Two Borel Probability Measure ($\mathbb{R^2}$) Agreeing on Open Balls

Question

In the problem it is required to prove that two Borel probability measures (in $\mathbb{R^2}$) agreeing on all open circles are equal.

I know about this question (Complex Measure Agreeing on Certain Balls) but there is a more general statement proved and in my case I think there is a more understandable and simple solution.

I have done the following:

Let $\mu_1$ and $\mu_2$ - the specified measures. Define $S = \{B \in \mathscr{B}(\mathbb{R^2}) : \mu_1(B) = \mu_2(B) \}$. I know that S is a monotone class also I know that $\mathbb{R^2} \in S$, $\ (A ,B \in S \: and \: A \subset B) \Rightarrow (B \setminus A \in S) $, $\: A_i \in S \: and \: $ $A_i$ - don't intersect in pairs then $\bigcup_i A_i \in S$. But then it all comes down to the fact that S is not closed with respect to the final intersections.

Maybe there is some way to get around this or come up with a better S system?

Answer 1

Your question is solved in this paper by Jens P.R. Christensen. Let $B(x,r)$ denote the open ball centered at $x$ with radius $r$.

Theorem: (Christensen) Let $(M, d)$ be a locally compact metric space for which there exists a uniform measure on $M$. If $m$ is a signed measure with $m(B(x,r))=0$ for all $x\in M$ and all $r>0$, then $m$ is the zero measure.

In the paper, Christensen defines the term "uniform measure". In the case of measures on metric spaces, a measure $u$ is uniform if $u(B(x,r))=u(B(y,r))$ for all $x,y\in M$ and all $r>0$. That is, all balls with the same radius have the same measure. Clearly, Lebesgue measure on $\mathbb R^2$ is uniform, so this theorem applies.

J. P. R. Christensen, On some measures analagous to Haar measure, Mathematica Scandinavica, 26, 1970, 103 - 106. https://doi.org/10.7146/math.scand.a-10969