Suppose $F$ is an ultrafilter on $I$ and $G$ is an ultrafilter on $J$. We let $F\otimes G$ be the collection of subsets of $I\times J$ given by $$F\otimes G=\{X\subseteq I\times J :\{j\in J:\{i\in I:\langle i,j\rangle\in X\}\in F\}\in G\}.$$ The definition of $F\otimes G$ may seem somewhat involved. The reader familiar with measure theory will see that it corresponds to the usual definition of the product measure.

This passage is quoted from Bell-Slomson: "Models and Ultraproducts", at the end of Ch.6,§2

Consider two measure spaces $(X,\Sigma_1,\mu_1)$ and $(Y,\Sigma_2,\mu_2)$, then a product measure $\mu_1\times\mu_2$ is defined on the $\sigma$-algebra $\Sigma_1\otimes\Sigma_2$ (which is the completion of the $\sigma$-algebra generated by the rectangles $A\times B$ with $A\in\Sigma_1$ and $B\in\Sigma_2$).
For $E\subseteq X\times Y$, $x\in X$ and $y\in Y$ we define $$E_x=\{y\in Y\mid(x,y)\in E\}\qquad E_y=\{x\in X\mid (x,y)\in E\}.$$ I read the passage quoted above as suggesting that if $E\subseteq X\times Y$ is such that $E_y$ is $\mu_1$-measurable for $\mu_2$-almost all $y$ then it must be the case that $E\in\Sigma_1\otimes\Sigma_2$, which is false, so I must be misinterpreting it.

In which way does $F\otimes G$ correspond to the usual definition of the product measure?

More specifically let $F$ and $G$ be ultrafilters on $I$ and $J$ and let $\mu_F:\mathcal{P}(I)\to\{0,1\}$ and $\mu_G:\mathcal{P}(J)\to\{0,1\}$ be the associated finitely additive measures, what is the relationship between $F\otimes G$ and $\mu_F\times\mu_G$?
I'm not familiar with products of finitely additive measures so we may assume tht $F$ and $G$ are $\omega$-complete ultrafilters so that $\mu_F$ and $\mu_G$ are measures in the classical sense if that makes things easier.


Well, if you restrict to sets $X$ of the form $A\times B$, then $X\in F\otimes G$ iff $A\in F$ and $B\in G$. This coincides with the usual "product measure" formula which would say that the measure of $X$ is the product of the measures of $A$ and $B$. But I don't think there's any deeper connection than this, and I would certainly consider it misleading (if not an error) to say $F\otimes G$ "corresponds to the usual definition of the product measure".

  • $\begingroup$ That's what I feared, a bit disappointing I must say! $\endgroup$ – Alessandro Codenotti Apr 20 '18 at 11:50

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