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This is a cool little result, the proof of which uses the machinery of local time. On p. 72, Prob 1 asks to show that $\int_0^1 dt/x(t)$ exists, where $x(t)$ is a continuous time brownian motion.

In the very first step, they stated,

$\int_{|x(t)| > \epsilon, t \leq 1} dt/x(t) = 2 \int_{\epsilon}^{\infty}[\tau(1,b) - \tau(1, -b)] \frac{db}{b}$

where $\tau$ is the local time up to time 1 at the two locations.

Can someone elucidate why this is true.

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This is a consequence of the occupation time formula satisfied by local time: If $f:\Bbb R\to\Bbb R$ is bounded and Borel measurable, then $\int_0^u f(x(t))\,dt =\int_{\Bbb R}\tau(u,b)f(b)\,db$, almost surely. Use this with $u=1$ and $f(b) =b^{-1}1_{(-\infty,\epsilon)\cup(\epsilon,+\infty)}(b)$.

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  • $\begingroup$ Is $f$ just $1_{(...) U (...)}/b$? what's the "$(b)$" for? is it just notation? $\endgroup$
    – horaceT
    Commented Apr 9, 2017 at 0:54
  • $\begingroup$ It's indicator function notation: for a set $A\subset \Bbb R$, $1_A(b)$ is equal to $1$ if $b\in A$ and $0$ if $b\notin A$. So $f(b)=b^{-1}$ if $|b|>\epsilon$ and $f(b)=0$ if $|b|\le\epsilon$. $\endgroup$
    – John Dawkins
    Commented Apr 9, 2017 at 15:08
  • $\begingroup$ Can you give a readable ref to the occupation time formula. Unfortunately I do not have access to a good library. $\endgroup$
    – horaceT
    Commented Apr 12, 2017 at 18:54

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