The signed surface x−y on the unit square has positive and negative regions. Their totals cancel, but cancellation is safe only because the absolute volume is finite. Compute the integral of abs (x−y) by splitting at the diagonal before computing the signed integral. A zero signed answer by itself does not certify that the positive and negative totals are finite.
Predict
Predict whether the existence of both iterated integrals forces them to agree. It does not. The surface (x²−y²)/(x²+y²)² near the origin will give two different finite iterated answers. The missing hypothesis is absolute integrability on the product space, not continuity away from one troublesome point.
Experiment
For that fixed counterexample, inspect the two orders.
Inner variable
Antiderivative
Inner integral
Outer integral
y
y/(x²+y²)
1/(1+x²)
π/4
x
−x/(x²+y²)
−1/(1+y²)
−π/4
Differentiate the antiderivatives yourself. Then check the region 0< y< x/2: the positive integrand is at least a positive constant times 1/x². Its y-width is x/2, so its area integral contains a divergent multiple of integral dx/x near zero.
Understand
Fubini’s theorem assumes sigma-finite factor measures and a product-measurable f with integral abs (f)< infinity. Tonelli applied to abs (f) gives a measurable section-size function x↦integral abs (f(x, y))dν(y), whose integral is finite. It is therefore finite for almost every x. On those sections the signed integral exists. Assign zero on the exceptional null set when forming an everywhere-defined representative.
Apply Tonelli separately to f⁺ and f⁻. Their product integrals are both finite because each is bounded by abs (f). Their section integrals are measurable, and their finite difference on the good set represents the signed section integral. Subtracting the two finite equalities proves the first iterated formula. Reverse the roles of the factors to prove the second. This argument explains both the absolute-integrability assumption and the almost-everywhere qualification.
In the counterexample, the cone estimate shows absolute integrability fails. More explicitly, for 0< y< x/2 the numerator is at least 3x²/4 and the squared denominator is at most 25x⁴/16, giving f≥12/(25x²). Integrating over a section of width x/2 yields at least 6/(25x). The divergence is real mass near the origin, even though each fixed positive coordinate admits a finite inner integral.
Completing product measures introduces an additional technical issue: arbitrary sections of a completed-measurable set need not all be measurable. Replace a completed-measurable function by a product-measurable version agreeing almost everywhere, and formulate section conclusions almost everywhere. This does not license a claim about every exceptional section.
Apply
For a series of signed functions, the condition Σ integral abs (g_k)< infinity is a useful certificate. Tonelli makes the absolute sum integrable; it dominates all partial sums. Dominated convergence then exchanges summation and integration. Compare this with merely knowing that the numerical series Σ integral g_k converges, which does not control cancellation inside the functions.
Continue
When changing integration order, write the certificate before the calculation: nonnegative for Tonelli, or finite absolute product integral for Fubini. If neither is available, calculate carefully without claiming an unrestricted interchange theorem. In applications, an exceptional null set is harmless for integrals but may still matter if a specific deterministic section is requested.
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Integrate x−2y over [0,2]×[0,1].
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Integrate abs (x−y) over the unit square.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Deduce Fubini from Tonelli when integral abs (f) is finite.
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For f=(x²−y²)/(x²+y²)² on (0,1)², compute both iterated integrals and explain the failure.
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Can an L1 product function have a nonintegrable section? Give an example.
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State and prove a sufficient condition for interchanging an integral and a signed series.
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Reflect and continue
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