Uncountable disjoint union of measure spaces Let $(a,b)$ be an interval. Let $(A_i, \Sigma_i, \mu_i)$ be a measure space for each $i \in (a,b)$. Is it possible to put a measure space on the disjoint union 
$$\bigcup_{i \in (a,b)}\{i\}\times A_i?$$
If the $A_i = \Omega_i$ were open subsets of $\mathbb{R}^n$, we can think of this disjoint union as a non cylindrical domain. Will the measure on the disjoint union coincide with the Lebesgue measure on this non cylindrical domain description? In this case $(a,b)$ would be a time interval. 
I know a countable version of this is true, where the disjoint measure is a sum of the measures $\mu_i$. Maybe we can replace this sum by an integral in this case.
Moreover, can I view this measure space as a product measure space? (See my previous thread). 
 A: Let $A = \bigcup_{i \in (a,b)} \{i\} \times A_i$, and for a set $X \subset A$ let $X[i]$ denote the projection to the $i$-th component, i.e. $$
  X[i] = \{x \in A_i \,:\, (i,x) \in X \} \text{.}
$$
A natural definition of a measure $\mu$ on $A$ is then $$
  \mu(X) := \int_{a}^b \mu_i(X[i]) \,di \text{.}
$$
This assumes that $X \in \Sigma \subset \mathcal{P}(A)$ for some (yet to be determined) $\sigma$-Algebra $\Sigma$ on $A$, which has to satisfy


*

*$X[i] \in \Sigma_i$ for all $i \in (a,b)$,

*$f_X \,:\, (a,b) \to \overline{\mathbb{R}}\,:\, i \to \mu_i(X[i])$ is a measurable function with respect to the lebesgue measure on $\mathbb{R}$ for all $X \in \Sigma$.


Satisfying (1) is straight-forward - we just pick those sets whose projections are all measurable, and observe that they form a $\sigma$-Algebra.
Satisfying (2) is harder, because we have to take the $\mu_i$ into account. Without any restriction on the $\mu_i$, not even the trivial $\sigma$-Algebra $\{0,A\}$ on $A$ necessarily satisfies (2).
