# Fubini-Tonelli theorem and absolutely Lebesuge integrable functions

As far as I know, a measurable function is Lebesgue integrable if and only it is absolutely integrable. It is simply because the definition of the integrability requires each of the positive part and the negative part has a finite integral. However, some theorems explicitly state that a function need to be "absolutetly integrable". For example, Fubini-Tonelli theorem says that if one of the iterative integrals or the double integral is absolutely integrable they have the same value. What's the point / importance of the absolute value here? Can I just replace the condition of absolute integrability with just integrability?

The point in this case is that for $$\int \int f(x,y) \, dx \, dy$$ to be a well-defined integral, we only need (in addition to measurability requirements) that for (almost) all $y$ $$\int |f(x,y)| \, dx < \infty$$ and then that $$\int \bigg| \int f(x,y) \, dx\bigg| \, dy< \infty.$$
Note that this is in general a very different condition than $$\int \int |f(x,y)| \, dx \, dy<\infty.$$
• Thank you for the answer. I understand your point. But my main concern is the absolute value in the double integral. The statement in the Wikipedia page requires $\int_{XY} |f(x,y)| d(x,y) < \infty$. I don't get what's the difference between $\int_{XY} |f(x,y)| d(x,y) < \infty$ and $\int_{XY} f(x,y) d(x,y) < \infty$ here. – leeto Jan 30 '16 at 11:23
• In this case, there is no difference (if $f$ is measurable with respect to the product sigma algebra). Only, I would write "$\int f (x,y)d (x,y)$ exists" instead of "$\int f (x,y)d (x,y)<\infty$" – PhoemueX Jan 30 '16 at 14:17