You want to compute ∫∫ f(x,y) d(μ×ν) for a function f of indefinite sign by switching the order of integration. What is the correct two-step procedure?
AApply Fubini directly — it always allows switching order as long as f is measurable
BApply Tonelli to f, then use the result to apply Fubini
CApply Tonelli to |f| to verify finiteness of the iterated integral, then apply Fubini to f
DApply Fubini to |f| to check absolute integrability, then apply Tonelli to f
The correct strategy is Tonelli first on |f|, then Fubini on f. Tonelli applies to non-negative functions without any integrability precondition — if the iterated integral of |f| is finite, then f is integrable over the product space, and Fubini applies. Option A is wrong because Fubini requires f to be integrable as a precondition — you cannot use it to establish integrability. Option B inverts the roles. Option D confuses the two theorems.
Question 2 Multiple Choice
Why can Tonelli's theorem be applied to a non-negative measurable function even when its integral equals +∞, whereas Fubini's theorem cannot?
ATonelli uses a different definition of the integral that avoids infinity entirely
BNon-negative functions have integrals valued in [0,∞], so iterated integrals are always well-defined — there is no risk of the indeterminate form ∞ − ∞
CTonelli's theorem only applies when the integral is finite; the statement about +∞ is a misstatement
DFubini can also handle +∞ integrals, so there is no real difference
The key is the sign constraint. For f ≥ 0, every integral is a non-negative extended real number in [0,∞]. Adding or comparing such values never produces the indeterminate form ∞ − ∞, so iterated integrals are always well-defined — even when they diverge. Functions of indefinite sign can produce ∞ − ∞ when integrated, which is why Fubini requires absolute integrability (∫|f| < ∞) as a precondition.
Question 3 True / False
Tonelli's theorem applies to any non-negative measurable function on a product measure space, even if its double integral is +∞.
TTrue
FFalse
Answer: True
This is exactly Tonelli's scope. For f ≥ 0, the iterated integrals and the double integral are all equal in [0,∞] — finite or infinite — with no integrability precondition. This is what makes Tonelli useful: it applies unconditionally to non-negative functions, allowing you to compute or check integrability by iteration.
Question 4 True / False
To check whether a function f (of indefinite sign) is integrable over a product space, you should apply Fubini's theorem to f directly.
TTrue
FFalse
Answer: False
Fubini's theorem requires you to *already know* f is integrable — it cannot be used to establish integrability. The correct approach is to apply Tonelli to |f| (which is non-negative and so Tonelli applies unconditionally). If the iterated integral of |f| is finite, then f is integrable, and you can then invoke Fubini. Attempting Fubini first without this check can produce invalid results when f is not actually integrable.
Question 5 Short Answer
Explain the 'chicken-and-egg' problem Tonelli solves, and describe the standard two-theorem strategy for computing double integrals of functions with indefinite sign.
Think about your answer, then reveal below.
Model answer: Fubini's theorem lets you switch integration order, but only if you already know the function is integrable. Yet computing the double integral to check integrability seems to require switching the integration order — a circularity. Tonelli breaks this by handling non-negative functions with no precondition: apply Tonelli to |f| to compute the iterated integral; if it is finite, |f| is integrable over the product space. This licenses applying Fubini to f itself for the actual computation.
The two-theorem strategy — Tonelli on |f| to establish integrability, Fubini on f to compute — is one of the most repeated proof patterns in measure theory. Without Tonelli, there would be no principled way to verify the precondition that Fubini requires, making double integrals of indefinite-sign functions difficult to handle rigorously.