Let $f:\Omega \to \mathbb{R}$, where $\Omega = [0,1] \times [0,1] \subseteq\mathbb{R}^2$ defined as $$ f(s,t) = \chi_S $$ with $ S = [0,\tfrac{1}{2}] \times (\tfrac{1}{2},1] \cup (\tfrac{1}{2},1] \times [\tfrac{1}{2},1] $
I would like to compute $\partial_1 f \in \mathcal{M}(\Omega)$, i.e. the distrbutional partial derivative w.r.t. the first variable, which must be (as far as I know) a measure on the square $\Omega$.
I have an idea of what I should get, although I am not sure: a measure $\nu$ concentrated on the subset $ \{ s=\tfrac{1}{2} \} $ such that, named $T_1 = \{\tfrac{1}{2}\} \times [0,\tfrac{1}{2})$ and $ T_2 = \{\tfrac{1}{2}\} \times (\tfrac{1}{2},1]$
$$ \nu = \mathcal{H}_{T_1} - \mathcal{H}_{T_2} $$
where $\mathcal{H}_T$ stands for the Hausdorff measure on the set $T$.
My first question is: is this correct? Does something special happen in the point $(\tfrac{1}{2},\tfrac{1}{2})$ (for example some Dirac delta pops up)?
My second question is much more vague: if not, is anyone aware of some natural way to make a Dirac delta pop up?
So for these kind of situations, it is normally sufficient to come back to the definitions, that is for a distribution $f$, its gradient $\nabla f$ is defined as the distribution acting on test functions (smooth and compactly supported) $\varphi$ through the formula $$ \langle \nabla f,\varphi\rangle = -\langle f,\nabla\varphi\rangle. $$ In the particular case of $f = \chi_S$, since $\chi_S$ is a locally integrable function, its action as a distribution is given by an integral, that is $$ \langle \nabla f,\varphi\rangle = -\iint_\Omega f\,\nabla\varphi \\ = -\int_0^{1/2}\!\!\!\!\int_{1/2}^1 \nabla\varphi -\int_{1/2}^1\!\!\int_{1/2}^1 \nabla\varphi \\ = -\int_0^{1}\!\!\!\!\int_{1/2}^1 \nabla\varphi(x,y)\,\mathrm d y\,\mathrm d x. $$ Since $\varphi$ is compactly supported on $[0,1]^2$, it is in particular $0$ when $x$ or $y$ is close to $0$ or $1$, so integrating the derivatives gives only values near $1/2$, and more precisely, $$ \langle \nabla f,\varphi\rangle = \left(0, \int_0^1 \varphi(x,1/2)\,\mathrm d x\right). $$ This can be interpreted as some kind of "Dirac line" on the line $y=1/2$, pointing towards $y=1$ (where it goes up). Notice that $\chi_S = \chi_A$ a.e. where $A = [0,1]\times[1/2,1]$, so they define the same distribution. A more formal way to write this would be to define the distribution $\delta_{[0,1]\times\{1/2\}}$ as the distribution acting on tests functions through the formula $\langle \delta_{[0,1]\times\{1/2\}},\varphi\rangle = \int_0^1 \varphi(x,1/2)\,\mathrm d x$, then $$ \langle \nabla f,\varphi\rangle = \left(0, \delta_{[0,1]\times\{1/2\}}\right). $$ This can also be written $$ \langle \nabla f,\varphi\rangle = \left(0, 1\otimes\delta_{\{1/2\}}\right). $$