Let $B_t$ be a Brownian motion and $B(x,r)$ the ball of x with radius r. Define $\mu_t(A):=\int_0^t 1_A(B_s)ds$ for $A\in \mathcal{B}(\mathbb{R})$.
Now I'm not sure if the following equations are correct, so I would be really grateful for any input.
$$\mathbb{E}\left[\int\mu_t(B(x,r))d\mu_t(x)\right] =\mathbb{E}\left[\int_{0}^{t}\int1_{B(x,r)}(B_{s_1})d\mu_t(x)ds_1\right] =\mathbb{E}\left[\int_{0}^{t}\int\mu_t(\{B_{s_1}\in B(x,r)\})dxds_1\right] =\mathbb{E}\left[\int_{0}^{t}\int\mu_t(\{x\in B(B_{s_1},r)\})dxds_1\right] =\mathbb{E}\left[\int_{0}^{t}\int_{0}^{t}1_{B(B_{s_1},r)}(B_{s_2})ds_2ds_1\right]$$