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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.

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    $\begingroup$ If $A \subseteq \mathbb{R}$ is a measurable set such that $|(a, b) \cap A| > 0$ and $|(a, b) \setminus A| > 0$ whenever $a < b$, and $F(x) = -1 + 2 \times \mathbb{1}_A(x)$, then, if I am not mistaken, there is no solution $\varphi$. $\endgroup$ – Mateusz Kwaśnicki Feb 12 at 6:16
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    $\begingroup$ @MateuszKwaśnicki I think this example can be easily generalized to higher dimensions by taking $F$ on each component with "separated variables". Please change your comment into an answer including higher dimensions. This is very nice. $\endgroup$ – Piotr Hajlasz Feb 12 at 12:24
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    $\begingroup$ The example mentioned by @MateuszKwaśnicki is also mentioned in the paper "A Necessary and Sufficient Condition for Existence of Measurable Flow of a Bounded Borel Vector Field" by Nikolay A. Gusev, Moscow Mathematical Journal 18(1):85-92, 2018. $\endgroup$ – Skeeve Feb 12 at 19:06
  • $\begingroup$ @Skeeve Thank for the reference! I will certainly read it. $\endgroup$ – Piotr Hajlasz Feb 12 at 19:21
  • $\begingroup$ @Skeeve: If you like, feel free to write a more detailed answer based on Gusev's article, I'll be happy to read it. And I am too busy at the moment to turn my comment into an answer, sorry. $\endgroup$ – Mateusz Kwaśnicki Feb 12 at 21:05
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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 $F(x) = (-1 + 2 \cdot 1_{A}(x_1), 1, \ldots, 1)$.

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    $\begingroup$ In higher dimensions you need to modify your construction. The set $A$ is one dimensional so $1_A(x)$ makes no sense. Did you mean $1_A(x_1)$? $\endgroup$ – Piotr Hajlasz Feb 17 at 17:36
  • $\begingroup$ @PiotrHajlasz fixed, thanks! $\endgroup$ – Skeeve Feb 17 at 18:30

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