Questions: Product Measures and Fubini's Theorem

Question 1

A function f on [0,1] × [0,1] has one iterated integral equal to +π/4 and the other equal to −π/4. What does this tell you about f?

• A. f is discontinuous, so the Riemann integral cannot be applied
• B. f is not Lebesgue integrable — its absolute value has infinite integral, so Fubini's theorem does not apply
• C. The σ-algebras on the two copies of [0,1] are incompatible
• D. A computational error was made — Fubini's theorem guarantees both iterated integrals are equal for any measurable f

Answer: (B)

This is the classic counterexample: f(x,y) = (x²−y²)/(x²+y²)² on [0,1]×[0,1]. The two iterated integrals give +π/4 and −π/4 because f is not Lebesgue integrable — ∫|f| d(μ×ν) = ∞. Fubini's theorem only guarantees equality of iterated integrals when the function is integrable. Option D is the most dangerous misconception: Fubini does NOT apply to all measurable functions, only to integrable ones.

Question 2

You want to compute ∫∫f(x,y) dμ dν but are unsure whether f is integrable. What is the correct strategy?

• A. Apply Fubini's theorem directly — it works for all σ-finite measure spaces
• B. Average the two iterated integrals to obtain the correct value
• C. Apply Tonelli's theorem to ∫∫|f| first: since |f| ≥ 0, Tonelli guarantees the iterated integrals of |f| are equal, allowing you to check whether ∫|f| d(μ×ν) < ∞ before invoking Fubini
• D. Differentiate under the integral sign to simplify the integrand

Answer: (C)

Tonelli's theorem is the safety check that unlocks Fubini. For non-negative functions, Tonelli guarantees the two iterated integrals always agree (possibly being ∞), so you can compute ∫(∫|f(x,y)|dν)dμ in whichever order is convenient. If this equals a finite number, f is integrable and Fubini applies to f itself — you can switch orders freely. If it equals ∞, Fubini is inapplicable and switching orders may give different answers. Tonelli → Fubini is the standard workflow.

Question 3

Fubini's theorem guarantees that for any measurable function f on a product of σ-finite measure spaces, the two iterated integrals are equal.

• T. True
• F. False

Answer: False

Fubini requires not just measurability but integrability: ∫|f| d(μ×ν) < ∞. Without this condition, switching integration order can yield different values — as demonstrated by f(x,y) = (x²−y²)/(x²+y²)² on [0,1]². The σ-finiteness condition is needed for the product measure construction, but it alone does not save you from the ordering problem. The integrability condition is the binding constraint for safe order-switching.

Question 4

Lebesgue measure on ℝ² is the product of two copies of Lebesgue measure on ℝ, in the sense that the measure of a rectangle A × B equals the product of the measures of A and B.

• T. True
• F. False

Answer: True

This is exactly how the product measure construction works: (μ × ν)(A × B) = μ(A) · ν(B) on measurable rectangles, then extended via Carathéodory to the full product σ-algebra. Lebesgue measure on ℝ² is precisely this product — the measure of a rectangle is width times height, consistent with the geometric notion of area. This concrete case motivates the abstract construction.

Question 5

Why can switching the order of integration change the result for some functions, and what condition prevents this from happening?

Model answer: When a function f has both large positive and large negative regions, the 'cancellation' between them can depend on which variable is integrated first — the partial integrals along one variable may diverge or accumulate differently. The condition that prevents this is absolute integrability: ∫|f| d(μ×ν) < ∞. When the integral of |f| is finite, the positive and negative parts of f each have finite integrals independently, so their contributions are unambiguous regardless of integration order.

This is the reason Fubini has a hypothesis rather than being a universal rule. In calculus courses, students switch integration order freely because the functions involved are typically continuous on bounded domains — integrability is implicit. In measure theory, the condition must be stated explicitly. The standard proof of Fubini proceeds by decomposing f = f⁺ − f⁻ into positive and negative parts, then applying Tonelli to each (both are non-negative). Integrability of f guarantees both f⁺ and f⁻ have finite integrals, making the decomposition valid and order-independent.