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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).

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  • $\begingroup$ I think you need some "compatibility" condition on your intended measure. For instance, could you not pick any point of the disjoint union, assign to it measure one, and assign measure zero to all the others? $\endgroup$ – John Hughes May 12 '14 at 11:47
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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. $$ X[i] = \{x \in A_i \,:\, (i,x) \in A \} \text{.} $$

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

  1. $X[i] \in \Sigma_i$ for all $i \in (a,b)$,
  2. $f_X \,:\, (a,b) \to \overline{\mathbb{R}}\,:\, i \to \mu_i(X[i])$ 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).

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  • $\begingroup$ Thanks. When $\mu_i$ is the Lebesgue measure on noncylindrical domain, we need $t \mapsto |\Omega(t)\cap X|$ to be measurable, which should be true if the domains evolve in a measurable way. I can write a point of $A$ as $(i,a)$ where $a \in A_i$. Do you know if it makes sense to think of this as a product? (I want to check measurability of a composition of a function) $\endgroup$ – riem May 12 '14 at 13:56
  • $\begingroup$ @riem They only way I see to view this as a product is if the $A_i$ are all the same, and the $\Sigma_i$ are "compatible" enough to replace them by a global $\Sigma$ and extending the $\mu_i$ accordingly. You can then view $A$ as the product of $(a,b)$ and $A$. The $\mu_i$ would then be called markov kernels. $\endgroup$ – fgp May 12 '14 at 19:17

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