equality of two Borel measures Assume we have a measurable space $(X, \mathscr{B})$, where $X$ is a separable metric space and $\mathscr{B}$ is the Borel sigma algebra. Then, since $X$ is separable, then, $\mathscr{B}...--prophetes.ai

equality of two Borel measures Assume we have a measurable space $(X, \mathscr{B})$, where $X$ is a separable metric space and $\mathscr{B}$ is the Borel sigma algebra. Then, since $X$ is separable, then, $\mathscr{B}$ equals to the sigma algebra generated by open balls. The question: assume that two probability measures on $\mathscr{B}$ are such that they agree on every open ball (just ball!) of $X$. Is it true that they are equal?

There is a counter-example to the general result in Theorem II of [1]. For results in the positive direction, you may be interested in [2].

**References**

[1] _Measures not approximable or not specifiable by means of balls._ Roy O. Davies, Mathematika, Volume 18, Issue 2 December 1971, pp. 157-160.

[2] _Measures which agree on balls._ J. Hoffmann-Jørgensen, Math. Scand. 37 (1975), no. 2, 319–326.