scaling invariance of brownian local time

Question

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.