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Suppose we have a progressively measurable process $X$ taking values in $\mathbb{R}^d$.

What are sufficient conditions on a function $$f( x, t, \omega ) \colon \mathbb{R}^d \times [0,\infty) \times \Omega \rightarrow \mathbb{R},$$ such that the process $f( X(t), t )$ is progressively measurable?

I feel that it should be sufficient to demand that $f( ., t, \omega)$ is a continuous function on $\mathbb{R}^d$ for all $(t, \omega) \in [0,T] \times \Omega$. Unfortunately, I failed to establish a rigorous proof.

Can you provide a proof or point me to a relevant result in the literature?

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    $\begingroup$ Continuity alone can't be sufficient; since $f$ is allowed to depend on $\omega$, you need some condition to ensure that it doesn't look into the future. For example if $Y(t)$ is any process that isn't progressively measurable, you could simply set $f(x,t,\omega)=Y(t,\omega)$. Then $f$ is certainly continuous in $x$ (even constant), and no matter what $X$ is, we have $f(X(t), t) = Y(t)$ which is not progressively measurable. $\endgroup$
    – Nate Eldredge
    Jan 27, 2015 at 19:26

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