Taking the derivative of a set function Lets say $f(A)$ is a set function $A \to \mathbb{R}$ for a collection of sets $\mathcal{A}$.
Is is possible to differentiate such a function, and if so, what theory is used to give this notion a rigorous foundation? Or maybe, is it only possible if $\mathcal{A}$ is a $\sigma-$field with measure $|\mu(A)|<\infty\;\forall A\in \mathcal{A}$.
Here are my attempts and thoughts about this
Let $\mathcal{A}$ be the collection of sets upon which $f(A)$ is defined. To make this easier, assume that $\mathcal{A}$ is a $\sigma$-Field.
If I define a decreasing sequence of sets $A_i \in \mathcal{A}: A_1 \supset A_2 \supset A_3 ... \supset A$ so that $\lim\limits_{n \to \infty} A_i = A$, then I have something like the univariate $x\to \infty$.
However, the problem is that $A \in \mathcal{A}$ need not define a metric space, so I can't just assign an absolute distance to each set in the decreasing sequence, right? If I could, that would essentially just translate this to univariate calculus. I think if I define a measure over the field $\mathcal{A}$ I could also make this easier, but that would be the Radon-Nikodym derivative (unless that is the derivative of a set function).
I thought maybe the following could work too. Let's define distance between two sets $A,B$ in the domain of $f(\cdot)$ as:
$$d(A,B) := \left|\sup_{x \in A, \;y \in B\cap A^c } |f(x)-f(y)| - \sup_{x \in B, \;y \in A\cap B^c } |f(x)-f(y)|\right|$$
With this, I thought we could define a derivative from this as:
$$f'(A) := \lim_{n\to \infty} \frac{f(A_{i+1})-f(A_{i})}{d(A_{i+1},A_{i})}$$
 A: OK, I did a bit of digging and it appears that the derivative of a set function is not defined for an arbitrary family of sets $\mathcal{A}$. The sets in $\mathcal{A}$ need to be measurable to admit a limiting process. 
This means that the general expression is, in fact, the Radon-Nikodym theorem. I consulted my Advanced Calc book (Advanced Calculus by Buck, 3ed), and found a similar discussion restricted to the class of Jordan-measurable sets (they are defined in $\mathbb{R}^n$ and have an "area").
If we let $m(A)$ be the Jordan measure of a set $A\in \mathcal{A}$, then we can define the derivative of a set function $f(A)$ defined for $A \in \mathcal{A}$ along the "line" of sets $A_I:=A_1 \supset A_2 \supset ... \supset A$ simply as:
$$\frac{df}{dm}(A_I):= \lim_{n \to \infty} \frac{f(A_{i+1})-f(A_{i})}{m(A_i\setminus A_{i+1})} $$
Similar approaches could be used for more exotic spaces, so long as they are measurable. Let $(\Omega,\mathcal{F},\mu)$ be a measure space, with measure $\mu \geq 0$ and $f(A)$ be a set function defined for $A\in \mathcal{F}$, then for a decreasing sequence of sets $A_I:=A_1 \supset A_2  ... \supset A: \lim_{n\to \infty} A_n = A$, we have:
$$\frac{df}{d\mu}(A_I):= \lim_{n \to \infty} \frac{f(A_{i+1})-f(A_{i})}{\mu(A_i\setminus A_{i+1})} $$
Maybe we can also get something like "unique derivative" $\frac{df}{d\mu}(S)$ of a set function evaluated at some set $S \in \mathcal{F}$ iff:
$$\lim B_i = \lim A_i = S \in \mathcal{F}\implies \frac{df}{d\mu}(B_I) = \frac{df}{d\mu}(A_I)\;\forall B_I,A_I \in \mathcal{F}$$
Anyway, that's as far as I've gotten. I'll happily accept another's answer if it can expand and/or correct this post.
