Lebesgue Limit Stochastic

Question

What happens to $\lim_{\epsilon \downarrow 0} \frac{1}{\epsilon} \int_0^t 1_{B_s \in [-\epsilon, \epsilon]} ds$?

Answer 1

If you've seen the Ito integral: Consider the function $$ F_\epsilon(x) := \cases{x^2/2\epsilon,&$|x|\le\epsilon$,\cr x-\epsilon/2,&$x\ge\epsilon$,\cr -x-\epsilon/2,&$x\le-\epsilon$.\cr} $$ This function is $C^1$ and close enough to being $C^2$ that you can apply (a slight extension of) Ito's formula to see that $$ F_\epsilon(B_t) =F_\epsilon(B_0)+\int_0^t F'_\epsilon(B_s)\,dB_s+{1\over 2\epsilon}\int_0^t 1_{\{|B_s|<\epsilon\}}\,ds. $$ Let's assume that $B_0=0$. Now $F_\epsilon$ converges pointwise to the absolute value function $x\mapsto |x|$, and so it's easy to see that $F_\epsilon(B_t)\to |B_t|$ in $L^2$. By the Ito isometry, the stochastic integral term converges in $L^2$ to $\int_0^t sign(B_s)\,dB_s$. (Where $sign(x) = 1,0,-1$ according as $x$ is $>0$, $=0$, or $<0$.) The remaining term ${1\over 2\epsilon}\int_0^t 1_{\{|B_s|<\epsilon\}}\,ds$ therefore converges in $L^2$ as well.