Is there a Borel-measurable function which maps every interval onto $\mathbb R$?

Using AC, one easily defines a function $$F:\mathbb R\to \mathbb R$$ such that the $$F$$-image of any real interval $$(a,b)$$ ($$a) is equal to $$\mathbb R$$.
(Equivalently, the $$F$$-preimage of any real singleton has to be a dense set in $$\mathbb R$$.) Does there exist a Borel-measurable $$F$$ with this property?

This is Exercise 9.M from A. C. M. van Rooij, W. H. Schikhof: A Second Course on Real Analysis.$$\newcommand{\dcc}[1]{\lfloor#1\rfloor}$$

Exercise 9.M. (Another function that maps every interval onto $$[0,1]$$) For $$x\in[0,1]$$ let
$$0.x_1x_2x_3\dots$$ be the standard dyadic development of $$x-\dcc x$$:
$$x_n=\dcc{2^n x}-2\dcc{2^{n-1}x}$$
where $$\dcc x$$ is the entire part of $$x$$. Define $$\phi\colon{\mathbb R}\to{\mathbb R}$$ by
$$\phi(x)=\limsup\limits_{n\to\infty}\frac{x_1+x_2+\dots+x_n}n$$
Show that $$\phi$$ maps every interval onto $$[0,1]$$. (Hint: First show that $$\phi(x)=\phi(y)$$ if there exist $$p,q\in\mathbb N$$ such that
$$x_p=y_q$$, $$x_{p+1}=y_{q+1}$$, $$x_{p+2}=y_{q+2}$$, etc., so that it suffices to show that $$\phi$$ maps $$[0,1]$$ onto $$[0,1]$$. Now let $$t\in[0,1]$$, $$t\ne1$$. Find an $$x\in[0,1]$$ such that $$x_1+\dots+x_n=\dcc{nt}$$ for every $$n$$ and prove that $$\phi(x)=t$$. Finally, find an $$x$$ with $$\phi(x)=1$$.)

The same function appears as Problem 1.3.29 in Kaczor, Nowak: Problems in Mathematical Analysis Vol II and it is given also in an answer here: Can we construct a function $$f:\mathbb{R} \rightarrow \mathbb{R}$$ such that it has intermediate value property and discontinuous everywhere?. The same function was also used by Andrés E. Caicedo as an example of a function which is of Baire class 2 but not of Baire class 1: Examples of Baire class 2 functions. (See also his blog post: 414/514 Examples of Baire class two functions)

As all functions $$x_n$$ are Borel measurable, so is the function $$\phi$$.

For this function, the image of a non-trivial interval is only the interval $$[0,1]$$. But we can get function which maps this onto reals by composition with some continuous Borel surjection from a unit interval to reals.