Let $X$ be a random variable in $\mathbb{R}^n$ with distribution $\mu$ and characteristic function $\varphi$ (i.e. $\varphi(t)=\mathbb{E} e^{i\langle t,X\rangle}$). The standard inversion formula asserts that $$\mu \big(\{a<x<b\}\big) = \frac{1}{(2\pi)^n} \lim_{T_1\to\infty}\cdots\lim_{T_n\to\infty} \int\limits_{-T_1\leq t_1\leq T_1} \cdots \int\limits_{-T_n \leq t_n \leq T_n} \prod_{k=1}^n\left(\frac{e^{-it_ka_k}-e^{-it_kb_k}}{it_k}\right)\varphi_X(t)\lambda(dt_1 \times \cdots \times dt_n),$$ where the set $\{ a<x<b\}$ is the rectangle $(a_1,b_1)\times\cdots \times (a_n,b_n).$

Instead of working with rectangles I am interested in working with open Euclidean balls. Is the corresponding integral relationship similar to the one above known in this case? I would be also very grateful for related references.

  • $\begingroup$ Are you assuming that $\mu$ is absolutely continuous? $\endgroup$ Mar 29, 2021 at 13:44
  • $\begingroup$ No, I am particularly interested in the case where $\mu$ is discrete and supported on a lattice. $\endgroup$
    – TOM
    Mar 29, 2021 at 14:47
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    $\begingroup$ Not sure it is related but Fefferman showed things can go wrong when looking at convergence of Fourier series in 2d on squares versus disks. See math.stackexchange.com/questions/1849763/… $\endgroup$ Mar 29, 2021 at 18:02
  • $\begingroup$ A thank you to @MichaelHardy for the clarity mods. $\endgroup$ Mar 30, 2021 at 3:25

2 Answers 2


A general way to view these formulae is as a Parseval identity. If your measure were of the form $d\mu(x)=f(x)\,dx$, where $f\in L^2$, then you would have, for any set $E$ of finite positive measure, $$ \mu(E)=\int_{\mathbb{R}^n}\mathbf{1}_Ef=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \hat{\mathbf{1}}_E\hat{\mathbf{f}}=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} \hat{\mathbf{1}}_E\varphi_X. $$ In the case $E$ is the unit ball, the Fourier transform is $\hat{\mathbf{1}}_E(\xi)=(2\pi)^{\frac{n}{2}}\lVert \xi\rVert^{-n/2}J_{n/2}(\lVert\xi\rVert)$, where $J$ is the Bessel function.

Of course, if $\mu$ is less regular, then the RHS does not need to converge absolutely. There are several ways to make sense of the formula nevertheless, namely:

  • replace $\mathbf{1}_E$ with a test function, then pass to the limit;
  • approximate $\mu$ with a more regular measure, then pass to a limit (as in the answer by Iosif Pinelis);
  • replace the integral by its principal value, i. e., integrate over a box or ball of size $T$ centered about the origin and send $T$ to infinity.

The reason the last approach often works is because cutting out the high frequencies has a smoothening effect on both $\mathbf{1}_E$ and $f$, so it is essentially a combination of the first two. In particular, it always works when $\mu$ is supported away from the boundary of $E$; indeed, by writing $\mu=\mu\star\delta$ and exchangind integrals, it suffices to check the formula for delta-measures, in which case it is just the Fourier inversion formula at a point of smoothness of the original function.

If, on the other hand, $\mu$ is allowed to be supported on the boundary, then the v. p. integration may fail, but so does your formula for rectangles, as far as I can see. For example, in dimension 1, RHS is additive under the union $(a,b)\cup(b,c)$, while the LHS is not if $\mu(\{b\})\neq 0$.


$\newcommand\ep\varepsilon\newcommand\de\delta\newcommand\De\Delta\newcommand\vp\varphi\newcommand\R{\mathbb R}$For real $\ep>0$, let $\mu_\ep$ be the distribution of $X+\ep Z$, where $Z$ is a standard Gaussian random vector in $\R^d$ independent of $X$. Then the pdf $p_\ep$ of $\mu_\ep$ is given by $$p_\ep(x)=\frac1{(2\pi)^d}\int_{\R^d}dt\,e^{-it\cdot x-\ep^2|t|^2/2}\vp_X(t)$$ for $x\in\R^d$.

Note that $X+\ep Z$ converges to $X$ almost surely (as $\ep\downarrow0$), and hence $\mu_\ep$ converges to $\mu$ weakly.

So, for any ball $B$ in $\R^d$ with $\mu(\partial B)=0$, $$ \begin{aligned} \mu(B)&=\lim_{\ep\downarrow0}\mu_\ep(B) \\ &=\lim_{\ep\downarrow0}\int_B dx\, p_\ep(x) \\ &=\nu(B):=\frac1{(2\pi)^d}\lim_{\ep\downarrow0}\int_B dx\, \int_{\R^d}dt\,e^{-it\cdot x-\ep^2|t|^2/2}\vp_X(t). \end{aligned} $$

Take now any open ball $B$ of some radius $r\in(0,\infty)$ and let $B_{-\de}$ denote the open ball concentric with $B$ of radius $r-\de$, where $\de\in(0,r)$. Let $\De:=\{\de\in(0,r)\colon\mu(\partial(B_{-\de}))=0\}$. Then $$\mu(B)=\lim_{\de\downarrow0}\mu(B_{-\de})=\lim_{\de\downarrow0,\,\de\in\De}\mu(B_{-\de}),$$ in view of the monotonicity of $\mu(B_{-\de})$ in $\de$ and because the set $(0,r)\setminus\De$ is at most countable. Therefore and in view of the monotonicity of $\nu(B_{-\de})$ in $\de$, $$\begin{aligned} \mu(B)&=\lim_{\de\downarrow0,\,\de\in\De}\nu(B_{-\de}) \\ &=\lim_{\de\downarrow0}\nu(B_{-\de}) \\ &=\frac1{(2\pi)^d}\lim_{\de\downarrow0}\lim_{\ep\downarrow0}\int_{B_{-\de}} dx\, \int_{\R^d}dt\,e^{-it\cdot x-\ep^2|t|^2/2}\vp_X(t). \end{aligned}$$

  • $\begingroup$ Thank you, this is most useful! $\endgroup$
    – TOM
    Mar 29, 2021 at 16:18
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    $\begingroup$ There seems to be a problem with the formula (or maybe I do not see something trivial). Consider a random variable $X$ such that $\mathbb{P}(X=\pm 1)=1/2$. Then the probability that $X$ hits the interval $(-1,1)$ is 0, but the smoothened version of $X$ assigns probability roughly $1/2$ to this interval (with limit $1/2$ as $\varepsilon \rightarrow 0$). Am I missing something here? $\endgroup$
    – TOM
    Mar 30, 2021 at 13:56
  • $\begingroup$ @TOM : Thank you for your comment. This is now fixed. $\endgroup$ Mar 30, 2021 at 14:53
  • $\begingroup$ In the paragraph startin So, for any ball $B$ in $\mathbb R^d$ with you probably mean $\mu(\partial B)=0$. $\endgroup$ Mar 31, 2021 at 5:06
  • $\begingroup$ @JochenWengenroth : Yes, of course. :-) Thank you for your comment. $\endgroup$ Mar 31, 2021 at 6:27

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