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Fubini's theorem says that a double integral equals an iterated integral if the double integral is absolutely integrable.

My question is: is the absolute integrability a necessary condition or merely a sufficient condition? To my intuition, a function is not even measurable if it is not absolutely integrable, so I assume that there should be an `only if' part, like this:

A double integral equals an iterated integral if and ONLY IF the double integral is absolutely integrable.

Otherwise, the double integral is not even well defined. Am I missing something?

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    I think it's also true in cases of nonnegative functions where the integral diverges to $\infty$.2011-09-29
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    I suppose that settles the question. So that' why it is not phrased as 'if and only if' ?2011-09-29
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    *A function is not even measurable if it is not absolutely integrable*: Sorry but this is not so.2011-09-29
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    Agreed, I should not have used the words "not even measurable"... But I am still not clear on the following: can we assign a unique well defined value to the double integral if it is not absolutely integrable? If the positive and negative parts of the integrals are not finite, how can we assign a unique value to the double integral? My doubt is: does Fubini's theorem even make sense when the double integral is not absolutely integrable?2011-09-29
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    @Gandhi: Yes it does. There exist examples of functions of two variables $f(x, y)$ such that both $\int\, dx\int f(x, y)\,dy$ and $\int\, dy\int f(x, y)\,dx$ make sense but are different. Of course in this case $f$ cannot be integrable with respect to $x$ and $y$ jointly.2011-09-29
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    One such example is (in what follows assume $x, y \in [0, 1], n \in \mathbb{N}$) $$f(x, y)=\begin{cases} 2^{2n} & \frac{1}{2^n} < x \le \frac{1}{2^{n-1}}\ \frac{1}{2^n}$$\int_0^1\left( \int_0^1f(x, y)\, dx\right)\, dy=0;$$ and $$\int_0^1\left( \int_0^1f(x, y)\, dy\right)\, dx=1.$$2011-09-29
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    @Giuseppe: If it's not integrable with respect to the two variables jointly then that's the same as saying the double integral doesn't make sense. The iterated integrals may still make sense, but the double integral doesn't.2011-09-29
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    @Michael, Gandhi: Whoops, sorry. I've answered to another question, which surely you knew already.2011-09-29
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    I thank you for your patience. So let me get this straight: ABsolute integrability is sufficient for the equality of double and iterated integrals. BUT AT THE SAME TIME, the absolute integrability is a NECESSARY condition for the equality to hold, simply because a Lebesgue integrable function is always absolutely integrable. If this is what you are saying, then the answer to the OP is a YES but for rather trivial reasons. Is this right?2011-09-29

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One more point:

I had asked this question several months ago and since then I have a much better understanding. My confusion was caused by the Tonelli and Fubini theorems both being called Fubini's theorem. In practice, one uses both of them together.

One can use Tonelli's theorem on $|f|$ to check integrability of a real or complex $f$, since every absolutely integrable function is also integrable. Having established integrability via Tonelli's theorem, one can use Fubini's theorem, which is a version of Tonelli's theorem for general functions. They are so closely related that they are sometimes known simply as the Fubini-Tonelli theorem or even simply as Fubini's theorem. (This is what caused my confusion.)

I am not deleting this question just in case somebody else experiences the same confusion.