Measure Theory 19 | Fubini's Theorem [dark version]

TL;DR

Fubini’s theorem applies when the component measures are σ-finite and the function is measurable on the product σ-algebra.

Briefing Cornell Notes

Briefing

Fubini’s theorem turns difficult integrals over a product space into easier, iterated one-variable integrals—provided the measures involved are σ-finite. In the setting of two measure spaces (X1, μ1) and (X2, μ2), their product measure μ on X1×X2 is the unique measure that behaves correctly on measurable rectangles. Once σ-finiteness holds, any nonnegative measurable function f on X1×X2 can be integrated by first integrating over one coordinate and then over the other, and the result does not depend on the order.

Concretely, for a measurable f≥0 (allowing the value ∞), the integral over the product space satisfies ∫_{X1×X2} f dμ = ∫_{X1} ( ∫_{X2} f dμ2 ) dμ1 = ∫_{X2} ( ∫_{X1} f dμ1 ) dμ2. The practical message is that integrating with respect to the product measure is “not harder” than repeatedly integrating with respect to the original measures. The theorem also extends beyond nonnegative functions: if f is integrable over the product measure (so f lies in L1(μ), meaning the integral has a finite value), then the same iterated-integral identity holds once the inner integrals are defined appropriately.

A worked example illustrates how the choice of integration order can make the computation straightforward. The measure is taken as the 2D Lebesgue measure on the unit square, built as a product of two 1D Lebesgue measures. The region A⊂R^2 is defined by the points between the curve y=x^2 and the line y=x, restricted to the unit square. Using the characteristic (indicator) function 1_A, the target integral becomes an integral over R^2 of 1_A(x,y)·(2xy), which is nonnegative and therefore fits directly into Fubini’s theorem.

The key step is selecting the better order: integrate with respect to y first. For a fixed x in [0,1], the variable y ranges from x^2 up to x. This makes the inner integral clean: ∫_{x^2}^{x} 2xy dy. Carrying out the y-integration yields a polynomial in x, and the remaining outer integral over x runs from 0 to 1. Evaluating that final one-dimensional integral gives the result 1/12.

Although Fubini guarantees both orders produce the same final value, the example emphasizes that one order can align with the geometry of the region so the bounds become immediate. The same logic scales to higher dimensions: with σ-finite product measures, iterated integration can be applied repeatedly, turning an n-dimensional product integral into n one-dimensional integrals—again with careful attention to the order of integration from the start.

Cornell Notes

Fubini’s theorem provides a reliable way to compute integrals over product measure spaces by reducing them to iterated one-variable integrals. For σ-finite measures μ1 on X1 and μ2 on X2, any nonnegative measurable function f on X1×X2 satisfies ∫_{X1×X2} f d(μ1×μ2) = ∫_{X1}(∫_{X2} f dμ2)dμ1 = ∫_{X2}(∫_{X1} f dμ1)dμ2. If f is integrable over the product measure (f∈L1), the same equality holds with properly defined inner integrals. A concrete example integrates 2xy over the region between y=x^2 and y=x inside the unit square, choosing y-first because it makes the bounds simple, yielding 1/12.

What conditions make Fubini’s theorem applicable to a product integral?

The measures μ1 and μ2 must be σ-finite, and the function f on X1×X2 must be measurable with respect to the product σ-algebra. For the basic form, f should be nonnegative (allowing ∞). For the extended form, f must be integrable over the product measure (finite integral), i.e., f∈L1(μ1×μ2), so the iterated integrals are well-defined and equal to the product integral.

Why does the theorem guarantee the same value no matter which variable is integrated first?

Under σ-finiteness, the product integral of a nonnegative measurable function can be computed as an iterated integral in either order: first integrate over X2 then X1, or first over X1 then X2. Fubini’s theorem asserts these two iterated integrals coincide, so the final value is order-independent even though intermediate expressions can differ.

How does the characteristic function 1_A help in the example?

The region A⊂R^2 is handled by inserting its indicator function into an integral over the whole space R^2. Because 1_A(x,y)=1 on A and 0 outside, the integral of 1_A(x,y)·(2xy) over R^2 becomes exactly the integral of 2xy over A. This converts a “region integral” into a standard product-measure integral suitable for Fubini.

Why was integrating with respect to y first the “best order” in the worked region example?

For points in A, y is constrained between x^2 and x. Fixing x first makes the y-bounds immediate: y∈[x^2, x] for x∈[0,1]. Integrating with respect to y first therefore produces a straightforward inner integral ∫_{x^2}^{x} 2xy dy, leaving a simple one-variable outer integral in x.

What is the final numerical result of the example integral, and what does it come from?

The integral of 2xy over the region between y=x^2 and y=x inside the unit square evaluates to 1/12. After performing the inner y-integration, the remaining x-integral reduces to a polynomial expression in x, which is then evaluated from 0 to 1 to obtain 1/12.

How can Fubini’s theorem be extended beyond two dimensions?

With σ-finite product measures, the same iterated-integration idea applies repeatedly. A four-dimensional product integral can be computed as four one-dimensional integrals by choosing an order that matches the region’s constraints, again relying on the theorem’s order-independence for the final value.

Review Questions

What role does σ-finiteness play in ensuring Fubini’s theorem works for product measures?
In the region defined by y between x^2 and x, which variable bounds become simplest when integrating first, and why?
How does the indicator function 1_A transform an integral over a region A into an integral over a larger space?

Key Points

1
Fubini’s theorem applies when the component measures are σ-finite and the function is measurable on the product σ-algebra.
2
For nonnegative measurable f, the product integral equals either iterated integral order: integrate over X2 then X1, or X1 then X2.
3
For integrable functions (f∈L1 of the product measure), the same equality holds once inner integrals are properly defined.
4
Using an indicator function 1_A converts an integral over a region A into an integral over the full product space.
5
Choosing the integration order that matches the region’s geometry can make bounds and algebra dramatically simpler, even though the final answer is the same.
6
The method generalizes: higher-dimensional product integrals can be reduced to iterated one-dimensional integrals by applying the theorem repeatedly.

Highlights

Fubini’s theorem reduces a product-space integral to iterated one-variable integrals, with the final value independent of the integration order under σ-finiteness.

The region between y=x^2 and y=x inside the unit square becomes easy to integrate when integrating with respect to y first, since y-bounds are [x^2, x].

The worked example computes ∫_A 2xy d(x,y) and gets 1/12 after two one-dimensional integrations.

Indicator functions 1_A let region integrals fit directly into product-measure integration frameworks. 

Topics

Fubini's Theorem
Product Measure
Iterated Integrals
Indicator Function
Lebesgue Measure