Editing Fourier transform (section)

=== Fourier–Stieltjes transform on measurable spaces <span class="anchor" id="Fourier-Stieltjes transform"></span>===
{{see also|Bochner–Minlos theorem|Riesz–Markov–Kakutani representation theorem|Fourier series#Fourier-Stieltjes series}}
The Fourier transform of a [[finite measure|finite]] [[Borel measure]] {{mvar|μ}} on {{math|'''R'''<sup>''n''</sup>}} is given by the continuous function:{{sfn|Pinsky|2002|p=256}}
<math display="block">\hat\mu(\xi)=\int_{\mathbb{R}^n} e^{-i 2\pi x \cdot \xi}\,d\mu,</math>
and called the ''Fourier-Stieltjes transform'' due to its connection with the [[Riemann–Stieltjes integral#Application to functional analysis|Riemann-Stieltjes integral]] representation of [[Radon measure|(Radon) measures]].{{sfn|Edwards|1982|pp=53,67,72-73}} If <math>\mu</math> is the [[probability distribution]] of a [[random variable]] <math>X</math> then its Fourier–Stieltjes transform is, by definition, a [[Characteristic function (probability theory)|characteristic function]].<ref>{{harvnb|Katznelson|2004|p=173}} {{pb}} The typical conventions in probability theory take {{math|''e''<sup>''iξx''</sup>}} instead of {{math|''e''<sup>−''i''2π''ξx''</sup>}}.</ref> If, in addition, the probability distribution has a [[probability density function]], this definition is subject to the usual Fourier transform.{{sfn|Billingsley|1995|p=345}} Stated more generally, when <math>\mu</math> is [[Absolute continuity#Absolute continuity of measures|absolutely continuous]] with respect to the Lebesgue measure, i.e., 
<math display="block"> d\mu = f(x)dx,</math>
then
<math display="block">\hat{\mu}(\xi)=\hat{f}(\xi),</math>
and the Fourier-Stieltjes transform reduces to the usual definition of the Fourier transform. That is, the notable difference with the Fourier transform of integrable functions is that the Fourier-Stieltjes transform need not vanish at infinity, i.e., the [[Riemann–Lebesgue lemma]] fails for measures.{{sfn|Katznelson|2004|pp=155,164}}

[[Bochner's theorem]] characterizes which functions may arise as the Fourier–Stieltjes transform of a positive measure on the circle.

One example of a finite Borel measure that is not a function is the [[Dirac measure]].{{sfn|Edwards|1982|p=53}} Its Fourier transform is a constant function (whose value depends on the form of the Fourier transform used).