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Well I am working something, which deals with the following problem: For example, I want to compute an integral $\int_{B(0,R)}f(x)dx$, where $B(0,R)=\{x\in\mathbb R^n:\;|x|\leq R\}$ and $S(0,R)=\{|x|=R\}$. Now we have the following formula

$$ \int_{B(0,R)}f(x)dx=\int_0^Rdr\int_{S(0,r)}f(y)dS(y) $$

My question is: I assume only that $f(x)$ is a measurable respect to Lebesgue measure and non-negative. Does the above formula hold? How do understand exactly the integral $\int_{S(0,r)}f(y)dS(y)$? Could I consider $\int_{S(0,r)}f(y)dS(y)$ as the Lebesgue integral of a Lebesgue measurable function $f$ on a Lebesgue measurable $S(0,r)$?

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The RHS is not what you should want to write. – Did Apr 11 '12 at 7:27
I have edited it – Jie Fan Apr 11 '12 at 12:24
up vote 2 down vote accepted

There is a coarea formula from which we can derive the following corollary (see p. 142 Krantz Parks, "Geometric Integration Theory"):

If $f \colon \mathbb{R}^{M} \to \mathbb{R}^{N}$ is a Lipschitz function and $M \geqslant N$, then $$ \int\limits_{A} g(x) J_{N}f(x) \, d\mathcal{L}^{M}(x) = \int\limits_{\mathbb{R}^N} \int\limits_{A \cap f^{-1}(y)} g \, d \mathcal{H}^{M-N} \, d \mathcal{L}^{N}(y) $$ holds for each Lebesgue measurable subset $A$ of $\mathbb{R}^{M}$ and each non-negative $\mathcal{L}^{M}$-measurable function $g \colon A \to \mathbb{R}$.

Here $\mathcal{L}^{N}$ is the Lebesgue measure, $\mathcal{H}^{M-N}$ is the $(M-N)$-dimensional Hausdorff measure. $J_{N}$ is the $N$-dimensional Jacobian (you can see more in Krantz, Parks book). If $N=1$, then $J_{N}f = |\nabla f|$.

Let's take $f(x) = |x|$ and $A = B(0,R)$. We will have $J_{N}f(x) = 1$ and the above formula will reduce to your formula. Hence your formula is valid for nonnegative Lebesgue measurable functions and $dS$ in this case is the $(n-1)$-dimensional Hausdorff measure.

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As Nimza stated above, you can interpret the integral over $S(0,R)$ to be taken with respect to Hausdorff measure $\mathcal H^{n-1}$. I would add, though, that it is not enough to assume Lebesgue measurability of $f$, in order for the integral to make sense. You need something stronger, like Borel measurability. Otherwise, your function might not be $\mathcal H^{n-1}$-measurable.

For example, let $E\subset S(0,1)$ be any subset that is not $\mathcal H^{n-1}$-measurable, and let $f=\chi_E\colon\mathbb R^n\rightarrow \mathbb R$ be the characteristic function of $E$. Then $f$ vanishes $\mathcal L^n$-almost everywhere, and is thus Lebesgue measurable, but the inner integral on the right does not make sense for $R=1$.

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