I don't seem to be able to find my mistake in the following:
Consider Sobolev spaces in $\mathbb{R}^2$.
We know that $W^{1,3}(\mathbb{R}^2)\subset C^{0,\alpha}(\mathbb{R}^2)$ for $\alpha=\frac{1}{3}$ because then we have $k=1,p=3,r=0,n=2$ and hence. $$ \frac{k-r-\alpha}{n}=\frac{1}{p} $$ So long story short: Every function in $W^{1,3}(\mathbb{R}^2)$ should have a continuous representative.
Now consider the function $u(x)=|x|$ inside the unit disk and $u(x)=0$ outside the unit disk. It's derivative (which is defined almost everywhere in the classical sense, so we can just take that derivative) fulfills $|\nabla u|=1$ inside the unit disk an $|\nabla u|=0$ outside the unit disk.
And now I get the following contradiction which must be wrong:
$|u|$ has compact support and is therefore in $L^3$. $|\nabla u|$ has compact support and is therefore in $L^3$. Therefore $u\in W^{1,3}(\mathbb{R}^2)$
But $u$ can't be made continuous by adjusting it on a set of measure zero. So $u$ can't have a continuous representative.
This would contradict $W^{1,3}(\mathbb{R}^2)\subset C^{0,\alpha}(\mathbb{R}^2)$.
Since it is unlikely that I just disproved Sobolev embedding theory :) I must be making some very stupid mistake in my line of thought...
Differentiability almost everywhere does not imply that the function belongs to a Sobolev space. In your example, your function does not belong to $W^{1,3}(\mathbb R^2)$.
To show this, suppose it belongs to $W^{1,3}(\mathbb R^2)$. Let also $\phi$ be a smooth cutoff, supported in $Β(0,3)$, with $\phi(x)=1$ in $B(0,1)$, and $\psi(x)=x_1^3$. Then $\phi\psi\in C_c^{\infty}(\mathbb R^3)$, therefore $$\int_{\mathbb R^3}u\cdot\partial_1(\psi\phi)=-\int_{\mathbb R^3}\partial_1u\cdot\psi\phi.$$ Since $\partial_1|x|=x_1|x|^{-1}$ for $x\neq 0$, and $\psi=\psi\phi$ on the support of $u$, the previous equality implies that $$\int_{\mathbb R^3}|x|\cdot 3x_1^2\,dx=-\int_{\mathbb R^3}x_1|x|^{-1}\cdot x_1^3\,dx.$$ But, this is a contradiction, since the left hand side is positive, while the right hand side is negative.