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Is there a generalized form of the differentiable change of variables theorem for Lebesgue integrals? That is, if we consider the well known change of variables theorem: If $\phi : X \rightarrow X$ is a diffeomorphism of open sets in $\mathbb{R}^n$, $X \subseteq \mathbb{R}^n$ is measurable, and $f : X \rightarrow \mathbb{R}$ is measurable, then:

$$ \int_X f(y) dy = \int_X f(\phi(x)) d(\phi(x)) = \int_X f(\phi(x)) |\det D\phi(x)| dx $$

I'de like to weight between some countable set of (simple) diffeomorphic mappings instead of just one, that is, let $\Phi = \{\phi_i \ | \ i \in \mathbb{N}, |\det D\phi_i(x)| = 1\}$. Additionally, I'de like to weight between these transformations as a convex-combination, so I define a weighting function, $w : X \times \mathbb{N} \rightarrow \mathbb{R}$, where: $\sum_i w(\phi_i^{-1}(y),i)$ = 1. Then, I'de like to show:

$$ \int_X \sum_i w(x,i) f(\phi_i(x)) dx = \int_X f(x) dx $$

My attempt at a proof is: \begin{align} \int_X \sum_i w(x,i) f(\phi_i(x)) dx & = \sum_i \int_X w(x,i) f(\phi_i(x)) dx\\ & = \sum_i \int_X w(x,i) f(\phi_i(x)) |\det D\phi_i(x)|dx\\ & = \sum_i \int_X w(\phi_i^{-1}(y),i) f(y) dy\\ & = \int_X f(y) \sum_i w(\phi_i^{-1}(y),i) dy \\ & = \int_X f(y) dy \end{align}

Trouble is, I'm not all that familiar with measure theory and I would need to show that my weighting function is measurable in order to invoke the single-mapping change of variables theorem mentioned above. Perhaps I cannot do this without being more explicit about what this function actually is, but at the same time, it's just a simple weight vector, normalized in some unique way, over a countable set it would be nice if I could say something at this level of generality. Also, perhaps it makes more sense to start at the case where $\Phi$ is finite, which is fine, but I have the same issues with the proof with this assumption.

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I could see some sort of relation between the Borel measure and differential structure giving you the measurable conditions you need. Then again requiring the weight functions to be measurable would not be an undue burden on this sort of theory. – Nate Iverson Jul 31 '12 at 23:26
yes, I could assume that $w$ is measurable for each $i$, the trouble is every time I want to use a different weighting function I would need to prove it is measurable. It would just be nice if I could say something a little more general about the type of weighting functions that would guarantee this. – anonymous_21321 Aug 1 '12 at 4:15

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