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I have two measures $\mu$ and $\nu$ supported on compacts in $\mbox{int } \mathbb{R}^{n}_+$. Are there some sufficiently general classes of such measures for which $$ \int\limits_{\mathbb{R}^n_+} \frac{\mu(dx)}{x_1^{z_1} x_2^{z_2}\cdots x_n^{z_n}} = \int\limits_{\mathbb{R}^n_+} \frac{\nu(dx)}{x_1^{z_1} x_2^{z_2}\cdots x_n^{z_n}} $$ that holds for any $\Re z_1 > R$, $\Re z_2 > R$, ..., $\Re z_n > R$ and $R>1$ implies $\mu = \nu$?

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Do you know such a "sufficiently general class" in the case $n=1$? – GEdgar Feb 10 '12 at 15:47
Maybe, a class of finite sums of delta functions and its natural generalisation. More nontrivial classes are given by the Mellin inversion theorem. I don't know, it there a Mellin inversion theorem for measures in $\mathbb{R}^n_+$? Actually, my question is about it. – Nimza Feb 10 '12 at 16:06

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It's true for any regular complex Borel measures with compact support. Let $K$ be the union of the supports of the two measures. The linear span $V_0$ of the functions $1/(x_1^{z_1} \ldots x_n^{z_n})$ for $\Re z_j > 0$ is dense in $C(K)$ by the Stone-Weierstrass Theorem. Since your functions are of the form $f/(x_1 \ldots x_n)^R$ for $f \in V_0$ and $(x_1 \ldots x_n)^R$ has no zeros on $K$, their linear span is also dense in $C(K)$. So $\mu$ and $\nu$, corresponding to continuous linear functionals on $C(K)$, agree on a dense set and therefore are equal.

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Thank you, but the Stone-Weierstrass Theorem requires subalgebra to contain a nonzero constant. I think that we can get rid of this problem adding the equality $\mu(\mathbb{R}^n_+) = \nu(\mathbb{R}^n_+)$. And what about the finite measures with a noncompact support? Can we use some analogues of the Stone-Weierstrass Theorem? – Nimza Feb 10 '12 at 18:45
$V_0$ doesn't contain a constant, but it contains functions that approximate a constant arbitrarily closely, so Stone-Weierstrass still applies (or if you prefer you could change $\Re z_j > 0$ to $\Re z_j \ge 0$). For finite measures $\mu$ and $\nu$ with noncompact support, there is a generalization of Stone-Weierstrass: if $X$ is locally compact, a subalgebra $V$ of $C_0(X)$ (the continuous functions that vanish at $\infty$) is dense in $C_0(X)$ if it separates points and there is no $x \in X$ where all members of $V$ vanish. – Robert Israel Feb 10 '12 at 20:19
Thank you, it's clear. – Nimza Feb 11 '12 at 11:42

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