# Fubini's theorem and the existence of integrals…

Suppose I am given some integral $$\int\int_{E} f(x,y) \, \mathscr{d}x \mathscr{d}y$$ and I want to show that the iterated integrals exist and are equal, but that the double integral does not exist.

To show that the iterated integrals exist and are equal, does it suffice to show by Tonelli's theorem that $f$ is measurable? How would I show that the double integral does not exist? It seems to me that even if this integral diverges to infinity we can still say that it "exists".

I am looking for general strategies with these types of questions, so I have omitted many of the specifics.

I am confused with these concepts and the application of the Fubini-Tonelli theorem. Any input would be useful.

• If the iterated integrals exist and are equal then the double integral exists, and Fubini's theorem tells you that you can obtain the value of the double integral by integrating (either of) the iterated integrals. It's possible for the double integral to exist while one of the iterated integrals doesn't though, so have you perhaps mis-stated your question? For Tonelli's theorem, the measurability of f also requires that the product measure be $\sigma$-finite; measurability alone isn't enough – postmortes Nov 3 '15 at 12:23
• @postmortes No I do not believe the question has been misstated, as it is possible for the iterated integrals to exist, for their values to coincide, and for the double integral not to exist (according to the questions I am looking at) – möbius Nov 3 '15 at 12:31
• So how are you defining the double integral then? – postmortes Nov 3 '15 at 13:11
• Are we talking about Lebesgue integrals, Riemann integrals, or some other integral? – Daniel Fischer Nov 3 '15 at 15:18
• If the assumptions of Fubini/Tonelli are satisfied, it always follows that the iterated integral exists if and only if the double integral exists. Thus, to show that the iterated integrals exist (and are equal), you don't have much of a choice but to explicitly compute the two iterated integrals. To show that the double integral does not exist, show $\int \int |f| \, dx \, dy = \infty$. By Tonelli, this shows that the double integral does not exist (if $f$ is measurable w.r.t. the product $\sigma$ algebra and your measure spaces are $\sigma$-finite) – PhoemueX Nov 3 '15 at 17:47