I am trying to show that the expectation of
$\int_{\tau_n}^{\tau_{n+1}}\int_{\tau_n}^{s_2}dWs_1ds_2$
is 0. Is it possible to take the expectation inside the time integral, so that
$\mathbb{E}\bigg[\int_{\tau_n}^{\tau_{n+1}}\int_{\tau_n}^{s_2}dWs_1ds_2\bigg] = \int_{\tau_n}^{\tau_{n+1}}\mathbb{E}\bigg[\int_{\tau_n}^{s_2}dWs_1\bigg]ds_2 = \int_{\tau_n}^{\tau_{n+1}}\mathbb{E}\bigg[W_{s_2}-W_{\tau_n}\bigg]ds_2 = 0$
since the expectation of the Brownian motion increments are $N(0,s_2-\tau_n)$? I feel that Fubini's theorem might come into play somewhere, but I do not know if Fubini's theorem applies to Brownian motions. Any help would be much appreciated. Thanks!
Yes, this is an application of the Fubini theorem: Take the probability space: $$(\Omega \times [t_1, t_2], \mathbb{P}_W \otimes \lambda)$$ where $\mathbb{P}_W $ is the Wiener measure (so implicitly $\Omega = C[0, T]$) and $\lambda$ is the Lebesgue measure. Now take the map $f(\omega, s) =\omega(s)= W_s.$ Then an application of Fubini's theorem on this map will give you the required result.
Note that for stopping times the result will not be true in general.