Fubini's Theorem

Analysis, Measure & PDE · Undergraduate · Mathematics Resources editorial compilation (AI-assisted)

For an absolutely integrable function on a product of sigma-finite measure spaces, the product integral equals either iterated integral.

Overview

Fubini's theorem turns an integral on a product space into repeated one-variable integrations. It justifies changing integration order for signed or complex functions when the total absolute integral is finite. The same principle includes sums, mixed sum-integral expressions, and many manipulations of convolution and Fourier transforms.

The central issue is cancellation. Separate one-variable integrations may hide positive and negative mass that cannot be combined into a finite Lebesgue integral on the product. Existence of both iterated integrals alone therefore does not justify equating them.

Prerequisites

Know measurable functions, absolute integrability, and sigma-finite measures. A measure is sigma-finite if the space is a countable union of measurable sets of finite measure. The product sigma-algebra $\mathcal A\otimes\mathcal B$ is generated by measurable rectangles. For sigma-finite factors, the product measure is uniquely determined by its values $\mu(A)\nu(B)$ on these rectangles.

Definition and conventions

Let $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ be sigma-finite measure spaces. If $f:X\times Y\to\mathbb C$ is product-measurable and

\int_{X\times Y}|f|\,d(\mu\otimes\nu)<\infty,

then $f(x,\cdot)$ is integrable for almost every $x$ , and $f(\cdot,y)$ is integrable for almost every $y$ . Their integrals define integrable functions of the remaining variable, after setting them to zero on the exceptional measurable null sets, and

\int_{X\times Y}f\,d(\mu\otimes\nu) =\int_X\left(\int_Yf(x,y)\,d\nu(y)\right)d\mu(x) =\int_Y\left(\int_Xf(x,y)\,d\mu(x)\right)d\nu(y).

Tonelli's theorem is the companion for nonnegative measurable functions: the corresponding equalities hold with $+\infty$ allowed. Applying Tonelli to $|f|$ is often the way to establish Fubini's hypothesis before manipulating $f$ .

Intuition

Integrating by slices amounts to assembling the same mass in a different order. Nonnegative mass cannot cancel, so approximation by increasing simple functions handles it. For signed functions, finite total absolute mass ensures that separating and reassembling positive and negative contributions does not create an undefined infinity minus infinity.

Examples and non-examples

On $(0,1)^2$ , take $f(x,y)=x+2y$ . It is bounded and therefore absolutely integrable. Integrating in $y$ first gives $x+1$ , whose integral is $3/2$ . Integrating in $x$ first gives $1/2+2y$ , again with integral $3/2$ . The finite measure of the square is part of the quick integrability check.

For a failure example on the same square, set

h(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}.

For fixed $x>0$ , this is the $y$ derivative of $y/(x^2+y^2)$ , so $\int_0^1h(x,y)\,dy=1/(1+x^2)$ . The outer integral is $\pi/4$ . For fixed $y>0$ , it is minus the $x$ derivative of $x/(x^2+y^2)$ , giving the reverse iterated integral $-\pi/4$ .

In polar coordinates near the origin, $|h|=|\cos(2\theta)|/r^2$ . On any angular sector where $|\cos(2\theta)|$ stays bounded below, its area integral dominates a positive multiple of $\int_0^\varepsilon dr/r$ , which diverges. The function is not absolutely integrable, so there is no contradiction with Fubini's theorem.

Important properties

A proof sketch establishes the slicing identity for indicators of measurable rectangles, extends it to measurable sets by a monotone-class argument, and then to nonnegative functions using simple approximation and monotone convergence. Applying this to positive and negative parts, or real and imaginary parts, proves the absolutely integrable case. Finiteness prevents undefined subtraction.

With counting measure on $\mathbb N$ , the theorem permits exchanging two sums when $\sum_{m,n}|a_{mn}|<\infty$ . For convolution, $f,g\in L^1(\mathbb R^n)$ give an absolute double integral bounded by $\|f\|_1\|g\|_1$ , justifying the relevant order changes.

Completed measures need care. The product of complete measure spaces need not itself be complete. For Lebesgue-measurable functions on $\mathbb R^{m+n}$ , a completed-product version supplies the same almost-everywhere conclusions; one should not strengthen them to assertions about every section of every representative.

Connections and further reading

Fubini handles integration across variables, while dominated convergence handles a limiting operation inside an integral. Together they support parameter-dependent integrals, Fourier identities, and weak formulations. Hunter, §§5.4–5.7 gives the product-measure and completed Lebesgue formulations.