characteristic function determines the distribution

Question

I noticed that some probability and stats books talk about the characteristic function of a probability distribution of $\mathbb{R}^n$ and make the claim that the characteristic function uniquely determines the distribution even if the distribution does not have a density wrt to lebesgue measure (else this follows from injectivity of the fourier transform). I assume this is not true for all Borel probability measures? What are the correct general conditions on the measure? And can I somehow use the proof I already know that the fourier transform is injective (namely by convolving $f$ with a sequence of $\phi_k$ so that $$f\ast \phi_k \rightarrow f$$ in $L^1$ and using properties of the Fourier transform to write $f\ast \phi_k$ in terms of $\hat{f}$) to prove the statement about characteristic functions?

I notice, in trying to adapt the proof, for instance, that i would at least need translation to be continuous with respect to the $L^1_P$ norm, where $P$ is the probability distribution. I'm not sure whether or not this in itself implies $dP$ is absolutely continuous with respect to Lebesgue measure...

Answer 1

It is true for all measures, once you decide what it means to say "uniquely determines." The correct formulation is that the characteristic functions of two measures are equal if and only if they have the same cumulative distribution function. In probability, this means that the two associated random variables are equal in distribution. One fairly explicit way to prove this is by the inversion formula.

Answer 2

If you know basic facts about the Fourier transform for functions on $\mathbb{R}^n$, here is a simple way to show injectivity of the map $$\mu \mapsto \hat{\mu}(\xi):= \int_\mathbb{R^n}e^{-ix\xi}d\mu(x)$$ sending a Borel probability measure to its characteristic function:

You can easily check that for any $\varphi\in C_c^\infty(\mathbb{R^n})$ we have

$$\int_\mathbb{R^n}\varphi(\xi)\cdot\hat{\mu}(\xi)~~d\lambda(\xi)= \int_\mathbb{R^n}\hat{\varphi}(x)~~d{\mu}(x)$$ where $\lambda$ is Lebesgue measure and $\hat{\varphi}$ is the Fourier transform of $\varphi$. But every smooth compactly supported function is of the form $\hat{\varphi}$, and $\mu$ is determined by the integrals of these functions. Thus $\mu$ is determined by $\hat{\mu}$.