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Suppose $f:\mathbb{R}\to\mathbb{R}$ is Borel. Let $\text{dim}_{\text{H}}(\cdot)$ be the Hausdorff dimension, and $\mathcal{H}^{\text{dim}_{\text{H}}(\cdot)}(\cdot)$ be the Hausdorff measure in its dimension on the Borel $\sigma$-algebra.

Question: If $G$ is the graph of $f$, is there an explicit $f$ such that:

  1. The function $f$ is everywhere surjective (i.e., $f[(a,b)]=\mathbb{R}$ for all non-empty open intervals $(a,b)$)
  2. $\mathcal{H}^{\text{dim}_{\text{H}}(G)}(G)=0$

Note, not all everywhere surjective $f$ satisfy 2. of the question. For example, consider the Conway base-13 function. Since it's zero almost everywhere, $\text{dim}_{\text{H}}(G)=1$, and $\mathcal{H}^{\text{dim}_{\text{H}}(G)}(G)=+\infty$.

Optional: If such an $f$ exists, does $f$ have other interesting properties?

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    $\begingroup$ Does "explicit" mean that it should not use the axiom of choice? Or that it has to be given by a nice formula/it cannot be defined in several parts? $\endgroup$
    – Saúl RM
    Commented Aug 8 at 12:16
  • $\begingroup$ @SaúlRM I was hoping we wouldn't need axiom of choice. Either way, I want it to be given by a nice formula. $\endgroup$
    – Arbuja
    Commented Aug 8 at 16:37
  • $\begingroup$ If anyone is still interested, here is the motivation behind finding such a function. $\endgroup$
    – Arbuja
    Commented Oct 1 at 22:22

1 Answer 1

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Here is a way to do it without the axiom of choice, but it isn't a nice formula either.

Consider a Cantor set $C\subseteq[0,1]$ with Hausdorff dimension $0$. Now consider a countable disjoint union $\cup_{n\in\mathbb{N}}C_n$ such that each $C_n$ is the image of $C$ by some affine map and every open set $O\subseteq[0,1]$ contains $C_n$ for some $n$. Such a countable collection can be obtained by e.g. at letting $C_n$ be contained in the biggest connected component of $[0,1]\setminus(C_1\cup\dots\cup C_{n-1})$ (with the center of $C_n$ being the middle point of the component).

Note that $\cup_nC_n$ has Hausdorff dimension $0$, so $\left(\cup_nC_n\right)\times[0,1]\subseteq\mathbb{R}^2$ has Hausdorff dimension $1$.

Now, let $g:[0,1]\to\mathbb{R}$ be such that $g|_{C_n}$ is a bijection $C_n\to\mathbb{R}$ for all $n$ (all of them can be constructed from a single bijection $C\to\mathbb{R}$, which can be obtained without choice, although it may be ugly to define) and outside $\cup_nC_n$ let $g$ be defined by $g(x)=h(x)$, where $h:[0,1]\to\mathbb{R}$ has a graph with Hausdorff dimension $2$ (this doesn't require choice either).

Then the function $g$ has a graph with Hausdorff dimension $2$ and is everywhere surjective, but its graph has Lebesgue measure $0$ because it is a graph (so it admits uncountably many disjoint vertical translates).

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    $\begingroup$ In general it would be enough to find an everywhere surjective function with a graph of dimension $2$, maybe that is not hard to construct using some construction with decimal representations $\endgroup$
    – Saúl RM
    Commented Aug 9 at 0:08
  • $\begingroup$ Note, I want to find an explicit function that's hard to meaningfully average (e.g., the expected value of the function, w.r.t any satisfying generalization of the Hausdorff measure, doesn't exist.) $\endgroup$
    – Arbuja
    Commented Aug 9 at 0:27
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    $\begingroup$ I think you can make the construction with the union of the $C_n$ rather explicit as follows. Split the binary expansion of $x$ as strings of size a power of two, say $x=0.110100010\ldots$ becomes $(s_0,s_1,s_2,\ldots)=(1,10,1000,\ldots)$. If this sequence eventually contains only strings of the form $0\cdots0$ or $1\cdots1$, say after $s_k$, then send it to $y=\sum_{i>0}\epsilon_i2^{-i}$, where $s_{k+i} = \epsilon_i\cdots\epsilon_i$. Otherwise, send it to the explicit continuous function $h$ given by the linked article. This will give you something from $[0,1)$ to $[0,1)$. $\endgroup$
    – Pierre PC
    Commented Aug 9 at 12:38
  • $\begingroup$ @PierrePC Sorry for the late response. If your comment gives a function from $[0,1)$ to $[0,1)$, how would you change it from $[0,1]$ to $\mathbb{R}$? (Is it possible to add an answer expanding on this.) $\endgroup$
    – Arbuja
    Commented Aug 30 at 21:02
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    $\begingroup$ Compose by an explicit (reasonable) bijection from $[0,1)$ to $\mathbb R$. In your case, the construction can be easily adapted so that the $[0,1]$ or $[0,1)$ target space is actually a $(0,1)$, then compose with $t\mapsto(1-2x)/(x^2-x)$. $\endgroup$
    – Pierre PC
    Commented Sep 5 at 9:16

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