Let us consider a bounded, Borel function $F\colon \mathbb R^d  \to \mathbb R^d$. Assume it satisfies the following 
[Osgood-like][1] 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?

Beside the link above, something (mildly) related can also be found [here][2].

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)][3]
  
>  [...] *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.* 

I honestly do not see how to show rigorously this claim in the case when $\rho$ is linear. How to use [Korn's inequality][4] (I suspect a variant of this $L^2$ inequality is needed, with estimates in $L^\infty$ but do not know)? 

 
  [1]: https://www.encyclopediaofmath.org/index.php/Osgood_criterion
  [2]: https://math.stackexchange.com/questions/158888/does-log-lipschitz-regularity-imply-h%C3%B6lder-continuity
  [3]: https://arxiv.org/pdf/0807.1592.pdf
  [4]: https://en.wikipedia.org/wiki/Korn%27s_inequality