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Let $B$ be a standard Brownian motion and the local time of $B$ at $0$ defined by Tanaka's formula \begin{equation} L_t = |B_t| - \int_0^t sgn(B_s) dB_s \ . \end{equation} Now I want to prove that $L_t$ only increases on the zero set of $B$. I was given the pointer to explain why it suffices to show that for fixed rationals $r < r'$, $B \neq 0$ on $[r,r']$ implies $L_r = L_{r'}$. I don't really see how looking at rationals will help me here.

Edit: What I tried: Let $p<q$ be rationals and $B_s \neq 0$ a.s. for all $s \in [p,q]$. By continuity of $L$ we have that either $B_s >0$ or $<0$ for all $s \in [p,q]$. Then, \begin{equation} L_{q} - L_p = | B_{q} | - | B_p |- \int_p^q sgn(B_s) dB_s = 0, \end{equation} because $sgn(B_s) = 1$ or $-1$.

Is this approach correct? And why wouldn't it work for any reals $u<v \in \mathbb{R}$?

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    $\begingroup$ Seems that you could have just used $p,q$ reals here $\endgroup$
    – fGDu94
    Nov 25, 2019 at 3:54
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    $\begingroup$ For each pair of rationals $p<q$, your calculation shows that the event $$ A_{p,q}:=\{B \hbox{ is zero free in }[p,q], L_p<L_q\} $$ has probability zero. Thus so does the countable union $A:=\cup_{p<q\in\Bbb Q}A_{p,q}$. As both $L$ and $B$ are continuous on an event $C$ of probability 1, the desired claim holds on $C\setminus A$, and this event has probability 1. $\endgroup$ Nov 27, 2019 at 0:14

1 Answer 1

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For each pair of rationals $p<q$, your calculation shows that the event $$ A_{p,q}:=\{B \hbox{ is zero free in }[p,q], L_p<L_q\} $$ has probability zero. Thus so does the countable union $A:=\cup_{p<q\in\Bbb Q}A_{p,q}$. As both $L$ and $B$ are continuous on an event $C$ of probability 1, the desired claim holds on $C\setminus A$, and this event has probability 1.

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