ODE with a measurable vector field Suppose we have a bounded Borel measurable vector field $F:\mathbb{R}^n\to\mathbb{R}^n$. To make the question non-trivial, assume that $F\neq 0$ eveywhere.

Question. Does there exist at least one Lipschitz integral curve? That is a Lipschitz function $\varphi:(a,b)\to\mathbb{R}^n$ such that
  $\varphi'(t)=F(\varphi(t))$ for almost all $t\in (a,b)$.

This question is related to:
Set of integral curves of a vector field.
 A: For completeness let me add some details (as suggested by @MateuszKwaśnicki).
Let $A\subseteq \mathbb R$ be a Borel set of positive but not full measure in each interval, set $F(x) = -1 + 2 \cdot 1_{A}(x)$. Let $\varphi\colon (a,b) \to \mathbb R$ be a Lipschitz integral curve of $F$. 
The set $R = \varphi((a,b))$ is an interval (by the intermediate value property) with positive Lebesgue measure (because $\varphi'(t)\ne 0$ for a.e. $t\in(a,b)$). Consider the set
$$
\tilde R = \{x \in R : \forall t \in \varphi^{-1}(\{x\}) \; \exists \varphi'(t) = F(\varphi(t)) \}
$$
The set $R\setminus \tilde R$ is Lebesgue negligible, being the image of a Lebesgue negligible set $\{t \in (a,b) : \not\exists \varphi'(t) \text{ or } \varphi'(t) \ne F(\varphi(t))\}$ under Lipschitz map $\varphi$.
Note that for any $x\in \tilde R$ there exists just one $t\in (a,b)$ such that $\varphi(t)=x$. Indeed, otherwise one could find distinct points $t_1, t_2 \in (a,b)$ such that $\varphi(t_1) = \varphi(t_2) = x \in \tilde R$ and $\varphi(t) - x$ has constant sign on $(t_1, t_2)$. Then $\varphi'(t_1)$ and $\varphi'(t_2)$ have to have different signs, a contradiction with the definition of $\tilde R$. Therefore $\varphi$ is injective on $\varphi^{-1}(\tilde R)$. Then it is easy to show that $\varphi$ is injective on the whole $(a,b)$, using the fact that $\varphi'(t)\ne 0$ for a.e. $t\in(a,b)$. 
By continuity and injectivity the function $\varphi$ is strictly monotone, hence one of the sets $R \cap A$ and $R \setminus A$ has measure zero (because $\varphi$ cannot map a negligible set to a set with positive measure). This contradicts the definition of $A$.
In higher dimensions one could take $F(x) = (-1 + 2 \cdot 1_{A}(x_1), 1, \ldots, 1)$.
