Brownian occupation time

The occupation time of a measurable set $A$ through time $t$ is ${\int }_0^t{1}_{\{B_s\in A\}}ds$. The occupation-times formula expresses such integrals through local time of a semimartingale. Recurrence and the Strong Markov property imply that one-dimensional Brownian motion spends an unbounded total time in every nonempty open interval.

A one-dimensional Brownian motion spends an infinite total time in every nonempty open interval almost surely. To prove this for $(-1,1)$ after first hitting zero, alternate exits from that interval with returns to zero. By recurrence of one-dimensional Brownian motion all these stopping times are finite, and the Strong Markov property makes the successive exit durations independent and identically distributed random variables that are strictly positive. Some fixed positive duration is exceeded with positive probability, so the Borel-Cantelli lemma makes the sum of the durations infinite. Translate and scale for other intervals. In particular, if a nonnegative function is bounded below by a positive constant on such an interval, its infinite-horizon Brownian occupation time integral diverges. Recurrent visits alone, without this duration argument, would not prove divergence of an integral.

For any real constant $c$, one-dimensional Brownian motion repeatedly returns to zero. Consider successive segments from zero until exiting $(-1,1)$, separated by returns to zero. Their capped durations are independent identically distributed positive bounded random variables by the Strong Markov property, so their sum diverges by the strong law of large numbers. On each such segment $e^{c{B}_s}\ge e^{-|c|}$. Thus the exponential clock has infinite lifetime, despite arbitrarily deep negative excursions.

For a continuous semimartingale $X$ with local time of a semimartingale $L_t^a(X)$,for every nonnegative measurable function $f$.