Uniqueness theorem for Fourier–Stieltjes transforms

Remarks

Equivalently, two measures in $M(G)$ with the same transform are equal.

Proof idea: fix a Haar measure $\nu$ on $\widehat{G}$, and for $\varphi \in L^1(\widehat{G})$ let $\check{\varphi}(x) = \int_{\widehat{G}} \varphi(\chi) \chi(x)\, d\nu(\chi)$. By Fubini's theorem, $\int_G \check{\varphi}\, d\mu = \int_{\widehat{G}} \varphi(\chi)\, \widehat{\mu}(\overline{\chi}) \, d\nu(\chi) = 0$. The functions $\check{\varphi}$ form a dense subspace of $C_0(G)$, by the Stone–Weierstrass theorem. By the Riesz representation theorem, a measure that integrates every function of $C_0(G)$ to $0$ is the zero measure.

Example: for $G = \R$ and a probability measure $\mu$, $\widehat{\mu}$ is the characteristic function of $\mu$ up to a change of variable. So the theorem says that the characteristic function determines the distribution. For $G = \mathbb{T}$ it says that a measure on the circle with all Fourier coefficients $0$ is zero.

Going deeper

For a student

The theorem makes the Fourier–Stieltjes transform an injective algebra homomorphism from $M(G)$ into the bounded continuous functions on $\widehat{G}$. It contains the injectivity of the Fourier transform on $L^1(G)$ as the case $d\mu = f\, dx$. In probability it is the reason why characteristic functions identify distributions, which is the first step of the Lévy continuity theorem and of the proof of the central limit theorem.