Suppose we have two probability spaces $(\Omega_1, \mathscr{F}_1, \{\mathcal{F}^1_t\},\mathbb{P})$ and $(\Omega_2, \mathscr{F}_2, \{\mathcal{F}^2_t\},\mathbb{P}_2)$, if we take product space $$\Omega = \Omega_1 \times \Omega_2, \quad \quad \mathscr{F} = \mathscr{F}_1 \otimes \mathscr{F}_2$$ $$\mathcal{F}_t = \underset{s>t}{\cap}\mathcal{F}^1_t \otimes \mathcal{F}^2_t,\ \ \forall t\geq 0, \quad \mathbb{P} = \mathbb{P}_1 \times \mathbb{P}_2.$$
If $X_1$ and $X_2$ are random objects(variables, adapted stochastic processes etc.) which defined on $(\Omega_1, \mathscr{F}_1, \{\mathcal{F}^1_t\},\mathbb{P})$ and $(\Omega_2, \mathscr{F}_2, \{\mathcal{F}^2_t\},\mathbb{P}_2)$, respectively. If we naturally extend definitions of $X_1$ and $X_2$ as $X_1(\omega_1,\omega_2) = X_1(\omega_1)$ and $X_2(\omega_1,\omega_2) = X_2(\omega_2)$. Can we say $X_1$ and $X_2$ are independent on $(\Omega, \mathscr{F}, \{\mathcal{F}_t\},\mathbb{P})$ automatically because of the construction of product spaces?
Yes, just compute for $A \in \mathcal{F}_1$ $B \in \mathcal{F}_2$
$$\Bbb{P}[X_1(\omega_1,\omega_2) \in A, X_2(\omega_1,\omega_2)\in B)] = \Bbb{P}[X_1(\omega_1) \in A, X_2(\omega_2)\in B)] = \Bbb{P}[\omega_1 \in X_1^{-1}(A), \omega_2\in X_2^{-1}B)] =\Bbb{P}_1[\omega_1 \in X_1^{-1}(A)] \Bbb{P}_2[\omega_2\in X_2^{-1}B)] = \Bbb{P}[X(\omega_1,\omega_2) \in A]\Bbb{P}[X(\omega_1,\omega_2) \in B] $$
You can see that independence follows from the structure of the product space construction.