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Let $B$ be a standard Brownian motion, and $A$ a process of finite variation on compacts almost surely, not necessarily adapted to the Brownian filtration.

Question: Denoting by $\mathcal L$ the Lebesgue measure, is it true that $$\mathcal L(\{t \, | \, B_t = A_t \}) = 0$$

almost surely?

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Fix a finite interval $I$.It suffices to show that almost surely, $$\mathcal L(\{t \in I \, | \, B_t = A_t \}) = 0 \,.$$ Brownian motion restricted to $\{t \in I \, | \, B_t = A_t \}$ has bounded variation, so a positive answer is implied by the following stronger result, a special case of Theorem 1.3 of [1]:

Theorem[1] Let $\{B(t): t\in [0,1]\}$ be a standard Brownian motion. Then, almost surely, for all $S\subset [0,1]$, if $B|_{S}$ is of bounded variation, then $\overline{\dim}_M S\leq 1/2$.

Here $\overline{\dim}_M$ is the upper Minkowski dimension (a.k.a. upper box dimension.)

[1] Angel, Omer, Richárd Balka, András Máthé, and Yuval Peres. "Restrictions of Hölder continuous functions." Transactions of the American Mathematical Society 370, no. 6 (2018): 4223-4247. https://www.ams.org/journals/tran/2018-370-06/S0002-9947-2018-07126-4/S0002-9947-2018-07126-4.pdf http://wrap.warwick.ac.uk/84253/7/WRAP-restrictions-continuous-functions-Peres-2018.pdf

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    $\begingroup$ Ah, I wonder if I had missed this in your Brownian motion book… $\endgroup$
    – Nate River
    Commented Feb 4, 2022 at 7:17
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    $\begingroup$ This is not discussed in the book- the relevant work was done well after the book was published. $\endgroup$
    – Yuval Peres
    Commented Feb 5, 2022 at 16:43
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    $\begingroup$ I see! Though there is actually one subtlety - the set on which $B_t = A_t$ is random, while in the theorem $S$ is deterministic. Does this affect anything at all? $\endgroup$
    – Nate River
    Commented Feb 6, 2022 at 5:11
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    $\begingroup$ No, the set $S$ in the Theorem is allowed to be random (and depend on the BM). $\endgroup$
    – Yuval Peres
    Commented Feb 6, 2022 at 19:49
  • $\begingroup$ I have read the linked paper, and have a question about the result for Brownian motion - isn’t it true that Theorem 4.3 implies that the result for Brownian motion may be extended to the critical $\beta = \alpha$ case? Specifically, using the fact that the BM is in $\mathcal A_n (\gamma, 2)$ for all $\gamma < \alpha$ and large enough $n$, and sending $\gamma \to \alpha^-$, we deduce that almost surely there exists no set $A$ on which the BM is $\alpha$ Hölder continuous and $\bar{\text{dim}}_M (A) > 1 - \alpha$. I may be making a mistake though… $\endgroup$
    – Nate River
    Commented May 29, 2023 at 2:22

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