# Complex Measure Agreeing on Certain Balls

I came across this problem and am lost as to how to solve it.

Let $r>0$ be fixed. Suppose $\mu, \nu$ are complex Borel measures on $\mathbb{R}^d$ such that for each open ball B of radius $r$, $\mu(B)=\nu(B)$. Then $\mu=\nu$.

I thought this might be an application of $\pi-\lambda$ theorem, but I realized we wouldn’t have a $\pi$-system with the collection of open balls $B$ since intersecting such balls might not necessarily be open or have radius $r$ still. Any help would be greatly appreciated.

#### Solutions Collecting From Web of "Complex Measure Agreeing on Certain Balls"

In which context did you come across this claim/exercise? The claim is indeed true, but it seems to me (see below) to be more of a Fourier-analytic theorem than a theorem of measure theory. If you do not know much about the Fourier transform, the following proof will probably be (almost) useless to you 🙁

By considering $\gamma := \mu – \nu$ it suffices to show $\gamma \equiv 0$ if $\gamma (B_r (x)) = 0$ for all $x \in \Bbb{R}^d$.

Let $f := \chi_{B_r (0)}$ be the indicator function/characteristic function of the ball of radius $r$ around the origin. We consider the convolution (cf. also http://en.wikipedia.org/wiki/Convolution#Convolution_of_measures) of the complex measure $\gamma$ and the function $f$, i.e.

$$\gamma \ast f (x) = \int f(y-x) \, d\gamma(y) = \gamma(B_r (x)) = 0 \text{ for all x } \in \Bbb{R}^d.$$

But by the convolution theorem for the Fourier transform, we have

$$0 \equiv \mathcal{F} ( \gamma \ast f) = \widehat{\gamma} \cdot \widehat{f} \qquad (\dagger)$$

Since $\widehat{f}$ has compact support, the Paley-Wiener theorem (or an explicit differentiation under the integral) shows that the Fourier transform $\widehat{f}$ extends to an entire function $\Bbb{C}^d \to \Bbb{C}$, which we again denote by $\widehat{f}$. Since $f \not\equiv 0$, we also have $\widehat{f} \not\equiv 0$.

As $\widehat{f}$ is analytic, this implies that the zero-set

$$\{\xi \in \Bbb{R}^d \mid \widehat{f}(\xi) = 0\}$$

has empty interior, i.e. $\{\xi \mid\widehat{f}(\xi) \neq 0\}$ is dense in $\Bbb{R}^d$.

Together with the continuity of $\widehat{\gamma}$ and with $(\dagger)$, this yields $\widehat{\gamma} \equiv 0$ and thus $\gamma \equiv 0$, completing the proof.

EDIT: Regarding the zero sets of analytic functions, see e.g. here Zeros set of analytic functions over complex plane with several variables. But this does not really cover our case, since we are considering a real analytic function. Let us assume that we already know that real-analytic functions in one variable have only isolated zero sets. (This follows easily by considering the power series $\sum a_n x^n$ and letting $n$ minimal with $a_n\neq0$, …).

For the multidimensional case, assume $f\equiv 0$ on some open ball $B_r (x_0)$. Let $x\in \Bbb{R}^d$ be arbitrary and consider the analytic(!) single variable function

$$t \mapsto f((1-t)x_0 + txt)$$

This function vanishes on an open neighborhood neighborhood of $0$, so that the 1-dimensional case implies that it vanishes identically. Hence, $f$ vanishes identically.

EDIT 2: It should also be observed that the claim is not true for measures of infinite (but still locally finite) measures: Consider for example

$$d\mu(x)= (1+\cos(\pi x))\,dx$$
on $\Bbb{R}$ with $r=1$. What happens here is that $\hat{\gamma}$ is not a continuous function anymore, but a finite sum of diracs. Thus, the argument above fails, since $\hat{f}$ has some (isolated) zeros.