---
title: "Product Measure and Fubini's Theorem"
description: "Two measure spaces combine into one, and an integral over the product can be done one variable at a time. The hypotheses of the theorem say exactly when the order of integration is free."
canonical: https://lightmysky.com/learn/mathematics/product-measure-and-fubinis-theorem-mt_1VzzVmMjTB
source: https://lightmysky.com/learn/mathematics/product-measure-and-fubinis-theorem-mt_1VzzVmMjTB.md
retrieved: 2026-09-12
---

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# Product Measure and Fubini's Theorem

Two measure spaces combine into one, and an integral over the product can be done one variable at a time. The hypotheses of the theorem say exactly when the order of integration is free.

Subject: Mathematics · Area: Calculus & Analysis · Ages 22 to 24
Page: https://lightmysky.com/learn/mathematics/product-measure-and-fubinis-theorem-mt_1VzzVmMjTB

## Ready when they can

- State the difference between the Tonelli and Fubini hypotheses
- Exchange the order of a double integral and justify the exchange
- Give an example where the two iterated integrals differ, and name the hypothesis that fails

## Lesson: Integrate one variable at a time

You combine two measure spaces into one product space, then integrate over both variables at once. Tonelli says you may split the double integral into iterated integrals when the function is nonnegative and measurable. Fubini says the same for a sign changing function, but only when the integral of its absolute value is finite.

**Example.** Take f(x, y) = x times y on the rectangle with x from 0 to 2 and y from 0 to 3. Integrate in y first to get 4.5 times x, then in x to get 9, and the other order gives 9 too. The exchange is licensed because a continuous function on a closed rectangle is bounded, so its absolute integral is finite and Fubini applies. A constant function is even easier: its integral is the constant times the area.

Some sign changing functions have two iterated integrals that both exist yet give different answers. In each case the integral of the absolute value is infinite, so Fubini does not apply, and Tonelli cannot help because the function takes negative values. Treat any such mismatch as the signal that absolute integrability failed.

**Tip.** Before you exchange the order, run this check. If the function is nonnegative, Tonelli covers you. Otherwise compute the integral of the absolute value and confirm it is finite, and then Fubini covers you. Defined iterated integrals alone never license the exchange.

**Recap.** Tonelli needs nonnegativity, Fubini needs a finite absolute integral, and either one lets you integrate one variable at a time.

## Practice

17 questions on this page, each with its working shown.

## Needs first

- [Monotone Convergence, Fatou and Dominated Convergence](https://lightmysky.com/learn/mathematics/monotone-convergence-fatou-and-dominated-convergence-mt_7n6ZDkHeBe)
- [Double Integrals over General Regions](https://lightmysky.com/learn/mathematics/double-integrals-over-general-regions-mt_UKS9_AhAWw)

## Opens up

- [Lp Spaces and the Inequalities They Rest On](https://lightmysky.com/learn/mathematics/lp-spaces-and-the-inequalities-they-rest-on-mt_Qd8jU26qgE)
