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!
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.