Here is the theorem statement:
Let $B$ and $C$ be two independent standard Brownian motions. If $\phi$ is square integrable on the unit square ($\phi \in L^2([0,1]^2)$ ), by suitable filtrations, give a meaning to the integrals $$\int_0^1 dB_u \int_0^1 dC_s \phi(u,s) \quad \text{and} \quad \int_0^1 dC_s \int_0^1 dB_u \phi(u,s)$$ and prove that they are almost surely equal.
(This is the first part of Exercise 2.19 in Chapter IV of Revuz and Yor)
What I tried: There are always so many measure-theoretic assumptions and results implicit in the theory of stochastic processes that are always brushed under the rug that I basically have no idea what is going on. Nevertheless, here is my pitiful attempt at a solution:
In order to give meaning to the integrals, we need to first choose a suitable filtration. We can simply use the natural filtration generated by $B$ and $C$ on $\mathbb{R}^2$ though -- this is because under such a filtration, both processes are adapted, and since they are also continuous, they also generated the predictable sigma algebra, which implies they are also progressive(ly measurable). The inner integrals, being continuous and adapted, will again be suitably measurable.
The almost sure equality follows from the progressive measurability of the processes on $\mathbb{R}^2$ allowing us to switch the order of integration for the simple predictable integrals/stochastic Riemann sums (by the regular Fubini theorem) and then follows by the uniqueness of limits in $L^2$ (using the dominated convergence theorem to get the double stochastic integrals as limits).