Fubini's theorem

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

$\int _{A\times B}|f(x,y)|\,d(x,y)<\infty ,$

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

$\int _{A}\left(\int _{B}f(x,y)\,dy\right)\,dx=\int _{B}\left(\int _{A}f(x,y)\,dx\right)\,dy=\int _{A\times B}f(x,y)\,d(x,y),$

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,

$\int _{A}f(x)\,dx\int _{B}g(y)\,dy=\int _{A\times B}f(x)g(y)\,d(x,y)$

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.

Formal Statement

Let (X, A, μ) and (Y, B, ν) be complete measure spaces, and let (X×Y, C, μ×ν) be their product measure space. Then for any measurable function f mapping X×Y to the extended real numbers, if f is μ×ν integrable, that is

$\int _{X\times Y}|f|d(\mu \times \nu )<\infty$

The the following three conclusions hold.

1. For almost all x in X, the function f_x mapping y (in Y) to f(x, y) is integrable, similarly for f_y

2. The function defined by

$F_{X}(x)=\left\{{\begin{array}{ll}\int _{Y}f_{x}d\nu &{\text{if }}f_{x}{\text{ is integrable}}\\0&{\text{otherwise}}\end{array}}\right.$

is integrable, similarly for F_Y

3. These integrals satisfy

$\int _{X}F_{X}d\mu =\int _{Y}F_{Y}d\nu =\int _{X\times Y}fd(\mu \times \nu )$

Some debate is present as to whether the requirement that the measure spaces be σ-finite forms part of "Fubini's Theorem", with Rudin and Williams arguing 'yes'. The requirement is not necessary, as is made clear in .

Proof

A complete proof of this theorem is beyond the scope of this article, see Royden for details. An outline is as follows:

1. First assume that f is the indicator function for some measurable rectangle, that is for some set of the form A₁×B₁, where A₁ is in A and B₁ is in B. The result then follows as the integral of this function is simply the product of the measures.

2. Next allow f to be the indicator function for the union of some measurable rectangles, which can (without loss of generality) be taken to be disjoint. Then the result follows by the additivity of the measures.

3. Next suppose f is the indicator function for some set of measure zero. Expressing this set as the limit of a decreasing sequence of measurable rectangles and applying Fatou's Lemma implies that the function f_x is zero almost everywhere for almost all x in X. By completeness, we then have that this function is measurable for almost all x in X. Similarly for f_y. Then applying the definitions of F_X and F_Y, we can show that the integrals are all zero, and so are equal.

4. Next suppose f is the indicator function of a set which can be expressed as the countable intersection of the countable unions of (without loss of generality, disjoint) measurable rectangles. That is

$f=\chi _{E},\qquad E=\cap _{i=1}^{\infty }\cup _{j=1}^{\infty }R_{ij}$

where R_ij are disjoint measurable rectangles. This is the decreasing limit of

$f_{N}=\chi _{(E_{N})},\qquad E_{N}=\cap _{i=1}^{N}\cup _{j=1}^{\infty }R_{ij}.$

We can also express E_N as a countable union of measurable rectangles. We can therefore express f as the decreasing limit of a sequence of functions, and hence f_x can be expressed in the same way. Therefore provided f is integrable, and so E has finite measure, f_x is measurable by decreasing continuity of measures. Similarly for f_y. The integrals for each term in this sequence must be equal, by Section 2. We now apply the dominated convergence theorem to show that the integrals are equal in limit.

5. Next suppose f is the indicator function of an arbitrary measurable set E. We can write E as the union of a set of measure zero and the countable intersection of a countable union of rectangles, and so we can write f as the sum of a function of the form discussed in Section 3 and a function of the form discussed in Section 5. By linearity, the integrals must be equal, etc...

6. Next suppose f is a nonnegative simple function. Then f can be expressed as the sum of a finite number of functions of the form discussed in Section 5. The results follow by linearity.

7. Next suppose f is an arbitrary nonnegative integrable function. Then f can be expressed as the increasing limit of functions of the form discussed in Section 6. The results follow by the Monotone Convergence Theorem.

8. Finally, suppose f is an arbitrary integrable function. Decomposing f into its positive and negative parts gives the difference of two functions of the form discussed in Section 7. The results follow by linearity.

Tonelli's theorem

Tonelli's theorem (named after Leonida Tonelli) is a predecessor 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 complete, 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).

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:

$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}.$

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 $|f(x,y)|$ for a complex valued function $f$.

It is useful to note that if

$\varphi (x)=\int |f(x,y)|\,dy$ and $\int \varphi (x)dx<\infty ,$

then

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

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

Rearranging a conditionally convergent iterated integral

The iterated integral

$\int _{0}^{1}\int _{0}^{1}{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\,dy\,dx$

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

$\int _{0}^{1}\int _{0}^{1}\left|{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\right|\,dy\,dx=\infty .$

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:

$\int _{0}^{1}\int _{0}^{1}\left|{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\right|\,dx\,dy=\int _{0}^{1}\left[\int _{0}^{y}{\frac {y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}}\,dx+\int _{y}^{1}{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\,dx\right]\,dy$

$=\int _{0}^{1}\left({\frac {1}{2y}}+{\frac {1}{2y}}-{\frac {1}{y^{2}+1}}\right)\,dy=\int _{0}^{1}{\frac {1}{y}}\,dy-\int _{0}^{1}{\frac {1}{1+y^{2}}}\,dy.$

Evaluation

Firstly, we consider the "inside" integral.

$\int _{0}^{1}{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\,dy$

$=\int _{0}^{1}{\frac {x^{2}+y^{2}-2y^{2}}{(x^{2}+y^{2})^{2}}}\,dy$

$=\int _{0}^{1}{\frac {1}{x^{2}+y^{2}}}\,dy+\int _{0}^{1}{\frac {-2y^{2}}{(x^{2}+y^{2})^{2}}}\,dy$

$=\int _{0}^{1}{\frac {1}{x^{2}+y^{2}}}\,dy+\int _{0}^{1}y\left({\frac {d}{dy}}{\frac {1}{x^{2}+y^{2}}}\right)\,dy$

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

$={\frac {1}{1+x^{2}}}.$

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

$\int _{0}^{1}{\frac {1}{1+x^{2}}}\,dx=\left[\arctan(x)\right]_{0}^{1}=\arctan(1)-\arctan(0)={\frac {\pi }{4}}.$

Thus we have

$\int _{0}^{1}\int _{0}^{1}{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\,dy\,dx={\frac {\pi }{4}}$

and

$\int _{0}^{1}\int _{0}^{1}{\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}\,dx\,dy=-{\frac {\pi }{4}}.$

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

Statement

When

$\int _{a}^{b}\int _{c}^{d}\left|f(x,y)\right|\,dy\,dx=\infty$

then the two iterated integrals

$\int _{a}^{b}\int _{c}^{d}f(x,y)\,dy\,dx\ {\mbox{and}}\ \int _{c}^{d}\int _{a}^{b}f(x,y)\,dx\,dy$

may have different finite values.

See also

• Clairaut's theorem