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I'm trying to dig in the details of the proof of Haar's theorem, and at some point I need to use Fubini's theorem, which requires that if we want to change the order of integration over the product space $X \times Y$, the spaces $X$ and $Y$ must be complete measure spaces. (In my case, they are topological measure spaces and my measure is also defined over the completion of their $\sigma$-algebras, but in order to have a topological measure space, I restrict the measure to the Borel $\sigma$-algebra).

However, I am working with a continuous function (the one I am integrating). I am wondering if the following holds : given a continuous function $h : X \times X \to \mathbb C$ where $(X, \mathcal T)$ is a topological space, $(X, \Sigma, \mu)$ is a complete measure space and $\sigma(\mathcal T) \subseteq \Sigma$ (the Borel sets are measurable), is it possible to show that $h$ is measurable with respect to $\overline{\sigma}(T) \times \overline{\sigma}(\mathcal T)$? (Here $\overline{\sigma}$ denotes the completion of the $\sigma$-algebra generated by $\mathcal T$.) This will imply I can use Fubini.

Note that my measure $\mu$ comes from Haar's theorem, so if it helps, it is non-zero and inner/outer regular.

Perhaps the following could be easier to prove : if $f : (X, \mathcal T_X) \to (Y, \mathcal T_Y)$ is a continuous function, is $f$ measurable with respect to $\overline{\sigma}(\mathcal T_X)$ and $\overline{\sigma}(\mathcal T_Y)$? I guess it's false if you put the zero measure on $Y$ because then the completion is just the trivial $\sigma$-algebra, but in my case I am working with $\mathbb C$ and the Lebesgue measure, so perhaps I can find my way around it.

All the proofs I've read seem to hide this tricky detail that the assumptions of Fubini must be satisfied...

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    $\begingroup$ Fubini doesn't require completeness of the measures... only $\sigma$-finiteness. $\endgroup$ – nullUser Jul 8 '13 at 5:15
  • $\begingroup$ @nullUser : I knew that, but I expected completeness to be the solution... I answered myself after thinking about it. Thanks for the comment though ; it gave me a spark. $\endgroup$ – Patrick Da Silva Jul 8 '13 at 5:22
  • $\begingroup$ If $h$ is continuous, and $\sigma(\mathcal{T}) \subseteq \Sigma$, then $h$ is $X\times X$ measurable. Don't we have $\sigma(\mathcal{T}) \times \sigma(\mathcal{T}) \subseteq \overline{\sigma}(\mathcal{T}) \times \overline{\sigma}(\mathcal{T})$? Completing the factor spaces only increases the number of measurable sets. Edit: Nevermind, this is only helpful if $\mathcal{B}(X\times X) = \mathcal{B}(X) \times \mathcal{B}(X)$. $\endgroup$ – nullUser Jul 8 '13 at 5:31
  • $\begingroup$ @nullUser : the problem is because you also take the completion of the $\sigma$-algebra in the image, therefore you have $f^{-1}(\sigma(\mathcal T_Y)) \subseteq \overline{\sigma}(\mathcal T_X)$ but there is nothing you can do with $f^{-1}(\overline{\sigma}(\mathcal T_Y))$ as far as I'm aware. $\endgroup$ – Patrick Da Silva Jul 8 '13 at 6:15
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I just realized I found my answer ; I can restrict the integral to the support of my function which is compact. Compact spaces are finite by a property of my Haar measure ; I can apply Fubini. Perhaps asking the question was the way to go.

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