You want to evaluate ∫∫f(x,y) d(μ⊗ν) by computing the inner integral first with respect to y, then the outer with respect to x. You have not verified that f is integrable. The correct approach is:
AProceed directly — Fubini's theorem allows iterated integration for any measurable f on a product space
BVerify that both component measures are sigma-finite, which is sufficient to apply Fubini to any measurable f
CApply Tonelli's theorem to |f| first to verify ∫|f| d(μ⊗ν) < ∞, then apply Fubini to f
DCompute both iterated integrals and check that they agree before trusting either result
Fubini's theorem requires the integrability hypothesis ∫|f| d(μ⊗ν) < ∞. The standard workflow when this is not known is to apply Tonelli to |f|: since |f| ≥ 0, Tonelli allows iterated integration without the L¹ hypothesis. If ∫(∫|f(x,y)| dν) dμ < ∞, then f is integrable and Fubini applies. Sigma-finiteness (option B) is needed for the product measure construction but does not alone make f integrable. Checking agreement of both orders (option D) doesn't help — non-integrable functions can produce equal iterated integrals that still don't equal the double integral.
Question 2 Multiple Choice
The key reason Fubini's theorem requires ∫|f| d(μ⊗ν) < ∞ is:
ATo ensure the product measure μ⊗ν assigns finite total measure to X × Y
BWithout integrability, swapping the order of iterated integration can yield different values — including disagreement with the double integral over the product space
CTo allow Tonelli's theorem to be applied as a preliminary step before Fubini
DTo guarantee that the slices f(x, ·) are continuous for μ-almost every x
The classic counterexample — a function on [0,1]×[0,1] where ∫(∫f dy) dx ≠ ∫(∫f dx) dy — shows iterated integration can be order-dependent without integrability. Functions with non-integrable positive and negative parts can 'cancel' differently depending on which direction you integrate first. The L¹ condition rules out this pathology: when |f| has finite integral over the product space, positive and negative contributions balance absolutely, making the result order-independent and equal to the double integral.
Question 3 True / False
For a non-negative measurable function, Tonelli's theorem allows iterated integration in either order even if both iterated integrals are infinite.
TTrue
FFalse
Answer: True
This is the key feature that makes Tonelli useful as a preliminary tool. For f ≥ 0 and σ-finite measures, iterated integration in either order always gives the same result — possibly both +∞. Non-negative functions cannot produce the cancellation pathology requiring Fubini's integrability hypothesis, because there are no negative parts to cancel in order-dependent ways. In practice: apply Tonelli to |f| to verify integrability, then apply Fubini to f itself.
Question 4 True / False
If both iterated integrals of a function f are finite and equal, Fubini's theorem guarantees that f is integrable over the product space.
TTrue
FFalse
Answer: False
Fubini's theorem is an implication in one direction: integrability of f implies all three quantities (both iterated integrals and the double integral) are equal. It is not a biconditional. There exist non-integrable functions for which both iterated integrals happen to be equal and finite, yet ∫|f| d(μ⊗ν) = ∞. Finite equal iterated integrals do not imply integrability. Tonelli applied to |f| is the correct tool for verifying the L¹ hypothesis.
Question 5 Short Answer
What does the existence of a function for which the two iterated integrals depend on the order of integration tell us about why Fubini's theorem requires a proof rather than being a tautology?
Think about your answer, then reveal below.
Model answer: Such a function demonstrates that iterated integration on a product space is not automatically order-independent — the equality of the two iterated integrals and the double integral is a non-trivial fact that requires conditions. Without integrability, a function with cancelling positive and negative parts can produce different sums depending on which variable is integrated first: the cancellation pattern depends on order. Integrability (finite total variation) prevents this by ensuring positive and negative parts each have finite integral separately, making the result stable across integration orders. The theorem says exactly that integrability is the condition under which all three quantities necessarily agree — a genuine theorem requiring proof, not a definition.
The canonical counterexample on [0,1]×[0,1] defines f so that integrating in one order gives 0 and in the other gives a nonzero value. The mechanism is that f oscillates between large positive and large negative values whose cancellation depends on the order of summation. Integrability (∫|f| < ∞) rules this out by requiring positive and negative parts to be each separately finite, so cancellation is stable regardless of integration order. This is why Fubini is a theorem: it identifies precisely the condition that makes the Lebesgue integral on product spaces coherent.