Direct Integral
Given a Borel space $\Omega$ with measure $\mu$.
Given Hilbert spaces $\mathcal{h}_x$ for $x\in\Omega$; set $\mathcal{h}:=\bigcup_{x\in\Omega}\mathcal{h}_x$.
Regard the function space $$\mathcal{F}(\Omega,\mathcal{h}):=\{\varphi:\Omega\to\mathcal{h}:\varphi(x)\in\mathcal{h}_x\}$$
with algebraic structure $$(\varphi+\psi)(x):=\varphi(x)+\psi(x)\quad(\lambda\varphi)(x):=\lambda\varphi(x)$$
as well as the structure $$\langle\varphi,\psi\rangle(x):=\langle\varphi(x),\psi(x)\rangle_x\quad\|\varphi\|(x):=\|\varphi(x)\|_x.$$
Suppose one has $$\mathcal{S}(\Omega,\mathcal{h})\leq\mathcal{F}(\Omega,\mathcal{h}):\quad\mathcal{\overline{\langle\pi_x\mathcal{S}\rangle}}=\mathcal{h}_x\bmod\mu$$
made of integrables $$\sigma\in\mathcal{S}(\Omega,\mathcal{h}):\quad\int_\Omega\|\sigma\|^2\mathrm{d}\mu<\infty\quad(\|\varphi\|\in\mathcal{B}(\Omega)).$$
Define the measurables $$\mathcal{B}(\Omega,\mathcal{h}):=\left\{\varphi\in\mathcal{F}(\Omega,\mathcal{h}):\varphi\in\overline{\mathcal{S}(\Omega,\mathcal{h})}\right\}.$$
Note for measurability: $$\|\varphi\|(x)=\|\varphi(x)\|=\|(\lim_n\sigma_n)(x)\|\\=\|\lim_n\sigma_n(x)\|=\lim_n\|\sigma_n(x)\|=\lim_n\|\sigma_n\|(x).$$
Define the integrables $$\mathcal{H}(\Omega,\mathcal{h}):=\left\{\varphi\in\mathcal{B}(\Omega,\mathcal{h}):\int_\Omega\|\varphi\|^2\mathrm{d}\mu<\infty\right\}.$$
Moreover it holds: $$\mathcal{S}\leq\mathcal{F}\implies\mathcal{B}\leq\mathcal{F}\implies\mathcal{H}\leq\mathcal{F}.$$
Concluding direct integral.
Dimension
Suppose one finds: $$\mathcal{A}\subseteq\mathcal{H}:\quad\overline{\pi_x\mathcal{A}}=\mathcal{h}_x\bmod\mu\quad(\#\mathcal{A}\leq\mathfrak{n})$$
Then one obtains: $$n\in\mathbb{N}:\quad\{\dim\mathcal{h}_x=n\}\in\mathcal{B}(\Omega)$$
How can I prove this? (Ideas?)