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I have a question about an integration method widely used in Analysis, namely the fact that $$ \int_{B(x_0,R)} { \hspace{-20pt} f(x)\,{\rm d} x } = \int_0^R { \hspace{-5pt} {\rm d}r \int_{\partial B(x_0,r)} { \hspace{-25pt} f(x)\,{\rm d} x } } $$ for every $f$ integrable over $B(x_0,R)\subseteq\mathbb R^N$ (the ball centered in $x_0$ with radius $r$). For the sake of simplicity ${\rm d}x$ in the first integral refers to the (Lebesgue) measure on $\mathbb R^N$, while in the second to that on $\mathbb R^{N-1}$.

I am well aware of the intuitive approach behind it, but I have never seen any formal proof of the property, nor do I know how (nor if) it is named. As a first question I would like to know how it is proved, or where I should look in order to find a detailed proof.

Besides, studying on a paper (pdf link: The Gaussian Free Field and Hadamard's Variational Formula, Eq. (3.2) page 3), I encountered what seems to be a generalization of the above formula, namely $$ \int_{\Omega(t)} { \hspace{-13pt} f(x)\,{\rm d} x } = \int_0^t { \hspace{-5pt} {\rm d}\tau \int_{\partial\Omega(\tau)} { \hspace{-15pt} f(x)\varrho(x)\,{\rm d} x } } $$ where $0<t\leq 1$ and $\{\Omega(\tau)\}_{0<\tau\leq 1}$ is a family of simply connected planar open sets satisfying the following conditions (the paper has also a fourth point which I removed by assuming, for simplicity, the simply connectedness of the sets; besides, condition no.3 is stated here in a simplified version for the same reason):

  1. $\overline{\Omega(\tau)}\subset\Omega(\tau')~$ if $~0<\tau<\tau'\leq t$
  2. $\partial\Omega(\tau)$ are $C^2$-smooth and vary in a $C^2$-fashion with $\tau$
  3. The intersection $\displaystyle{\bigcap_{0<\tau\leq 1} \!\!\!\Omega(\tau)}$ is made of a single point

and $\varrho(x)$ is the rate at which the boundary $\partial\Omega(\tau)$ moves at $x\in\partial\Omega(\tau)$ along the direction of the exterior unit normal vector as $\tau$ grows (note that such a $\tau$ is unique for property no. 1).

First I wonder what a mathematical formal definition of the rate $\varrho(x)$ would be (and I presume that if $\Omega(t)=B(x_0,t)\subseteq\mathbb R^2$, then $\varrho\equiv 1$ so that the formula applies to the case of the disk). My idea is that for any $t,t'$ there is a $C^2$-smooth isomorphism $f(\,\cdot\,,t,t'):\Omega(t)\rightarrow\Omega(t')$ and that, if $x\in\partial\Omega(t)$, then $\varrho(x)$ could be defined as $$ \varrho(x) = \left| \lim_{h\to 0^+} { \frac{f(x,t,t+h)-x}{h} } \right| = \left| \frac{\partial f}{\partial t'}(x,t,t) \right| $$ which seems to work fine with $\Omega(t)=B(x_0,t)$ and $f(x,t,t') = x_0 + \frac{t'}{t}(x-x_0)$ (the radial dilatation centered in $x_0$), since indeed it gives $\rho(x)=1$ $\forall x$. The problem is that I strongly doubt that such a definition does not depend on the chosen family of isomorphisms $\{f(\,\cdot\,,t,t')\}_{t,t'\in(0,1]}$.

Secondly I would like to know whether it is fundamental to work with planar domains or if the formula works in any dimension.

And lastly, then again, I would like to know where to look for a detailed proof.

Thanks in advance for your consideration, any hint is greatly appreciated.

Note: the article to which the paper refers for the matter was of no help (it either relates only to Green's functions or I did not understand it...)

.: EDIT :.

I am now wondering if both cases could be generalized involving abstract measures. Specifically, if $\lambda$ and $\sigma$ are measures on $\mathbb R$ and $\mathbb{R}^{N-1}$ respectively, and $\mu=\lambda\!\otimes\!\sigma$ is their product measure, I wonder if it is true that $$ \mu\big(B(x_0,R)\big) = \int_{B(x_0,R)} { \hspace{-20pt} {\rm d} \mu } = \int_{[0,R)} { \hspace{-15pt} {\rm d}\lambda(r) \int_{\partial B(x_0,r)} { \hspace{-25pt} \,{\rm d} \sigma } } $$ and $$ \mu\big(\Omega(t)\big) = \int_{\Omega(t)} { \hspace{-13pt} {\rm d} \mu } = \int_{[0,t)} { \hspace{-10pt} {\rm d}\lambda(\tau) \int_{\partial\Omega(\tau)} { \hspace{-15pt} \varrho\,{\rm d} \sigma } } $$ Involving general measures makes it possible to omit the function $f$ since by linearity it could be assumed to be non-negative, and therefore we could consider $\mu'=f\,{\rm d}\mu$ and $\sigma'=f\,{\rm d}\sigma$ as measures.

share|cite|improve this question
Can you do the case $N = 2$? – KCd Jan 19 '13 at 11:54
@KCd Actually I didn't find any way to make it simpler... – AndreasT Jan 22 '13 at 12:38

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