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.

  • $\begingroup$ I think there is a "not" missing from this definition somewhere. $\endgroup$
    – Buzz
    Jun 20, 2021 at 0:15
  • $\begingroup$ Ah corrected, thanks! $\endgroup$
    – Nate River
    Jun 20, 2021 at 0:40

1 Answer 1


Yes, the local time (at zero) maps the zero set of Brownian motion to an interval. See e.g. Lemma 6.9 page 159 in [1] for continuity.

[1] Brownian motion, by Peter Mörters and Yuval Peres. Cambridge University Press, 2010 https://people.bath.ac.uk/maspm/book.pdf

  • $\begingroup$ Oh, I was not aware that the local time was continuous. I somehow intuitively had in mind a jump process, but I guess it makes more sense that it would be continuous. $\endgroup$
    – Nate River
    Jun 20, 2021 at 2:42

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