Measurability of reminder term in Taylor expansion about a point in a Brownian motion Suppose that $f: \mathbb R \to \mathbb R$ is twice continuously differentiable and that $\mathrm B$ is a one-dimensional Brownian motion. Using Lagrange's form of the reminder in the Taylor expansion of $f(\mathrm{B}_t(\omega))$ about $f(\mathrm{B}_s(\omega))$ we have that for each $\omega \in \Omega$,
$$
f\left(\mathrm{B}_t(\omega)\right) = f\left(\mathrm{B}_s(\omega)\right) + f'\left(\mathrm{B}_s(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right) + \frac{1}{2}f''\left(\xi(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega) \right)^2
$$
for some point
$$
\xi(\omega) \in \left [\mathrm{B}_s(\omega) , \mathrm{B}_t(\omega)\right].
$$

I would like to show that the map
$$
\omega \mapsto \xi(\omega)
$$
is measurable.

In the case of a first order Taylor expansion it is possible to prove measurability as follows [following an argument by saz https://math.stackexchange.com/questions/896394/mean-value-theorem-inside-the-expectation)]. For each $\omega$ we have from Taylor's theorem
$$
f\left(\mathrm{B}_t(\omega)\right) - f\left(\mathrm{B}_s(\omega)\right) = f'\left(\xi(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right)
$$
for some point
$$
\xi(\omega) \in \left [\mathrm{B}_s(\omega) , \mathrm{B}_t(\omega)\right].
$$
Now since
$$
f\left(\mathrm{B}_t(\omega)\right) - f\left(\mathrm{B}_s(\omega)\right) = \int_{\mathrm{B}_s(\omega)}^{\mathrm{B}_t(\omega)} f'(r) d r,
$$
we get that
$$
\xi(\omega) = \frac{1}{\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right)}\int_{\mathrm{B}_s(\omega)}^{\mathrm{B}_t(\omega)} f'(r) d r.
$$
Since
$$
s, t \mapsto \int_s^t f'(r) d r,
$$
subtraction and division are continuous maps, and composition of continuous and measurable maps are measurable, it follows that
$$
\omega \mapsto \frac{1}{\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right)}\int_{\mathrm{B}_s(\omega)}^{\mathrm{B}_t(\omega)} f'(r) d r
$$
is measurable.

Is it possible to do something similar with $\xi(\omega)$ above?

Most grateful for any help provided!
 A: As $$
f\left(\mathrm{B}_t(\omega)\right)= f\left(\mathrm{B}_s(\omega)\right) + f'\left(\mathrm{B}_s(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right) + \frac{1}{2}f''\left(\xi(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega) \right)^2
$$
it follows that
$$
f''\left(\xi(\omega)\right) = \frac{2 (f\left(\mathrm{B}_t(\omega)\right) - f\left(\mathrm{B}_s(\omega)\right) - f'\left(\mathrm{B}_s(\omega)\right)\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right) ) }{\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega) \right)^2}
$$
where $\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega) \right)^2 \ne 0$ a.s.
Hence $f''\left(\xi(\omega)\right)$ is measurable as a quotient of measurble expressions.
Also we may get that $\xi(\omega) = F(B_t, B_s)$, where $F$ may be found from $$
\xi(\omega) = \frac{1}{\left(\mathrm{B}_t(\omega)- \mathrm{B}_s(\omega)\right)}\int_{\mathrm{B}_s(\omega)}^{\mathrm{B}_t(\omega)} f'(r) d r.
$$
As $F$ is continious a.s. we get that $\xi$ is measurable.
