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Good morning,

I'm reading a paper of W. Stoll in which the author uses some implicit facts (i.e. he states them without proofs and references) in measure theory. So I would like to ask the following question:

Let $G$ be a bounded domain in $\mathbb{R}^n$ and $S^{n-1}$ the unit sphere in $\mathbb{R}^n.$ For each $a\in S^{n-1},$ define $L(a) = \{x.a~:~ x\in \mathbb{R}\}.$ Denote by $L^n$ the n-dimensional Lebesgue area. Is the following formula true? $$\int_{a\in S^{n-1}}L^1(G\cap L(a)) = L^n(G) = \mathrm{vol}(G).$$

Could anyone please show me a reference where there is a proof for this? If this formula is not true, how will we correct it?

Thanks in advance,

Duc Anh

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Could you provide a link to the paper? –  nullUser Jan 3 '13 at 20:49
    
link.springer.com/article/10.1007%2FBF01117123 my question comes from the page 160 of the paper where the author says that "the set of all $a\in S(1)$ where the divisor is degenerated is a set of measure zero on $S(1)$", so I guessed the above formula. –  Đức Anh Jan 3 '13 at 21:14
    
What does the notation $x.a$ mean? –  nullUser Jan 3 '13 at 21:25
    
it is the multiplication of the vector $a$ with the scalar $x.$ –  Đức Anh Jan 4 '13 at 2:05

1 Answer 1

up vote 1 down vote accepted

This formula seems to be false. Consider the case of the unit disk in $\mathbb{R}^2$, $D^2$. This is obviously bounded.

$L^1(D^2 \cap L(a)) = 2$ for any $a$ in $S^1$, as the radius of $D^2$ is 1, and the intersection of the line through the origin that goes through $a$ and $D^2$ has length 2. The integral on the left is therefore equal to $4\pi$ and $L^2(D^2)$ was known by the greeks to be $\pi$.

So your formula is like computing an integral in polar coordinates without multiplying the integrand by the determinant of the jacobian.

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thank you very much. I will search for another formula. –  Đức Anh Jan 26 '13 at 17:30

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