The title speaks of itself. How far is an arbitrary finite diffuse measure space from being **almost** isomorphic to a product of $[0;1]$ with another diffuse measure space? What would be reasonable sufficient conditions? I need only almost isomorphism because I am interested in analysis of $L^p$-functions. Thank you.



Diffuse measure space as a product of $[0;1]$ and another diffuse measure space
Here is a result that covers "most" finite measure spaces encountered in applications. Let $(\Omega,\Sigma)$ be a *standard Borel space*, that is $\Sigma$ is the Borel $\sigma$-algebra for some separable completely metrizable topology on $\Omega$. Let $\mu$ be any finite diffuse measure on $(\Omega,\Sigma)$, we can take it without loss of generality to be a probability measure (the trivial case of the zero-measure is excluded). By Theorem 16 in Section 5 of Chapter 15 of *Real Analysis* by Royden (3rd Ed.), such $(\Omega,\Sigma,\mu)$ must be isomorphic to $[0,1]$ with the usual Borel $\sigma$-algebra and Lebesgue measure. But by the same theorem and the fact that the product $\sigma$-algebra for the product of the Borel $\sigma$-algebra of two separable metrizable spaces is the Borel $\sigma$-algebra of the topological product, $[0,1]\otimes[0,1]$ with the Borel $\sigma$-algebra and the product of Lebesgue measure with itself is too isomorphic with $[0,1]$ and therefore with $(\Omega,\Sigma,\mu)$. This shows that the latter space can be written as a product with $[0,1]$.


To see how bad things might be for general measurable spaces, one can use the set-theoretic assumption that there is a diffuse probability measure defined on all subsets of [0,1]. By a result of Solovay, this is consistent if measurable cardinals are consistent. But since many subsets of $[0,1]$ are not Borel- or Lebesgue measurable, any product of $[0,1]$ in the usual sense with any other measure space will contain a lot of nonmeasurable sets. This precludes the existence of any almost isomorphism


