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 $N$?
 A: 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 $\dim_H \Gamma(\tilde u)\ge \dim_H \tilde u(\Omega)\ge \dim_H [0,1]^2=2$.
A: 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.
