Does the Krasovskii Regularization Preserve Measurability?

Question

The Krasovskii regularization for the flow map (set-valued) $F$ over the flow set $C$ is defined as $$\hat{F}(x)\triangleq\bigcap_{\delta >0}\overline{\text{co}}F((x+\delta\mathbb{B})\cap C),\quad\forall x\in \overline{C}.$$

Now, is it true that if $F$ is Lebesgue measurable, then so is $\hat{F}$?