Expressing the measure of a set in terms of the characteristic function of the measure

Question

Let $\mu$ be a discrete, finitely supported probability measure in $\mathbb{R}^d$ and denote by $\phi$ be the characteristic function of $\mu$, i.e. $\phi(t)=\mathbb{E}e^{i<t,X>}$, where $X$ is a random variable with distribution $\mu$. Given a bounded open set $A\subset \mathbb{R}^d$, how does one express $\mu (A)$ in terms of the characteristic function? Is it true that something like the following integral expression $$\mu(A)=\int f\phi,$$ holds? Here $f$ stood for the Fourier transform of the indicator function of the set $A$? I would be very grateful for a reference.

Answer 1

$\newcommand\R{\mathbb R}\newcommand\sn{\operatorname{sign}}$Let $n:=d$ and $f:=\phi$, so that $f(t)=Ee^{it\cdot X}$ for $t\in\R^n$. We have the following straightforward multivariate extension of (the special case, with $t=0$, of) formula (6) in this arXiv paper or its published version: $$K_af:=\frac1{(\pi i)^n}\, \int_{\R^n}e^{-iu\cdot a}f(u)\frac{du_1\cdots du_n}{u_1\cdots u_n}=E\prod_{j\in[n]}\sn(X_j-a_j),\tag{1}$$ where $a=(a_1,\dots,a_n)\in\R^n$, $[n]:=\{1,\dots,n\}$, $\sn$ is the sign function, and the integral is is understood in the principal-value sense.

For any real $A_j$ and $B_j$, $$\prod_{j\in[n]}(B_j-A_j)=\sum_{J\subseteq[n]}(-1)^{|J|}\prod_{j\in J}A_j\;\prod_{j\in [n]\setminus J}B_j.\tag{2}$$

Take now any $a=(a_1,\dots,a_n)\in\R^n$ and $h=(h_1,\dots,h_n)\in(0,\infty)^n$, and for each $J\subseteq[n]$ define $a^{h,J}\in\R^n$ by the formula $$a^{h,J}:=a_j\,1(j\in J)+(a_j-h_j)\,1(j\in[n]\setminus J)$$ for all $j\in[n]$. Note that for each real $t$ we have $$\sn(t+s)-\sn t\to1(t=0)\tag{3}$$ as $s\downarrow0$. So, by (1), (2), (3), and dominated convergence, \begin{align} &\sum_{J\subseteq[n]}(-1)^{|J|}K_{a^{h,J}}f \\ &=\sum_{J\subseteq[n]}(-1)^{|J|} E\prod_{j\in J}\sn(X_j-a_j)\;\prod_{j\in [n]\setminus J}\sn(X_j-a_j+h_j) \\ &=E\sum_{J\subseteq[n]}(-1)^{|J|} \prod_{j\in J}\sn(X_j-a_j)\;\prod_{j\in [n]\setminus J}\sn(X_j-a_j+h_j) \\ &=E\prod_{j\in[n]}(\sn(X_j-a_j+h_j)-\sn(X_j-a_j)) \\ &\to E\prod_{j\in[n]}1(X_j=a_j)=P(X=a), \end{align} where the convergence holds as $h\downarrow0$ -- in the sense that $h_j\downarrow0$ for all $j\in[n]$.

Thus, for each $a\in\R^n$ $$\mu(\{a\})=P(X=a)=\lim_{h\downarrow0}\sum_{J\subseteq[n]}(-1)^{|J|}K_{a^{h,J}}f.$$ This completely determines the discrete probability distribution $\mu$, since for any $A\subseteq\R^n$ $$\mu(A)=\sum_{a\in A}\mu(\{a\}) =\sum_{a\in A}\lim_{h\downarrow0}\sum_{J\subseteq[n]}(-1)^{|J|}K_{a^{h,J}}f.$$ (The general, not necessarily "discrete", case is similar.)