How to think of function derivatives in terms of differentiation of measures. Loosely speaking, given two measures $m_1$ and $m_2$, we can define
$$\frac {dm_1}{dm_2}(x) = \limsup_{r\rightarrow0}\frac{m_1(B(x, r)}{m_2(B(x, r))}$$
where $B(x, r)$ is an open ball around x with radius r.
How can we interpret/understand the (usual) derivative of a function in terms of differentiation of measures?
I am asking for intuitions here. Loose explanation of the idea or the thms involved suffice.
 A: $f'$ is the Radon-Nikodym derivative of the measure induced by $f$ with respect to Lebesgue measure.  In particular, define a measure $m_{f}$ on $\mathbb{R}$ by 
$$m_{f}((a,b]) = f(b^{+}) - f(a^{+}).$$
This construction works if $f$ is non-decreasing.  Moreover, if $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$, then the Radon-Nikodym derivative $\frac{d m_{f}}{d \lambda}$ of $m_{f}$ with respect to $\lambda$ satisfies $\frac{d m_{f}}{d \lambda}(x) = f'(x)$ almost everywhere.  For more, I would look at the chapter on "differentiation" in a real analysis textbook (e.g. Chapter 6 in Royden-Fitzpatrick or Chapter 14 in Bass.)
More generally, it's a straightforward exercise to show that if $\nu$ is a non-negative Radon measure on $\mathbb{R}$, $g : \mathbb{R} \to \mathbb{R}_{\geq 0}$ is continuous, and we define $f(x) = \int_{[0,x]} g(y) \, \nu(dy)$, then
$$g(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{\nu((x, x + h])}.$$  Observe that this is a statement about differentiating measures as much as functions since the measure $\nu_{g}(A) = \int_{A} g(y) \, \nu(dy)$ satisfies $\nu_{g}([0,x]) = f(x)$ and
$$g(x) = \lim_{\epsilon \to 0^{+}} \frac{\nu_{g}((x - \epsilon,x + \epsilon))}{\nu((x - \epsilon,x + \epsilon))} \quad \nu\text{-a.e.}$$
A less straightforward fact is this last statement about $\nu_{g}$ remains true if $g \in L^{1}(\mathbb{R},\nu)$ (not necessarily continuous).  (The proof of this statement is a consequence of the Besicovitch covering lemma and the assertion itself is called the Besicovitch differentiation theorem.  When $\nu$ is Lebesgue measure, it suffices to apply Vitali's covering lemma and the result is Lebesgue's differentiation theorem.)
From the point of view of measure theory on $\mathbb{R}$, differentiation is something you can do generally to measures (with respect to other measures) and the classical derivative $f'$ is the special case where you differentiate a function (understood in the appropriate way as a measure) with respect to Lebesgue measure specifically.
