Let us consider a bounded, Borel function $F\colon \mathbb R^d \to \mathbb R^d$. Assume it satisfies the following 
[Osgood-like](https://www.encyclopediaofmath.org/index.php/Osgood_criterion) condition: 
$$\tag{O}
\boxed{\vert \langle F(x) - F(y), x-y \rangle\vert \le \Vert x-y \Vert \rho( \Vert x-y \Vert)} \qquad \forall x,y \in \mathbb R^d,
$$
where $\rho \colon [0,1) \to \mathbb [0,+\infty) $ is an Osgood modulus of continuity, i.e. a continuous, non-decreasing function with $\rho(0)=0$ and 
$$
\int\_0^1 \frac{1}{\rho(s)} \, ds = +\infty. 
$$



> 
> **Q**. Is it true that $F$ is continuous? Or, more precisely, is it true that $F$ is equivalent, up to null sets, to a continuous function?
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Beside the link above, something (mildly) related can also be found [here](https://math.stackexchange.com/questions/158888/does-log-lipschitz-regularity-imply-h%C3%B6lder-continuity).


Of course the case $d=1$ is trivial and the question makes sense essentially for $d \ge 2$. Thanks.


**Update (the "linear" case)**. Quoting from [here (pag. 2)](https://arxiv.org/pdf/0807.1592.pdf)



> 
> [...] *in the case when the modulus $\rho$ is linear, (O) implies that the symmetric part of the distributional derivative is bounded, hence Korn’s inequality gives that $F$ is equivalent, up to Lebesgue negligible sets, to a continuous function.* 
> 
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> 


I honestly do not see how to show rigorously this claim in the case when $\rho$ is linear. How to use [Korn's inequality](https://en.wikipedia.org/wiki/Korn%27s_inequality) (I suspect a variant of this $L^2$ inequality is needed, with estimates in $L^\infty$ but do not know)? 



Does this Osgood-like condition imply continuity?
You do not need such heavy high-tech as Korn's inequality or even Lebesgue measure theory for an elementary geometry homework. Let's say $F(0)=0$. 


The first claim is that $F$ is bounded in some neighborhood of the origin. Indeed, just choose finitely many points $x\_j$ such that $x\_i-x\_j$ span the space. Then for any $x$ in some fixed ball we have $\langle F(x)-F(x\_j), x-x\_j\rangle$ bounded by some constant whence $\langle F(x), x-x\_j\rangle$ are bounded by a constant, whence $\langle F(x), x\_i-x\_j\rangle$ are bounded too and that is the end of this story.


Now suppose $x\_j\to 0$ with $|F(x\_j)|\ge\delta>0$. Then WLOG $F(x\_j)\to u\ne 0$. Take any $x\ne 0$ close to $0$. We have
$$
|\langle F(x)-0,x-0\rangle|\le |x|\rho(|x|)
$$
and 
$$
|\langle F(x)-F(x\_j),x-x\_j\rangle|\le |x-x\_j|\rho(|x-x\_j|)
$$
which, after passing to the limit, becomes 
$$
|\langle F(x)-u,x\rangle|\le |x|\rho(|x|)
$$
Subtracting, we get
$$
|\langle u,x\rangle|\le 2|x|\rho(|x|)
$$
but this is absurd if $x$ is collinear with $u$ and $2\rho(|x|)<|u|$.


