# Proving that occupation measure of Brownian Motion is absolutely continuous almost surely

I am reading the section on occupation measures from Morters and Peres. I need some help with the following.

$$\{B(t):t\geq0\}$$ denotes the standard Brownian Motion on the probability space $$(\Omega,\mathcal{F},{\Bbb P})$$. For $$\omega\in\Omega$$ we define the measure $$\mu_t$$ on $${\Bbb R}$$ as,

$$μ_t (A) =\int_0^t1_A(B(s))ds$$ for $$A ⊂ {\Bbb R}$$ Borel.

I am omitting the implicit $$\omega$$ in the definition of $$\mu_t$$.

We have to prove that $${\Bbb P}-$$a.s., $$\mu_t$$ is absolutely continuous with respect to the lebesgue measure $$L$$ on $${\Bbb R}$$.

To do this, we use the following sufficient condition:

$$\liminf_{r↓0} \frac{μ_t (B(x, r))}{L(B(x, r))} < ∞$$ for $$μ_t$$-almost every $$x ∈ R$$

To prove this condition, the book gives the following sequence of steps which it says are justified by Fatou's lemma and Fubini's theorem

$${\Bbb E}(\int\liminf_{r↓0}\frac{μ_t (B(x, r))}{L(B(x, r)) }dμ_t (x)) \\\leq \liminf_{r↓0}\frac{1}{2r}{\Bbb E}(\int μ_t (B(x, r)) dμt (x) \\= \liminf_{r↓0}\frac{1}{2r}\int_0^t\int_0^t {\Bbb P}\{|B(s_1) − B(s_2)|\leq r\}ds_1 ds_2 \\\leq \infty$$

I understand how the first inequality follows by Fatou's lemma. However, I am unable to get how to use Fubini to get the second step. I basically want a rigorous justification of this step:

$$\\\liminf_{r↓0}\frac{1}{2r}{\Bbb E}(\int μ_t (B(x, r)) dμt (x)) \\= \liminf_{r↓0}\frac{1}{2r}\int_0^t\int_0^t {\Bbb P}\{|B(s_1) −B(s_2)|\leq r\}ds_1 ds_2$$

I have been able to get the following:

$$\\D_r := \{(x,y):|x-y|

I think what I have gotten is correct but I would like someone to confirm it. And second, is it useful in proving what I want to prove?

If anyone needs more details, this is Theorem 3.26 in Morters and Peres.