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](https://mathoverflow.net/q/255525/121665).



ODE with a measurable vector field
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](https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-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 $F(x) = (-1 + 2 \cdot 1\_{A}(x\_1), 1, \ldots, 1)$.


