Building the product measure
Start with two σ-finite measure spaces (X, A, μ) and (Y, B, ν). The product σ-algebra A ⊗ B on X × Y is generated by the measurable rectangles A × B with A ∈ A, B ∈ B. The product measure μ × ν is the unique measure on A ⊗ B that assigns to each rectangle its expected area, (μ × ν)(A × B) = μ(A) · ν(B). σ-finiteness is exactly what makes this measure unique.
Tonelli then Fubini
The combined Fubini–Tonelli theorem comes in two halves. Tonelli: if f ≥ 0 is A ⊗ B-measurable, then the iterated integrals and the double integral all agree — no integrability assumption needed, the value just may be +∞. Fubini (the Fubini theorem proper): if f is integrable for μ × ν, the same three integrals agree and are finite. The standard recipe is to run Tonelli on |f| first to license Fubini on f.
Statement (σ-finite μ, ν; f measurable on X×Y).
Tonelli (f ≥ 0):
∫_{X×Y} f d(μ×ν)
= ∫_X ( ∫_Y f(x,y) dν(y) ) dμ(x)
= ∫_Y ( ∫_X f(x,y) dμ(x) ) dν(y),
where every expression lies in [0, +∞].
Fubini (f integrable, i.e. ∫|f| d(μ×ν) < ∞):
the same three quantities are equal AND finite;
moreover y ↦ f(x,y) is in L¹(ν) for μ-a.e. x,
and x ↦ ∫ f(x,y) dν(y) is in L¹(μ).
The usual workflow:
1. Apply Tonelli to |f| ≥ 0; check ∫∫ |f| dν dμ < ∞.
2. That finiteness is exactly the Fubini hypothesis.
3. Now swap the order of integration for f freely.Why the hypotheses are not negotiable
It is tempting to swap integration orders for any function. Don't. Here is the classic counterexample: a function that is not integrable, whose two iterated integrals exist but disagree. The lesson is that the integrability check via Tonelli is load-bearing, not bureaucratic.
Counterexample on [0,1] × [0,1] with Lebesgue measure.
Let f(x,y) = (x² − y²) / (x² + y²)² (and f(0,0) := 0).
Inner integral over y, then x:
∫₀¹ ( ∫₀¹ f dy ) dx = ∫₀¹ [ 1/(1+x²) ] dx = π/4.
Inner integral over x, then y (by the x↔y antisymmetry):
∫₀¹ ( ∫₀¹ f dx ) dy = − π/4.
The two iterated integrals are π/4 and −π/4: NOT equal.
No contradiction with Fubini, because
∫∫ |f| dx dy = ∞,
so f is NOT integrable and Fubini does not apply.
Tonelli on |f| would have caught this immediately. ∎