Product Measures and Fubini's Theorem — Mathematics Knowledge System

Overview

Given two sigma-finite measure spaces, the product measure is the unique measure on the product sigma-algebra that assigns to each measurable rectangle the product of the factor measures. Fubini's theorem then justifies interchanging the order of integration: if a function is integrable (or non-negative measurable), the double integral equals the iterated integrals in either order. Tonelli's theorem handles the non-negative case without integrability, while Fubini's theorem requires absolute integrability.

Intuition

A double integral over a rectangle [a,b] x [c,d] can be computed as an iterated integral: first integrate over y for each fixed x, then integrate the result over x. Fubini's theorem says this works for Lebesgue integrals whenever the function is integrable (or non-negative). The key insight is that the product measure lambda x mu assigns area/volume to rectangles consistently and extends to all measurable sets. Tonelli's theorem (for non-negative functions) lets you freely swap the order to check integrability.

Formal Definition

Definition

Let (X, M, mu) and (Y, N, nu) be sigma-finite measure spaces. The product sigma-algebra M ⊗ N is generated by measurable rectangles A x B (A in M, B in N). The product measure mu x nu is the unique measure on M ⊗ N satisfying (mu x nu)(A x B) = mu(A) * nu(B). Fubini's theorem: if f is M ⊗ N-measurable and integrable (or non-negative), then the iterated integrals equal the double integral.

\mathcal{M} \otimes \mathcal{N} = \sigma(\{ A \times B : A \in \mathcal{M},\, B \in \mathcal{N} \})

Product sigma-algebra

(\mu \times \nu)(A \times B) = \mu(A) \cdot \nu(B)

Product measure on rectangles

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

Fubini's theorem

Notation

Notation	Meaning
$\mu \times \nu$	Product measure on X x Y
$\mathcal{M} \otimes \mathcal{N}$	Product sigma-algebra
$f_x(y) = f(x,y)$	x-section of f

Theorems

Theorem 1: Existence and Uniqueness of Product Measure

\text{Let } (X,\mathcal{M},\mu) \text{ and } (Y,\mathcal{N},\nu) \text{ be sigma-finite. There exists a unique measure } \mu \times \nu \text{ on } \mathcal{M} \otimes \mathcal{N} \text{ such that } (\mu\times\nu)(A\times B) = \mu(A)\nu(B) \text{ for all } A\in\mathcal{M},\, B\in\mathcal{N}.

Theorem 2: Tonelli's Theorem

\text{If } f : X \times Y \to [0,\infty] \text{ is } \mathcal{M}\otimes\mathcal{N}\text{-measurable, then} \int_{X\times Y} f\,d(\mu\times\nu) = \int_X\!\int_Y f(x,y)\,d\nu(y)\,d\mu(x) = \int_Y\!\int_X f(x,y)\,d\mu(x)\,d\nu(y).

Theorem 3: Fubini's Theorem

\text{If } f \in L^1(\mu\times\nu), \text{ then for } \mu\text{-a.e. } x,\; y \mapsto f(x,y) \in L^1(\nu); \text{ for } \nu\text{-a.e. } y,\; x \mapsto f(x,y) \in L^1(\mu); \text{ and the iterated integrals equal the double integral.}

Worked Examples

Since f(x,y) = x^2 + y^2 >= 0 and is continuous (hence measurable), Tonelli's theorem applies.
$\int_{[0,1]^2} (x^2 + y^2)\,d(\lambda\times\lambda)$
Apply Fubini to iterate: first integrate over y.
$= \int_0^1 \left( \int_0^1 (x^2 + y^2)\,dy \right) dx = \int_0^1 \left[ x^2 y + \frac{y^3}{3} \right]_0^1 dx = \int_0^1 \left( x^2 + \frac{1}{3} \right) dx$
Integrate over x.
$= \left[ \frac{x^3}{3} + \frac{x}{3} \right]_0^1 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

Answer: 2/3.

Practice Problems

Difficulty 7/10

Use Fubini's theorem to compute integral_0^infty integral_0^infty e^{-x} * e^{-y} * min(x,y) dy dx.

Difficulty 8/10

Prove that if E in M ⊗ N and (mu x nu)(E) = 0, then for mu-a.e. x, the section E_x = {y : (x,y) in E} satisfies nu(E_x) = 0.

Difficulty 7/10

Fubini's theorem requires the function f to satisfy which condition?

f is continuous on X x Yf is boundedf is in L^1(mu x nu) (or f >= 0 for Tonelli)f is a product function f(x,y) = g(x)h(y)

Common Mistakes

Common Mistake

Fubini's theorem applies to any measurable function.

Fubini requires L^1 integrability. For non-negative functions, Tonelli's theorem applies without integrability. Without some assumption, the iterated integrals can differ or fail to equal the double integral.

Common Mistake

The product measure mu x nu is always unique.

Uniqueness requires sigma-finiteness of both factor measures. Without sigma-finiteness, the product measure need not be unique.

Common Mistake

The product sigma-algebra M ⊗ N contains all subsets of X x Y.

M ⊗ N is generated by measurable rectangles and is typically much smaller than the power set. On R^2, M ⊗ N equals B(R^2) = B(R x R), not all subsets of R^2.

Quiz

Historical Background

Guido Fubini proved his theorem in 1907 for continuous functions on rectangles. Leonida Tonelli extended it in 1909 to non-negative measurable functions. The modern abstract formulation using product measures and sigma-algebras was developed by Halmos, Saks, and others in the 1940s–1950s.

1907
Fubini proves iterated integral theorem for continuous functions
Guido Fubini
1909
Tonelli extends the result to non-negative measurable functions
Leonida Tonelli
1950
Halmos presents the abstract product measure theory in textbook form
Paul Halmos

Summary

The product measure mu x nu is the unique measure on M ⊗ N assigning mu(A)*nu(B) to each rectangle A x B, given sigma-finite factors.
Tonelli's theorem: for f >= 0 measurable, the double integral equals both iterated integrals (possibly infinite).
Fubini's theorem: for f in L^1(mu x nu), iterated integrals exist a.e., are integrable, and equal the double integral.
If the iterated integrals of |f| are finite, then f is in L^1(mu x nu) and Fubini applies.
Applications include computing joint distributions, convolutions, and justifying interchange of sum and integral in analysis.

References

BookFolland, G. B. (1999). Real Analysis: Modern Techniques and Their Applications (2nd ed.). Wiley. Chapter 2.
BookRudin, W. (1987). Real and Complex Analysis (3rd ed.). McGraw-Hill. Chapter 8.
WebsiteWikipedia — Fubini's theorem