_t\tag3\end{align} for all $B\in\mathcal E$ for all $t\ge0$. Assuming $(X^2_t)_{t\ge0}$ is $(\mathcal F^1_t)_{t\ge0}$-progressive, we see that $\tilde X^2_{t-\tilde\tau_1\wedge t}$ is $\mathcal F_t$-measurable (by the same simple proof yielding that $Y_{\zeta\wedge t}$ is $\mathcal G_t$-measurable whenver $(Y_t)_{t\ge0}$ is $(\mathcal G_t)_{t\ge0}$-adapted and $\zeta$ is an $(\mathcal G_t)_{t\ge0}$-stopping time) for all $t\ge0$. Thus, $$\left\{\left.X_t\right|_{F^c}\in B\right\}=\left\{\tilde X^2_{t-\tilde\tau_1}\in B\right\}\cap F^c=\left\{\tilde X^2_{t-\tilde\tau_1\wedge t}\in B\right\}\cap\left\{\tilde\tau_1\le t\right\}\in\mathcal F_t\tag4$$ for all $B\in\mathcal E$ and $t\ge0$.
