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Assume given a probability space $(\Omega,\mathcal{F},P)$ and a measurable space $(E,\mathcal{E})$. Let $(\nu_\omega)_{\omega\in\Omega}$ be a family of signed measures on $(E,\mathcal{E})$. Assume that $\omega\mapsto\nu_\omega(A)$ and $\omega\mapsto|\nu_\omega|(A)$ are $\mathcal{F}$ measurable for all $A\in\mathcal{E}$, where $|\nu_\omega|$ denotes the total variation measure of $\nu_\omega$. Further assume that $\int_\Omega |\nu_\omega|(E)dP(\omega)$ is finite. The family $(\nu_\omega)$ is basically a signed measure analogue of the Markov kernels from probability theory used for regular conditional distributions. Under these assumptions, it is possible to define the integration of $(\nu_\omega)$ with respect to $P$ as the unique signed measure $\lambda$ on $\mathcal{F}\otimes\mathcal{E}$ such that $\lambda(F\times A) = \int_F \nu_\omega(A) dP(\omega)$ for $F\in\mathcal{F}$ and $A\in\mathcal{E}$. This can be done simply by Jordan-Hahn decomposing each $\nu_\omega$ and carrying out the ordinary kernel integration construction for the positive and negative parts separately.

My question is the following: What is the total variation measure of $\lambda$?

My own guess is the following. Let $\mu$ denote the integration of $(|\nu_\omega|)_{\omega\in\Omega}$ with respect to $P$, that is, the unique measure such that $\mu(F\times A)=\int_F |\nu_\omega|(A)dP(\omega)$ for $F\in\mathcal{F}$ and $A\in\mathcal{E}$. I would think that $|\lambda|=\mu$. It is quite clear that $|\lambda|\le \mu$. However, I don't find the converse inequality as obvious.

An appealing line of argument would be to simply claim that with $\nu_\omega = \nu_\omega^+ - \nu_\omega^-$, the Jordan-Hahn decomposition of $\lambda$ is obtained by integrating the kernels $(\nu^+_\omega)$ and $(\nu^-_\omega)$. However, while this yields two nonnegative measures whose difference is $\lambda$, it is not obvious that the two measures are singular: Intuitively, these two measures would have mass on "glued together" strips of the $\mathcal{E}$-sets where each $\nu^+_\omega$ and $\nu^-_\omega$ are concentrated, but it is not obvious that such glued together sets are measurable.

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I guess, answers and links here may be of help - I asked there a very similar question. – Ilya Feb 20 '13 at 13:21
Thanks! That seems to be pointing in the right direction... – Alexander Sokol Feb 27 '13 at 11:00
Nice, please don;t forget to post an answer here if you find one, I'm interested as well. If you don't find one, you may wanna ask it on MO. – Ilya Mar 1 '13 at 11:57

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