I consider a special kind of sets in $\mathbb{R}^n\_+$ given by $G\_t = $ {$x \in \mathbb{R}^n\_+ \mid g(x) < t$}, where $\nabla g > 0$ entrywise. Let's consider an integral
$$
 f(t) = \int\limits\_{ G\_t } \mu(dx) 
$$
If measure $\mu$ is absolutely continious with the density $a(x)$ and $\Omega$ is a $(n-1)$-form such that $dg \wedge \Omega = dx$ than
$$
 f(t) = \int\limits\_{ G\_t } a(x) dx = \int\limits\_{ G\_t } a(x) dg \wedge \Omega
$$
and
$$
 \frac{\Delta f}{\Delta t} = \frac{1}{\Delta t} \int\limits\_{ t \leq g(x) < t + \Delta t } a(x) dg \wedge \Omega \to \int\limits\_{ g(x) = t } a(x) \Omega = \frac{df}{dt}
$$
Help me please with the generalisation of this result to the case of arbitrary measures. I think that currents (continious linear functionals on the set of differential forms) must be involved (I try to find a similar representation for the derivative $f'(t)$).



Derivative of the Lebesgue integral. Currents.
Denote by $\nu$ the pushforward of $\mu$ via the function $g$. More precisely $\nu$ is a measure on $\mathbb{R}$ and for any Borel subset $B\subset \mathbb{R}$ we have


$$ \nu(B)=\mu\bigl(\; g^{-1}(B)\;\bigr).$$


Then 


$$f(t)= \nu\bigl(\;(-\infty,t)\;\bigr).$$


Now invoke Radon-Nicodym Theorem. The "derivative" of $f(t)$ is a measure on $\mathbb{R}$, more precisely, the measure $\nu$.


