Let $(X,\mu)$ be a measure space, and $(H_x)_{x \in X}$ is a "field" of Hilbert spaces, one can form the direct integral $\int^{\oplus} H_x d\mu(x)$ which is a Hilbert space as well.
When $(H_x)_x$ is constant equal to $H$, then this direct integral is (or should be, I haven't checked properly) isomorphic to $L^2(X,H)$.
My question is: if $\mu_1$ and $\mu_2$ are two measures on $X$, $(H^1_x)_x$ and $(H^2_x)_x$ are two fields of Hilbert spaces, how can one write the direct sum $\int^{\oplus} H^1_xd\mu_1(x) \oplus \int^{\oplus} H^2_xd\mu_2(x)$ in the form of a direct integral?
Context: If $\pi : \mathbb{R} \rightarrow U(H)$ is a unitary representation, then $H$ should decompose as a direct integral and $\pi$ should arise as the direct integral of a field of irreducible representations.
If $\pi$ is cyclic, then this just follows from the spectral theorem (or variants of it): for any norm $1$ vector $\xi$, the map $x \mapsto \langle \xi, \pi(x)\xi\rangle$ is the Fourier transform of a measure $\mu$ on $\mathbb{R}$ and the closed span of $\{\pi(x)\xi \ \vert \ x \in \mathbb{R}\}$ is isometrically isomorphic to $L^2(\mathbb{R},\mu)$ and each $\pi(x)$ corresponds to the pointwise multiplication by $\xi \mapsto e^{ix\xi}$.
However, of course, not every unitary representation has a cyclic vector, though one can always decompose it as a direct sum of cyclic subrepresentations (by applying Zorn's lemma on the ordered space of subrepresentations that are direct sums of cyclic subrepresentations). I am now helpless in "summing" all of these direct integrals.
I also have a more general question: to my taste, direct integrals have the defect of not being very canonical (changing $\mu$ by another equivalent measure does not change the isomorphism class of the direct integral, for example); is there an alternative construction that is more canonical and is as useful as direct integrals?