Fourier–Stieltjes transform

Remarks

The integral exists because the integrand has modulus $1$ and $\mu$ has finite total variation $\norm{\mu}$. So $\abs{\hat\mu(\xi)} \le \norm{\mu}$, and $\hat\mu$ is uniformly continuous on $\R^n$.

Example: the Dirac measure $\delta_a$ at a point $a \in \R^n$ is a finite measure, and $\hat\delta_a(\xi) = e^{-2\pi i a \cdot \xi}$; in particular $\hat\delta_0 = 1$.

Example: if $d\mu = f\, dx$ with $f \in L^1(\R^n)$, then $\hat\mu$ is the usual Fourier transform $\hat f$.

Non-example: Lebesgue measure on $\R$ is not finite, and $\int_\R e^{-2\pi i x \xi}\, dx$ diverges for every $\xi$; its transform exists only as a tempered distribution.

Unlike the transform of an integrable function, $\hat\mu$ need not tend to $0$ at infinity, as the example $\delta_0$ shows. The transform is injective: $\hat\mu = 0$ implies $\mu = 0$. It turns convolution of measures into products, $\widehat{\mu * \nu} = \hat\mu\, \hat\nu$. For a probability measure $\mu$ on $\R$, the characteristic function $\varphi(t) = \int e^{itx}\, d\mu(x)$ equals $\hat\mu(-t / (2\pi))$. On $\R$, writing $\mu$ through a function $F$ of bounded variation, the transform is the Stieltjes integral $\int e^{-2\pi i x \xi}\, dF(x)$.