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I need this as lemma.

Given the Borel space $\mathcal{B}(\mathbb{R})$.

Consider a complex measure: $$\mu:\mathcal{B}(\mathbb{R})\to\mathbb{C}$$

Then one has: $$\int_{-\infty}^{+\infty}e^{it\lambda}\mathrm{d}\mu(\lambda)=0\implies\mu=0$$

How can I prove this?

C-star-W-star
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  • Would it be inappropriate to argue that if the Fourier transform of a tempered distribution is $0$, then the tempered distribution is $0$? – paul garrett Jun 08 '15 at 23:47
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    @paulgarrett this seems to be for a more general measure than ordinary Lebesgue measure, so classical Fourier analysis theorems won't apply. – Adam Hughes Jun 08 '15 at 23:57
  • To the op: in what sense do you mean that integral? PV, or are you assuming that $e^{it\lambda}$ is in $L^1(\Bbb R, \mathcal{B}(\mathbb{R}),\mu)$? – Adam Hughes Jun 08 '15 at 23:59
  • @paulgarrett: Yep thats the problem: Not necessarily Lebesgue measure nor absolutely continuous measure nor positive measure. – C-star-W-star Jun 09 '15 at 00:00
  • @AdamHughes: It exists as Lebesgue integral: $\int_{-\infty}^{+\infty}|e^{it\lambda}|\mathrm{d}|\mu|(\lambda)=|\mu|(\mathbb{R})<\infty$ – C-star-W-star Jun 09 '15 at 00:02
  • Perfect, thanks for clarifying. – Adam Hughes Jun 09 '15 at 00:03
  • @AdamHughes: No problem, you're welcome!! :) – C-star-W-star Jun 09 '15 at 00:03
  • In fact every complex Borel measure on $\mathbb{R}$ is regular. – zhw. Jun 09 '15 at 00:34
  • @AdamHughes, depending on context, since it's not necessary that a "measure" be comparable to Lebesgue measure to qualify as a "tempered distribution", I'm not at all sure what's going on... but only that it's not entirely illegal to have measures (Dirac deltas, to say the least) that aren't comparable to Lebesgue... But maybe I don't understand the context... – paul garrett Jun 09 '15 at 00:34
  • @paulgarrett sorry if I was unclear, I just meant that quoting the theorem that the transform is injective, is basically saying "it's true because it's true," since that theorem is only widely-known in the form $\mu$ is Lebesgue measure (the classical setting). – Adam Hughes Jun 09 '15 at 00:39
  • @AdamHughes, ah, ok, now I understand the/your intention. Thanks for clarifying for me! – paul garrett Jun 09 '15 at 00:43

1 Answers1

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Let $f\in C^2(\mathbb {R})$ have compact support. Integrating by parts twice shows $\hat f (x) = O(1/x^2)$ at $\infty;$ hence $\hat f \in L^1.$ By the Fourier inversion formula, $f$ is the inverse Fourier transform of $\hat f.$ Thus

$$\int_{\mathbb {R}} f(x)\,d\mu(x) = \int_{\mathbb {R}}\int_{\mathbb {R}}\hat {f}(t)e^{ixt}\,dt \,d\mu(x) = \int_{\mathbb {R}}\hat {f}(t)\int_{\mathbb {R}}e^{ixt} d\mu(x)\,dt = 0.$$

Because the set of such $f$'s is dense in $C_0(\mathbb {R)},$ and the space of finite Borel measures form the dual space of $C_0,$ we see $\mu = 0.$

zhw.
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