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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:

$\\D_r := \{(x,y):|x-y|<r\} \\\int_{\Bbb R} \mu_t(B(x,r))d\mu_t(x) = \int_{D_r}d\mu_t(x)d\mu_t(y)$

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.

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