scaling invariance of brownian local time

I am studying Brownian local time processes and several references mentioned the scaling invariance of local time. For example, page 10 of this reference (https://hal.archives-ouvertes.fr/hal-00091335/document) says "It is well known that the Brownian motion and its local time have the fol-lowing scaling property: $$(B_u,L_u)_{u \geq 0} \sim (\sqrt{t}B_{\frac{u}{t}},\sqrt{t}L_{\frac{u}{t}})_{u \geq 0}$$," where $$B$$ is a standard Brownian motion and $$L$$ is its local time. The definition of local time that I am familiar with is $$L(t,a)=\lim_{\epsilon \rightarrow 0}\frac{1}{2\epsilon}\mu(\{s \in [0,t]: |B_t-a| \leq \epsilon\})=\lim_{\epsilon \rightarrow 0} \frac{1}{2\epsilon} \int_{0}^{t}1_{Bs \in [a-\epsilon,a+\epsilon]}ds,$$ where $$\mu$$ is the Lebesgue measure. I tried to prove $$L(u,a) = \sqrt{t}L(\frac{u}{t},\frac{a}{\sqrt{t}})$$ in distribution using the scale invariance of Brownian motion and the definition: $$\int_{0}^{u/t}1_{B_s \in [\frac{a}{\sqrt{t}}-\epsilon,\frac{a}{\sqrt{t}}+\epsilon]}ds=\int_{0}^{u/t}1_{\frac{B_{ts}}{\sqrt{t}} \in [\frac{a}{\sqrt{t}}-\epsilon,\frac{a}{\sqrt{t}}+\epsilon]}ds=\int_{0}^{u/t}1_{B_{ts} \in [a-\epsilon \sqrt{t},a+\epsilon \sqrt{t}]}ds\\=\frac{1}{t}\int_{0}^{u}1_{B_x \in [a-\epsilon \sqrt{t},a+\epsilon \sqrt{t}]}dx,$$ but I'm not sure what to do next.

• So far so good. Now divide by $2\epsilon$ and let $\epsilon\to 0$. – zhoraster May 6 at 18:06
• But how does that give me $\frac{1}{\sqrt{t}} L(u,a)$? – user0617 May 6 at 21:46
• Since in the rhs you would need to have $\frac1{2\epsilon \sqrt{t}}$ in order to get the definition of local time. And then you would end up with $\sqrt t$. – zhoraster May 7 at 5:43
• So did you succeed? – zhoraster May 8 at 15:47
• @zhoraster I did! Thanks so much!!! – user0617 May 9 at 3:06