Derivative of the Lebesgue integral. Currents. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:14:36Z http://mathoverflow.net/feeds/question/87552 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87552/derivative-of-the-lebesgue-integral-currents Derivative of the Lebesgue integral. Currents. Nimza 2012-02-04T22:46:29Z 2012-02-18T17:10:28Z <p>I consider a special kind of sets in $\mathbb{R}^n_+$ given by $G_t =$ {$x \in \mathbb{R}^n_+ \mid g(x) &lt; t$}, 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 $$\frac{\Delta f}{\Delta t} = \frac{1}{\Delta t} \int\limits_{ t \leq g(x) &lt; t + \Delta t } a(x) dg \wedge \Omega \to \int\limits_{ g(x) = t } a(x) \Omega = \frac{df}{dt}$$ 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 $f'(t)$).</p> http://mathoverflow.net/questions/87552/derivative-of-the-lebesgue-integral-currents/88843#88843 Answer by Liviu Nicolaescu for Derivative of the Lebesgue integral. Currents. Liviu Nicolaescu 2012-02-18T17:10:28Z 2012-02-18T17:10:28Z <p>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</p> <p>$$\nu(B)=\mu\bigl(\; g^{-1}(B)\;\bigr).$$</p> <p>Then </p> <p>$$f(t)= \nu\bigl(\;(-\infty,t)\;\bigr).$$</p> <p>Now invoke Radon-Nicodym Theorem. The "derivative" of $f(t)$ is a measure on $\mathbb{R}$, more precisely, the measure $\nu$.</p>