Let $(x_i,y_i) \in S^{\mathbb{N}} \times S^\mathbb{N}$ where $S$ is the unit sphere in $\mathbb{R}^3$.
Assume the following result of weak convergence for measures :
$$\frac{1}{N} \sum_{i=1}^N \delta_{x_i,y_i} \underset{N \rightarrow + \infty}{\longrightarrow} f(x,y) dx dy$$ which means that for any continous function $\varphi \in C^0(S^2)$, we have :
$$\frac{1}{N} \sum_{i=1}^N \varphi(x_i,y_i) \underset{N \rightarrow + \infty}{\longrightarrow} \int_{S^2} \varphi(x,y) f(x,y) dx dy.$$ We ask that $f$ is in $W^{1,1}(S^2)$ (for exemple, any other kind of regularity would do) so that everything is well defined. Knowing this, I would like to find a measure $\mu_N$ such that :
$$\mu_N \underset{N \rightarrow + \infty}{\longrightarrow} \partial_n f(x,y) dx dy$$ where $n$ is the extern normal on $S^2$. More exactly, I want that, for any suitable test functions $\varphi \in C^0(S^2)$,
$$\int_{S^2} \varphi d\mu_N \underset{N \rightarrow + \infty}{\longrightarrow} \int_{S^2} \varphi(x,y) \partial_n f(x,y) dx dy.$$
My guess was the following :
$$\mu_N=-\frac{1}{N} \sum_{i=1}^N \delta'_{x_i,y_i} \cdot n(x_i,y_i),$$ where $\delta'_{x_i,y_i}$ is the distribution such that $\langle \delta'_{x_i,y_i}| \varphi \rangle := \nabla \varphi(x_i,y_i)$ for any $\varphi \in D(S^2)$. However, it is just an intuition using the definition of integration by parts through distribution, and I'm not even sure this is well defined since my $\mu_N$ need to be applied with function at least $C^1$, which is not the case for the weak convergence of measures.
Any help or advice are welcome.