# Largeness of the set of zeroes of a Brownian motion

Definitions:

A measurable subset $$S$$ of $$\mathbb R$$ is said to be mesoscopic if there exists a continuous function $$f: \mathbb R \to \mathbb R$$ such that $$f(S)$$ is Lebesgue measurable and has nonzero Lebesgue measure.

Question: Is the set of zeroes of a Brownian motion almost surely a mesoscopic set?

Remark: Note that there exist mesoscopic sets of Lebesgue measure zero - for example the Cantor set with $$f$$ being the Cantor staircase function.

• I think there is a "not" missing from this definition somewhere. – Buzz Jun 20 at 0:15
• Ah corrected, thanks! – Nate River Jun 20 at 0:40