Definitions

# Fubini's theorem

In mathematical analysis, Fubini's theorem, named after Guido Fubini, states that if

the integral being taken with respect to a product measure on the space over $Atimes B$, where A and B are complete measure spaces, then

$int_Aleft\left(int_B f\left(x,y\right),dyright\right),dx=int_Bleft\left(int_A f\left(x,y\right),dxright\right),dy=int_\left\{Atimes B\right\} f\left(x,y\right),d\left(x,y\right),$

the first two integrals being iterated integrals with respect to two measures respectively, and the third being an integral with respect to a product of these two measures. Also,

$int_A f\left(x\right), dx int_B g\left(y\right), dy = int_\left\{Atimes B\right\} f\left(x\right)g\left(y\right),d\left(x,y\right)$

the third integral being with respect to a product measure.

If the above integral of the absolute value is not finite, then the two iterated integrals may actually have different values. See below for an illustration of this possibility.

Another version of Fubini's theorem states that if A and B are sigma-finite measure spaces, not necessarily complete, and if either

## Tonelli's theorem

Tonelli's theorem (named after Leonida Tonelli) is a successor of Fubini's theorem. The conclusion of Tonelli's theorem is identical to that of Fubini's theorem, but the assumptions are different. Tonelli's theorem states that on the product of two σ-finite measure spaces, a product measure integral can be evaluated by way of an iterated integral for nonnegative measurable functions, regardless of whether they have finite integral.

In fact, the existence of the first integral above (the integral of the absolute value), can be guaranteed by Tonelli's theorem (see below).

A formal statement of Tonelli's theorem is identical to that of Fubini's theorem, except that the requirements are now that (X, A, μ) and (Y, B, ν) are σ-finite measure spaces, while f maps X×Y to [0, ∞].

## Applications

One of the most beautiful applications of Fubini's theorem is the evaluation of the Gaussian integral which is the basis for much of probability theory:

$int_\left\{-infty\right\}^infty e^\left\{-x^2\right\},dx = sqrt\left\{pi\right\}.$

To see how Fubini's theorem is used to prove this, see Gaussian integral.

Another nice use of Tonelli's theorem is to apply it to $|f\left(x,y\right)|$ for a complex valued function $f$.

It is useful to note that if

$varphi\left(x\right)=int |f\left(x,y\right)|,dy$ and

then

$int |f\left(x,y\right)| , d\left(x,y\right) < infty.$

This is often a useful way to check the conditions of Fubini's theorem.

### Rearranging a conditionally convergent iterated integral

$int_0^1int_0^1 frac\left\{x^2-y^2\right\}\left\{\left(x^2+y^2\right)^2\right\},dy,dx$

does not converge absolutely (i.e. the integral of the absolute value is not finite):

$int_0^1int_0^1$
left|frac{x^2-y^2}{(x^2+y^2)^2}right|,dy,dx=infty.

Fubini's theorem tells us that if the integral of the absolute value is finite, then the order of integration does not matter; if we integrate first with respect to x and then with respect to y, we get the same result as if we integrate first with respect to y and then with respect to x. The assumption that the integral of the absolute value is finite is "Lebesgue integrability". That the assumption of Lebesgue integrability in Fubini's theorem cannot be dropped can be seen by examining this particular iterated integral. Clearly putting "dx dy" in place of "dy dx" has the effect of multiplying the value of the integral by −1 because of the "antisymmetry" of the function being integrated. Therefore, unless the value of the integral is zero, putting "dx dy" in place of "dy dx" actually changes the value of the integral. That is indeed what happens in this case.

#### Proof

One way to do this without using Fubini's theorem is as follows:

$int_0^1int_0^1$
left|frac{x^2-y^2}{(x^2+y^2)^2}right|,dx,dy=int_0^1left[int_0^y frac{y^2-x^2}{(x^2+y^2)^2},dx+int_y^1frac{x^2-y^2}{(x^2+y^2)^2},dxright],dy
$=int_0^1left\left(frac\left\{1\right\}\left\{2y\right\}+frac\left\{1\right\}\left\{2y\right\}-frac\left\{1\right\}\left\{y^2+1\right\}right\right),dy=int_0^1 frac\left\{1\right\}\left\{y\right\},dy-int_0^1frac\left\{1\right\}\left\{1+y^2\right\},dy.$

#### Evaluation

Firstly, we consider the "inside" integral.

$int_0^1frac\left\{x^2-y^2\right\}\left\{\left(x^2+y^2\right)^2\right\},dy$
$= int_0^1 frac\left\{x^2 + y^2 - 2y^2\right\}\left\{\left(x^2 + y^2\right)^2\right\} , dy$
$= int_0^1 frac\left\{1\right\}\left\{x^2 + y^2\right\} , dy + int_0^1 frac\left\{-2y^2\right\}\left\{\left(x^2 + y^2\right)^2\right\} , dy$
$= int_0^1 frac\left\{1\right\}\left\{x^2 + y^2\right\} , dy + int_0^1 y left\left(frac\left\{d\right\}\left\{dy\right\} frac\left\{1\right\}\left\{x^2 + y^2\right\}right\right) , dy$
$= int_0^1 frac\left\{1\right\}\left\{x^2 + y^2\right\} , dy + left\left(left\left[frac\left\{y\right\}\left\{x^2 + y^2\right\}right\right]_\left\{y=0\right\}^1 - int_0^1 frac\left\{1\right\}\left\{x^2 + y^2\right\} , dyright\right)$ (by parts)
$= frac\left\{1\right\}\left\{1 + x^2\right\}.$

This takes care of the "inside" integral with respect to y; now we do the "outside" integral with respect to x:

$int_0^1frac\left\{1\right\}\left\{1+x^2\right\},dx$
=left[arctan(x)right]_0^1

# arctan(1)-arctan(0)

frac{pi}{4}.

Thus we have

$int_0^1int_0^1frac\left\{x^2-y^2\right\}\left\{\left(x^2+y^2\right)^2\right\},dy,dx=frac\left\{pi\right\}\left\{4\right\}$

and

$int_0^1int_0^1frac\left\{x^2-y^2\right\}\left\{\left(x^2+y^2\right)^2\right\},dx,dy=-frac\left\{pi\right\}\left\{4\right\}.$

Fubini's theorem implies that since these two iterated integrals differ, the integral of the absolute value must be ∞.

#### Statement

When

$int_a^bint_c^d left|f\left(x,y\right)right|,dy,dx=infty$

then the two iterated integrals

$int_a^bint_c^d f\left(x,y\right),dy,dx mbox\left\{and\right\} int_c^dint_a^b f\left(x,y\right),dx,dy$

may have different finite values.

## Strong versions of Fubini's theorem

The existence of strengthenings of Fubini's theorem, where the function is no longer assumed to be measurable but merely that the two iterated integrals are well defined and exist, is independent of the standard Zermelo–Fraenkel axioms of set theory. Martin's axiom implies that there exists a function on the unit square whose iterated integrals are not equal, while a variant of Freiling's axiom of symmetry implies that in fact a strong Fubini-type theorem for [0, 1] does hold, and whenever the two iterated integrals exist they are equal. See List of statements undecidable in ZFC.

## References

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