Borel Measures: Coproduct I need this thread as lemma!
(See the advice: SE: Q&A)
Given Borel spaces $\Omega_\lambda$.
Consider the coproduct:
$$\Omega:=\coprod_\lambda\Omega_\lambda:\quad\mu(A):=\sum_\lambda\mu_\lambda(\iota_\lambda^{-1}A)\quad(\iota_\lambda^{-1}\mathcal{B}(\Omega):=\mathcal{B}(\Omega_\lambda))$$
Then they are isomorphic:
$$\mathcal{L}^2(\Omega,\mu)\cong\bigoplus_\lambda\mathcal{L}^2(\Omega_\lambda,\mu_\lambda):\quad h_\lambda:=h\circ\iota_\lambda$$
How can I prove this?
 A: Additivity
Remind on summability:
$$a_{\kappa\lambda}\geq0:\quad\sum_\kappa\sum_\lambda a_{\kappa\lambda}=\sum_\lambda\sum_\kappa a_{\kappa\lambda}$$
So the measure is additive:
$$\sum_k\mu(A_k)=\sum_k\sum_\lambda\mu_\lambda(\iota_\lambda^{-1}A_k)\\
=\sum_\lambda\sum_k\mu_\lambda(\iota_\lambda^{-1}A_k)=\sum_\lambda\mu_\lambda\big(\iota_\lambda^{-1}\biguplus_kA_k\big)=\mu\big(\biguplus_kA_k\big)$$
(That is not clear a priori!)
Measurability
That follows from:
$$S\subseteq\mathbb{C}:\quad h_\lambda^{-1}(S)=\iota_\lambda^{-1}(h^{-1}(S))$$
(This extends to Bochner's theory!*)
Integrability
Remind on summability:
$$a_{\kappa\lambda}\geq0:\quad\sum_\kappa\sum_\lambda a_{\kappa\lambda}=\sum_\lambda\sum_\kappa a_{\kappa\lambda}$$
For simple functions:
$$s_\lambda(\omega_\lambda)=\sum_ka_k\delta_{\iota_\lambda(\omega_\lambda)\in A_k}=\sum_ka_k\delta_{\omega_\lambda\in\iota_\lambda^{-1}A_k}$$
So one obtains:
$$\int|s|^2\mathrm{d}\mu=\sum_k|a_k|^2\sum_\lambda\mu_\lambda(\lambda^{-1}A)\\
=\sum_\lambda\sum_k|a_k|^2\mu(\iota_\lambda^{-1}A)=\sum_\lambda\int|s_\lambda|^2\mathrm{d}\mu_\lambda$$
Remind on limits:
$$a_n\leq a_{n+1}:\quad\lim_n a_n=\sup_na_n$$
For arbitrary functions:
$$|s_n|\leq|s_{n+1}|:\quad h(\omega):=\lim_ns_n(\omega)$$
So by monotone convergence:
$$\int|h|^2\mathrm{d}\mu=\lim_n\int|s_n|^2\mathrm{d}\mu=\lim_n\sum_\lambda\int|s_{n;\lambda}|^2\mathrm{d}\mu_\lambda\\
=\sum_\lambda\lim_n\int|s_{n;\lambda}|^2\mathrm{d}\mu_\lambda=\sum_\lambda\int|h_\lambda|^2\mathrm{d}\mu_\lambda$$
Concluding integrability..
Properties
It is algebraic since:
$$(h+h')_\lambda=h_\lambda+h'_\lambda\quad(ah)_\lambda=ah_\lambda$$
$$(hh')_\lambda=h_\lambda h'_\lambda\quad(\overline{h})_\lambda=\overline{h_\lambda}$$
Also it is isometric:
$$\|h\|^2=\int|h|^2\mathrm{d}\mu=\sum_\lambda\int|h_\lambda|^2\mathrm{d}\mu_\lambda=\|(h_\lambda)_\lambda\|^2$$
Concluding properties.
Unitarity
Remind the embedding:**
$$\Omega=\bigcup_\lambda\iota_\lambda\Omega_\lambda:\quad\iota_\lambda(\omega_\lambda)=\iota_\lambda(\omega_\lambda')\implies\omega_\lambda=\omega_\lambda'$$
So one can rebuild:
$$h(\omega\in\iota_\lambda\Omega_\lambda):=h_\lambda(\iota_\lambda^{-1}(\omega)):\quad h_\lambda(\omega_\lambda)=h(\iota_\lambda(\omega_\lambda))$$
Concluding unitarity.
*Separabaility may fail though!
**That is special to the category of sets!
