I am reading a paper that uses the following fact without proof:
Let $\{ \mathcal{F}_u \}_{u \geq 0}$ be the filtration generated by a Brownian motion $\{B_u \}_{u \geq 0}$. Consider the stochastic process $\{ X^{t,x}_s \}_{s \in [t,T]}$ defined by $$ X^{t,x}_s = x + \int_t^s \sigma(r, X^{t,x}_r) \, dB_r + \int_t^s b(r,X^{t,x}_r) \, dr, \quad \quad s \in [t,T], $$ for some Lipschitz functions $b$ and $\sigma$ with bounded derivatives. By standard results, the pathwise derivative of this process is given by $$ \partial_x X^{t,x}_s = 1+ \int_t^s (\partial_x \sigma)(r, X^{t,x}_r) \partial_x X^{t,x}_r \,d B_r + \int_t^s (\partial_x b)(r, X^{t,x}_r) \partial_x X^{t,x}_r \,dr, \quad \quad s \in [t,T]. $$ The paper claims that for any $s \in (t,T]$, $\mathcal{F}_t$-measurable random variable $\xi$,
$$ \text{Since }\partial_x X^{t,x}_s \text{ is independent of } \mathcal{F}_t, \quad \quad \partial_x X^{t,x}_s \Big|_{x= \xi} \text{ is well-defined a.s..} $$
The argument is based on the fact that $$ \mathbb{E} \bigg[ \sup_{s \in [t,T]} \bigg( \partial_x X^{t,x}_s \bigg|_{x= \xi} \bigg)^2 \bigg| \mathcal{F}_t \bigg] \leq \sup_{x \in \mathbb{R}} \mathbb{E} \bigg[ \sup_{s \in [t,T]} |\partial_x X^{t,x}_s|^2 \bigg] \leq C,$$ for some bound $C$ depending only on the bounds of $\partial_x \sigma$ and $\partial_x b$, due to the independence condition.
I am not sure how this condition guarantees well-definedness of the process $ \partial_x X^{t,x}_s \bigg|_{x= \xi}$.