Let $f(x_1,x_2)=x_2^2\chi_E$ where $E=\{(x_1,x_2)\in \mathbb{R}^2:x_1\geq 0 \}$ and let $B$ be the unit ball in $\mathbb{R}^2$. Since $f$ is locally integrable, it can be interpreted as a distribution and I can take its distributional derivatives. I need to calculate $$\lim_{\varepsilon\to 0} \langle \partial_{x_1}f,\rho_\varepsilon \ast \chi_B \rangle$$ where $\{\rho_{\varepsilon}\}$ are the standard mollifiers. Intuitively I think that this limit should be $0$. Here I present an approach that a friend of mine suggested.
Attempt
The answer would be clearly $0$ if $f(x_1,x_2)=x_2^2$ (I'll call this the "trivial version"). By the simmetry of the problem, the answer $A$ is the same if we replace $E$ with its complement. Clearly $2A$ is the answer to the trivial version (by additivity of the integral). And since $2A=0$, we have $A=0$.
This seems really nice, but is it correct? And is there a more direct approach?
Using the definition of distributional derivatives, Fubini’s theorem, and the fundamental theorem of calculus, you have \begin{align} \langle \partial_{x_1}f,\rho_\varepsilon \ast \chi_B \rangle=&-\int_{\mathbb{R}^2}x_2^2\chi_E\partial_{x_1}(\rho_\varepsilon \ast \chi_B ) \,dx\\=&-\int_{\mathbb{R}}x_2^2\int_0^{\infty}\partial_{x_1}(\rho_\varepsilon \ast \chi_B ) \,dx_1dx_2=\\&=- \int_{\mathbb{R}}x_2^2((\rho_\varepsilon \ast \chi_B ) (\infty,x_2)-(\rho_\varepsilon \ast \chi_B )(0,x_2))\,dx_2 \end{align} The mollification converges pointwise at every continuity point and the function has compact support. So you can use the Lebesgue dominated convergence theorem to get that the limit is $$\int_{\mathbb{R}}0+x_2^2\chi_B (0,x_2)\,dx_2$$