I've noticed that step functions in two dimensions, for instance $u(x,y)= \mathbb{1}_{\{x,y > 0\}}(x,y)$ or $u(x,y)=\mathbb{1}_{\{ (x,y) \in B \setminus L \ : \ y >0\}}(x,y)$ where $B$ is the open unit ball and $L=\{(x,0) \in \mathbb{R}^2 : x \in \mathbb{R} \}$, admit first order weak derivatives in $L^p$ (we need second order derivatives to find a Dirac mass) and hence lie in $W^{1,p}$ for every $p$, if I didn't miss something.
I was then wondering if we can prove that, for example
$$ H^1(B \setminus L)=H^1({B}) $$
where $B,L$ were previously defined. This is false in one dimension, if $L$ is interpreted as just the origin; my intuition would say that this is false also in two dimensions. On the other hand, the above examples say that things might be different. Can someone give an insight into this one?
You can write $B^+ = B \cap \{(x,y) : y > 0\}$ and $B^- = B \cap \{(x,y) : y < 0\}$.
Let $u = 1$ on $B^+$ and $-1$ on $B^-$. Suppose (for the sake of getting a contradiction) that $u$ has weak partial derivatives in $B$. If $\phi \in C_0^\infty(B)$ is supported in $B^+$ you get $$\int_B u D\phi = \int_{B^+} D\phi = 0 = - \int_B 0 \phi$$ which tells you that $Du = 0$ in $B^+$. Likewise you can find that $Du = 0$ in $B^-$, so that $Du = 0$ almost everywhere in $B$.
Now let $\phi \in C_0^\infty (B)$. It comes from the definition that $$\int_B u D\phi = - \int_B Du \phi = 0$$
Since also $$\int_B D\phi = 0$$ you find algebraically that $$ \int_{B^+} D\phi = 0 \quad \text{and} \quad \int_{B^-}D\phi = 0$$ for every $\phi \in C_0^\infty(B)$. There are limitless counterexamples to this final claim, so $u$ can't have a weak derivative.
p.s. If you know the characterization of weak derivatives using absolute continuity on lines the proof is immediate: $u$ has a discontinuity on every vertical segment in $B$ that passes through the $x$-axis.