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 $f'(t)$).
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$\begingroup$ Why don't you try to see what happens when $\mu$ is a Dirac measure concentrated at a point. $\endgroup$– Liviu NicolaescuCommented Feb 11, 2012 at 16:17
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$\begingroup$ So, if we consider $f'(t)$ as a generalized function then for any measure we have $\langle f'(t), \varphi(t) \rangle = \int\limits_{\mathbb{R}^n_+} \varphi(g(x)) \mu(dx)$, so for Dirac measure $\delta_{a}$ we have $\langle f'(t), \varphi \rangle = \varphi(g(a))$. In the case of the Dirac's measure $f(t)$ is not differentiable in an ordinary sense. $\endgroup$– AppliquéCommented Feb 11, 2012 at 17:00
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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 $\nu$.
answered Feb 18, 2012 at 17:10