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Dec 23, 2011 at 0:06 comment added Lunasaurus Rex I thought that a simple function is one that has finite image on a set of full measure. Perhaps I am just quibbling.
Dec 14, 2011 at 7:44 comment added Kevin Buzzard A "simple function" is a very restrictive class of function -- for example, it has finite image. This is why I'm wondering whether we're really doing sums, not integrals.
Dec 14, 2011 at 5:10 comment added Junkie "Is the paper cited just saying, at the end of the day, that one can exchange the order of two finite sums?" I don't think, the conclusion is just that. If you look at 6.2, they say the integral of the product measure is the double integral over each measure separately (for simple functions). So my take on the category theory, is that it is concluding the product measure acts functorially, as expected by the suspected diagram chase?! So I guess $\int_{A\times B}f(x,y)(\mu_A\times\mu_B)(x,y)=\int_A\int_B f(x,y)\mu_A(x)\mu_B(y)$ is the Fubini theorem. I agree there is no analysis.
Dec 14, 2011 at 0:25 comment added Yemon Choi Further to my previous comment: "unlike Modo, Sergeant Colon did understand the meaning of the word 'irony'. He thought it meant 'something like iron'."
Dec 13, 2011 at 22:04 comment added Yemon Choi @Kevin: shush, everyone knows that these kind of technicalities are of no concern to Proper Mathematicians... ;)
Dec 13, 2011 at 21:25 comment added Kevin Buzzard I've only had a superficial look at the paper, but it seems to me that what people usually call Fubini's theorem is an analytical fact that under certain boundedness conditions one can exchange the order of two integrals. Is the paper cited just saying, at the end of the day, that one can exchange the order of two finite sums? Near the bottom of the first page the author says things "since we are not worried about convergence" (as he's in some finite world) -- but surely what people usually call Fubini's theorem is precisely an issue of convergence?
Dec 13, 2011 at 20:55 history answered Peter Arndt CC BY-SA 3.0