Laying the groundwork to study SDEs we discussed the following:
We consider the SDE $dX_t = f(t,X) dB_t$ where $f: \mathbb{R}_+ \times C(\mathbb{R}_+,\mathbb{R}) \to \mathbb{R}$. This means that we consider the entire sample path of $X$ in the SDE. Because we still only want dependence up to a time $t$, we define a filtration on $C(\mathbb{R}_+,\mathbb{R})$, namely $$ \mathcal{G}_t := \sigma(\pi_s:s \in [0,t]), $$ where $\pi_s$ maps a continuous function $w$ to $w(s)$ (evaluation map).
Using this definition we can consider $\mathbb{G}$-progressive maps $f$. We say that $f$ is $\mathbb{G}$-progressive, if for all $t \geq 0$, we have that $$ f: [0,t] \times C(\mathbb{R}_+,\mathbb{R}) \to \mathbb{R} $$ is $\mathcal{B}([0,t]) \otimes \mathcal{G}_t$ measurable.
We showed that, whenever X is an adapted and continuous process, then the map $\mathbb{R}_+ \times \Omega \ni (t,\omega) \mapsto f(t, X(\omega))$ is progressive in the usual sense.
Question: Now I wanted to show that the usual dependence on "only" $X_t$ is also progressive, i.e. show that for a continuous and adapted process $X$, the map $\mathbb{R}_+ \times \Omega \ni (t,\omega) \mapsto f(t, X_t(\omega))$ is progressive in the usual sense, when $f$ is $\mathbb{G}$-progressive.
Can someone give me a hint on how to proceed? Usually there is a clever decomposition of the map, but I can't seem to find it?
Thanks in advance!