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Tonelli's theorem
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(Theorem)
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Here denote  as the space of measurable functions
![$ X \to [0,\infty]$](http://images.planetmath.org:8080/cache/objects/5713/l2h/img2.png "$ X \to [0,\infty]$") . Furthermore all integrals are Lebesgue integrals.
Basically this says that you can switch the order of integrals, or integrate over the product space as long as everything is positive and the spaces are  -finite. Do note that we allow the functions to take on the value of infinity with the standard conventions used in Lebesgue integration. That is,
 , so that if a function is infinite on a set of measure 0, then this does not contribute anything to the value of the integral. See the entry on extended real numbers for further discussion.
If we take the counting measure on
 , then one can state the Tonelli theorem for sums.
Theorem 2 (Tonelli for sums) Suppose that
 for all
 , then
In the above theorem we have used
 as our index set for simplicity and familiarity of notation. If you would have an uncountable number of non-zero elements  then all the sums would be infinite and the result would be trivial. So the theorem for arbitrary index sets just reduces to the above case.
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- Gerald B. Folland. Real Analysis, Modern Techniques and Their Applications. John Wiley & Sons, Inc., New York, New York, 1999
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"Tonelli's theorem" is owned by jirka.
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Cross-references: number, uncountable, sums, counting measure, extended real numbers, measure, infinite, infinity, positive, product, product measure, functions, measure spaces, Lebesgue integrals, integrals, measurable functions
There are 2 references to this entry.
This is version 4 of Tonelli's theorem, born on 2004-03-15, modified 2005-03-07.
Object id is 5713, canonical name is TonellisTheorem.
Accessed 6767 times total.
Classification:
| AMS MSC: | 28A35 (Measure and integration :: Classical measure theory :: Measures and integrals in product spaces) |
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Pending Errata and Addenda
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![$\displaystyle \int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left[ \int_Y f... ...nu(y) \right] d\mu(x) = \int_Y \left[ \int_X f(x,y) \,d\mu(x) \right] d\nu(y) .$](http://images.planetmath.org:8080/cache/objects/5713/l2h/img12.png "$\displaystyle \int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left[ \int_Y f... ...nu(y) \right] d\mu(x) = \int_Y \left[ \int_X f(x,y) \,d\mu(x) \right] d\nu(y) .$")
