We know that the $\sigma$-algebra generated by open balls in $\mathbb{R}^n$ is the Borel $\sigma$-algebra (with the Euclidean metric).
In general, does the set of open balls of a metric space generate the Borel $\sigma$-algebra? If not, is there a sufficient condition for this to be the case?
Best Answer
No. See, for example, this question here or this question on MO. Separability, or equivalently second countability, is sufficient to guarantee the ball sigma-algebra is the same as the Borel sigma-algebra (since the topology is then countably generated by balls). Infinite-dimensional normed spaces are typically not separable, and the ball $\sigma$-algebra is smaller than the Borel $\sigma$-algebra.
The ball $\sigma$-algebra does have rather curious applications in nonparametric statistics and empirical process theory, mainly because many interesting functions (e.g. Banach-space-valued functions) are ball-measurable but not Borel measurable, and we still have a lot of nice things (regularity, weak convergence for measures whose support is separable, etc.). However, when they don't coincide, it meant that continuous functions are no longer necessarily measurable, which does create some oddities.