Finding a Borel measurable function Let $f :[0,\infty)\rightarrow \mathbb{R}$ be continuous and differentiable on $(0,\infty)$. Can we find a Borel measurable function $ g :(0,\infty) \rightarrow (0,\infty)$ such that $g(x)\in (0,x)$ and $f(x)=f(0)+f'(g(x))x$ for all $x\in (0,\infty)$? 
 A: Define a set-valued function $G:(0,\infty)\to \mathcal P[0,\infty)$ by
$$G(x)=\operatorname{argmax}\limits_{t\in[0,x]} \left|f(t)-f(0)-t\frac{f(x)-f(0)}{x}\right|.$$
The intuition is that $t\in G(x)$ means that $t$ is one of the solutions coming from the usual proof of the mean value theorem, though instead of picking an extreme point in Rolle's theorem we select all of them at this stage. It may also include $0$ and $x$ as a convenience.
Considered as a set, $G$ is relatively closed in $(0,\infty)\times [0,\infty)$ - if $t_n\in G(x_n)$ and $(t_n,x_n)\to (t,x)$ then $t\in G(x).$ By the proof of Rolle's theorem we know that for each $x$ there is some $0<t<x$ with $t\in G(x).$ The only problem is how to select a unique $t$ for each $x.$ We can get an at-most-two-valued function by defining
$$G'(x)=\operatorname{argmin}_{t\in G(x)}|t-x/2|.$$
One way to see this is Borel is to observe that $x\mapsto \min_{t\in G(x)}|t-x/2|$ is a lower semicontinuous function. Finally we can define the single-valued function
$$g(x)=G'(x)\setminus \{x-t\mid t\in G'(x), t>x/2\}.$$
to disambiguate the cases where $F'(x)=\{t, x-t\}.$
