# 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.

• How can there be a continuous surjection from the compact interval $[0,1]$ to the reals? Nov 30, 2019 at 17:06
• @GregoryArone You're right about that. (And I should have been more careful when writing the answer.) Luckily enough, a Borel function is enough for the requirements of the asker. (For example, we can take a continuous bijection $(0,1)\to\mathbb R$ and define values at $0$ and $1$ arbitrarily. Nov 30, 2019 at 17:11
• I have to thank everybody for comprehensive answers. A bit more complex question could be to require that the $F$-preimage of every singleton be a COUNTABLE dense set. Nov 30, 2019 at 18:33

Let $$\{I_n:n\in\mathbb N\}$$ be the set of all open intervals with rational endpoints. Construct pairwise disjoint sets $$A_n$$ $$(n\in\mathbb N)$$ such that each $$A_n$$ is homeomorphic to the Cantor set and $$A_n\subseteq I_n$$. For each $$n\in\mathbb N$$ define a continuous surjection $$f_n:A_n\to[-n,n]$$. Define $$f:\mathbb R\to\mathbb R$$ so that $$f(x)=f_n(x)$$ if $$x\in A_n$$ and $$f(x)=x$$ if $$x\notin\bigcup_{n\in\mathbb N}A_n$$. It's easy to see that $$f$$ is Borel measurable and maps every interval onto $$\mathbb R$$.

P.S. As Martin Sleziak pointed out in a comment the functions $$f_n$$ can be chosen to be at most two-to one, in which case $$f^{-1}(x)$$ will be a countable dense set for each $$x\in\mathbb R$$.

• If we choose the function $f_n$ in a manner similar to Devil's staircase, we get that $f^{-1}(x)$ is countable for each $x\ne0$, right? (For each $x$ we get at most two preimages inside $A_n$.) I am asking this since Vladimir Kanovei mentioned in a comment that they would also be interested in an example where the fibers are countable dense sets. Dec 1, 2019 at 11:26
• Sounds right. Hmm. What if we define $f(x)=x$ for $x\notin\bigcup_nA_n$?
– bof
Dec 1, 2019 at 12:06