# If the Hausforff dimension of the graph of a function $u$ is $N$ and $\tilde u = u$ a.e. then $\dim_H \mathrm{graph} \, \tilde u = N$ too

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 $$N$$?

• 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). – Skeeve Apr 6 at 22:19
• @Skeeve You're right, of course. – Riku Apr 7 at 11:35
• Can you do the case $N=M=1$? – Gerald Edgar Apr 9 at 11:56
• @GeraldEdgar No. – Riku Apr 9 at 12:14
• 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 ${}\gt 1$. – Gerald Edgar Apr 9 at 12:26

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.

• Thank you. Can you show that $u$ is not of bounded variation? – Riku Apr 10 at 17:53
• @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. – Skeeve Apr 10 at 18:06
• Yes, I was thinking about the limit function. Why does it not have bounded total variation? – Riku Apr 10 at 18:11
• 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). – Skeeve Apr 10 at 18:15
• I see; thank you. Also, what software&code did you use to draw that picture? – Riku Apr 11 at 11:45

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 $$\pi(x)$$ and $$\pi (y)$$ for each $$x,y\in \Gamma(\tilde u)$$), we have that $$\dim_H \Gamma(\tilde u)\ge \dim_H \tilde u(\Omega)\ge \dim_H [0,1]^2=2$$.