# Can the inverse image of the set of continuous functions under a stochastic process be measurable?

In Billingsley's probability and measure, theorem 38.2 roughly says

Let $T=[0, \infty)$.

Two stochastic processes $X: T \times (\Omega, \mathcal F, P) \to (\mathbb R, \mathcal B(\mathbb R))$ and $X': T \times (\Omega', \mathcal F', P') \to (\mathbb R, \mathcal B(\mathbb R))$ have the same finite-dimensional distributions.

If a subset $A \subseteq \mathbb{R}^T$ (not necessarily in the product sigma algebra on $\mathbb{R}^T$) satisfies certain conditions, $X^{-1}(A)$ lies in $\mathcal F$ and has $P$-measure 1, and $X'$ is separable, then $X'^{-1}(A)$ contains an $\mathcal F'$-set of $P'$-measure 1.

As an example,

Theorem 38.2 now implies that if a process has continuous paths with probability 1, then any separable process having the same finite-dimensional distributions has continuous paths outside a set of probability 0.

In particular, a Brownian motion with continuous paths was constructed in the preceding section, and so any separable process with the finite-dimensional distributions of Brownian motion has continuous paths outside a set of probability 0.

Let $A$ be the set of all continuous functions from $T$ to $\mathbb R$. It has been shown not measurable wrt the product sigma algebra on $\mathbb{R}^T$.

When $X$ has the finite dimensional distributions of a Brownian motion, how can we know $X^{-1}(A)$ lies in $\mathcal F$?

"a process has continuous paths with probability 1" doesn't imply $X^{-1}(A)$ lies in $\mathcal F$, does it?

Thanks and regards!

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