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Let $\Omega$ be an open (non empty) set and $u:\Omega \subset \mathbb{R}^N \to \mathbb{R}^M$ be a function such that the Hausdorff dimension of its graph is $N$.

Let $\tilde u = u$ a.e. Is it true that the Hausdorff dimension of the graph of $\tilde u$ is also $N$?

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    $\begingroup$ Maybe you would like $\Omega$ to have strictly positive Lebesgue measure, e.g. be nonempty and open. Otherwise the equality $\tilde u = u$ a.e. might always be true (and I think the Hausdorff dimensions might differ in this case). $\endgroup$
    – Skeeve
    Commented Apr 6, 2019 at 22:19
  • $\begingroup$ @Skeeve You're right, of course. $\endgroup$
    – Riku
    Commented Apr 7, 2019 at 11:35
  • $\begingroup$ Can you do the case $N=M=1$? $\endgroup$
    – Gerald Edgar
    Commented Apr 9, 2019 at 11:56
  • $\begingroup$ @GeraldEdgar No. $\endgroup$
    – Riku
    Commented Apr 9, 2019 at 12:14
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    $\begingroup$ OK maybe a counterexample can be done like this. $f : \mathbb R \to \mathbb R$, $f$ is zero except on the Cantor set, but on the Cantor set $f$ is so wild that its graph has dimension ${}\gt 1$. $\endgroup$
    – Gerald Edgar
    Commented Apr 9, 2019 at 12:26

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In addition to a very concise answer by Alex Ravsky let me address the case $N=M=1$, following the comment by Gerald Edgar. I am going to omit some technical details (which can be added if necessary).

The classical Cantor function is generated by the function $g$ which is defined by $g(0)=0$, $g(\frac13) = \frac12$, $g(\frac23) = \frac12$, $g(1)=1$ (and interpolated linearly). Now repeat the construction for a different function $g$, defined by $g(0)=0$, $g(\frac13) = \frac23$, $g(\frac23) = \frac13$, $g(1)=1$ (and interpolated linearly).

For instance, at the third step of Cantor's iterative construction we obtain the following function:

                                      

Let $u$ denote the limit function, redefined to be zero outside of the Cantor set $C$ (which explains why some pieces of the graph are dotted). Then for $\tilde u \equiv 0$ it is evident that $u = \tilde u$ a.e.

However $\alpha = \frac{\ln 4}{\ln 3} > 1$ is the Hausdorff dimension of the graph of $u$.

Indeed, it is possible to cover the graph $\Gamma_C = \{(x,u(x)) : x\in C\}$ with $4^n$ balls with of $3^{-n}$, which gives the upper estimate $\dim_H \Gamma_C \le \alpha$. For the lower estimate one can consider the image $\mu$ of Cantor measure on $C$ (which is the weak derivative of the classical Cantor function) under the mapping $x\mapsto (x,u(x))$. The measure $\mu$ is supported on $\Gamma_C$ and it is possible to show that there exists a constant $\kappa>0$ such that for any $x\in \Gamma_C$ it holds that $\mu(B_r(x)) \le \kappa \cdot r^\alpha$. Then by Lemma 1.2.8 from [1] it holds $\dim_H \Gamma_C \ge \alpha$.

Dimensions of graphs of some other functions are also discussed in [1].

[1] Bishop C.J., Peres Y. Fractals in probability and analysis. Cambridge; New York: Cambridge university press, 2017.

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  • $\begingroup$ Thank you. Can you show that $u$ is not of bounded variation? $\endgroup$
    – Riku
    Commented Apr 10, 2019 at 17:53
  • $\begingroup$ @Riku Strictly saying $u$ has zero total variation because it is zero a.e., read my answer more carefully. But if you refer to the limit function $f$ before redefining it to be zero outside of the Cantor set then it is easy to see that $f$ has infinite total variation. $\endgroup$
    – Skeeve
    Commented Apr 10, 2019 at 18:06
  • $\begingroup$ Yes, I was thinking about the limit function. Why does it not have bounded total variation? $\endgroup$
    – Riku
    Commented Apr 10, 2019 at 18:11
  • $\begingroup$ Well, simply by definition. The total variation is at least $2^n \frac{2^n}{3^n}$ for every $n\in\mathbb N$ (look at the blue pieces on my picture). $\endgroup$
    – Skeeve
    Commented Apr 10, 2019 at 18:15
  • $\begingroup$ I see; thank you. Also, what software&code did you use to draw that picture? $\endgroup$
    – Riku
    Commented Apr 11, 2019 at 11:45
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Put $N=1$, $M=2$, $\Omega=\Bbb R^N$, and $u(x)=(x,0)$ for each $x\in\Bbb R^N$. Then the graph of $u$ is a straight line, so it has Hausdorff dimension $1=N$. On the other hand, let $C\subset [0,1]$ be a Cantor set and $f:C\to [0,1]^2$ be a surjective map. For each $x\in\Bbb R^N $ put $\tilde u(x)=f(x)$, if $x\in C$ and $\tilde u(x)=u(x)$, otherwise. Since a projection $\pi$ of the graph $\Gamma(\tilde u)$ onto the image $\tilde u(\Omega)$ is a non-expanding map (that is the distance between $\pi(x)$ and $\pi (y)$ is not bigger than the distance between $x$ and $y$ for each $x,y\in \Gamma(\tilde u)$), we have that $\dim_H \Gamma(\tilde u)\ge \dim_H \tilde u(\Omega)\ge \dim_H [0,1]^2=2$.

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    $\begingroup$ It is worth mentioning that we can always take $f$ to be measurable (for example by composing the continuous surjective Peano function $[0,1]\to[0,1]^2$ with the Cantor staircase function that maps $C$ onto $[0,1]$). $\endgroup$
    – No-one
    Commented Sep 1, 2022 at 12:27

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