Election measurable in uniform continuity Let $f:[0,1]\times [0,1] \rightarrow \mathbb{R}$ borel measurable such that for all $x \in [0,1]$ $f(x,-):[0,1] \rightarrow \mathbb{R}$ is continuous, in particular uniformly continuous.
Then there $\delta(x)>0$ such that for $\vert w-s\vert < \delta(x)$ implies $\vert f(x,w)-f(x,s)\vert <1$ as is well known.
My question is if $\delta(x) $ may be chosen so that $x \mapsto \delta(x)$  is  Borel measurable ?
I appreciate any suggestions
 A: For $\delta > 0$, let
$$
F_\delta : [0,1] \to \Bbb{R}, x \mapsto \sup_{v,w \in \Bbb{Q} \cap [0,1], |v-w|<\delta} |f(x,v) - f(x,w)|
$$
and note that $F_\delta$ is measurable as a countable(!) supremum of measurable functions (and well-defined, as $f(x, \cdot)$ is a bounded function).
Thus, $\chi_{F_\delta \leq 1/2}$ (where $\chi_M$ is the indicator function of $M$) is also measurable for each $\delta > 0$.
Now define
$$
G : [0,1] \to \Bbb{R}, x \mapsto \sup_{\delta \in \Bbb{Q} \cap [0,1]} \delta \cdot \chi_{F_\delta \leq 1/2}(x).
$$
As above, $G$ is measurable.
Now, essentially as you noted, for each $x \in [0,1]$, there is some $\delta_x > 0$ so that $F_{\delta_x} (x) < 1/2$ holds. This implies $G(x) \geq \delta_x > 0$ and thus
$$
G(x) = \sup_{\delta \in \Bbb{Q} \cap [0,1] \text{ with } F_\delta (x) \leq 1/2} \delta.
$$
For $v,w \in \Bbb{Q} \cap [0,1]$ with $|v-w| < G(x)$, there is thus $\delta \in \Bbb{Q} \cap [0,1]$ with $|v-w| < \delta$ and
$$
|f(x, v) - f(x,w)| \leq \sup_{v',w' \in \Bbb{Q} \cap [0,1], |v'-w'| < \delta} |f(x, v') - f(x,w')| = F_\delta (x) \leq 1/2.
$$
Because $f(x, \cdot)$ is continuous and $\Bbb{Q} \cap [0,1]$ is dense in $[0,1]$, we easily see $|f(x, v) - f(x,w)| \leq 1/2 < 1$ for all $v,w \in [0,1]$ with $|v-w| < G(x)$.
Thus, $\delta(x) := G(x)$ does what you want.
EDIT: Also note that we did not use that $f$ is Borel-measurable in both variables. It actually suffices that $f(\cdot, \delta)$ is Borel-measurable for every $\delta \in \Bbb{Q} \cap [0,1]$.
