# Proof local time of BM only increasing on 0

Let $$B$$ be a standard Brownian motion and the local time of $$B$$ at $$0$$ defined by Tanaka's formula $$$$L_t = |B_t| - \int_0^t sgn(B_s) dB_s \ .$$$$ Now I want to prove that $$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 $$p be rationals and $$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]$$. Then, $$$$L_{q} - L_p = | B_{q} | - | B_p |- \int_p^q sgn(B_s) dB_s = 0,$$$$ because $$sgn(B_s) = 1$$ or $$-1$$.

Is this approach correct? And why wouldn't it work for any reals $$u?

• Seems that you could have just used $p,q$ reals here Nov 25, 2019 at 3:54
• 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. Nov 27, 2019 at 0:14

For each pair of rationals $$p, your calculation shows that the event $$A_{p,q}:=\{B \hbox{ is zero free in }[p,q], L_p has probability zero. Thus so does the countable union $$A:=\cup_{p. As both $$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.