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.



Expressing the measure of a set in terms of the characteristic function of the measure
$\newcommand\R{\mathbb R}\newcommand\sn{\operatorname{sign}}$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](https://arxiv.org/pdf/1309.5928.pdf) or its [published version](https://www.sciencedirect.com/science/article/pii/S0167715215002722):
$$K\_af:=\frac1{(\pi i)^n}\,
\int\_{\R^n}e^{-iu\cdot a}f(u)\frac{du\_1\cdots du\_n}{u\_1\cdots u\_n}=E\prod\_{j\in[n]}\sn(X\_j-a\_j),\tag{1}$$
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.)


