Proving the characteristic function is not weakly differentiable on $\mathbb{R}^n$. I'm trying to work out the following proof for why the characteristic function 
$\chi_{B_1(0)}$ is not weakly differentiable on $\mathbb{R}^n$.
As it goes:
Assume it is weakly differentiable. Then there exists a $g \in L^1_{\text{loc}}(R^n)$ such that for all $\psi \in C_c^{\infty}$,
\begin{align}
-\int_{\mathbb{R}^n}g \psi dV &=\int_{\mathbb{R}^n}\chi_{B_1(0)} \partial_i \psi 
 \, dV\\
&=\int_{B_1(0)} \partial_i \psi \, dV\\
&=\int_{\partial B_1(0)} \psi \, \nu_i dV^{n-1}
\end{align}
where $\nu_i$ is the outward pointing unit normal field of $\partial B_1(0)$.
From this it is meant to be clear that there exists no such function $g$ and no more detail is given, but I can't see how to deduce the result.
 A: Continuing from your work, suppose we have $g$ such that 
$$
-\int g\psi = \int_{\partial B_1(0)}\psi\nu_i.
$$
Since this should hold by definition for all compactly supported, smooth $\psi$, we can pick $\psi$ to be zero on the boundary of the unit ball. Then the right-hand side is zero. From this, we see that $\int g\psi  = 0$ for all $\psi$ which are compactly supported in $\mathbb{R}^n \setminus \partial B_1(0)$. Since the boundary is measure zero, such functions may be used to approximate the characteristic function of any measurable set. This shows that $g = 0$ a.e. in $\mathbb{R}^n$. We now show that 0, in fact, cannot be the weak derivative of $\chi_{B_1(0)}$, so that no weak derivative can exist. 
But this part is easier; if the weak derivative were zero, then $\int f\psi_{x_i} = \int_{B_1(0)}\psi_{x_i} = 0$ for all $\psi \in C_c^\infty(\mathbb{R}^n)$. But just pick some function $\psi$ which does not have this property.
A: For a test function $\psi$ with support disjoint from $\partial B_1(0)$, the integral on the right hand side vanishes. Thus $g(x) =0$ for almost every $x\in \mathbb{R}^n\backslash{\partial B_1(0)}$ by the fundamental lemma of variational calculus. But the unit sphere $\partial B_1(0)$ itself has measure $0$, thus $g=0$ a.e. on all of $\mathbb{R}^n$.
A: Here's another idea: if $Du = 0$ a.e. on a connected domain then $u$ is  equal to a constant a.e. (prove this using mollifiers). If the weak derivative exists, then it should agree with the actual derivative of $u$ (where it is defined), up to a null set, say. Now, you can conclude.
