Fubini's Theorem

Statement

Let (X, μ) and (Y, ν) be σ-finite measure spaces and f : X × Y → ℝ measurable on the product. If

∫_{X × Y} |f| d(μ × ν) < ∞

then f is integrable on X × Y, and:

∫_{X × Y} f d(μ × ν) = ∫_X (∫_Y f(x, y) dν(y)) dμ(x)
                    = ∫_Y (∫_X f(x, y) dμ(x)) dν(y)

Each of the two iterated integrals exists for μ-almost every x (and ν-almost every y), is itself integrable, and equals the double integral. Three equalities, one hypothesis.

The geometric picture

Imagine a positive function f(x, y) over a rectangle in the xy-plane. The double integral is the volume of the solid below the graph. You can compute this volume two ways:

• Slice along x. For each fixed x, ∫f(x, y) dy is the area of the cross-section at that x. Sum the cross-section areas over x: ∫(∫f dy) dx.
• Slice along y. For each fixed y, ∫f(x, y) dx is the area of the cross-section at that y. Sum over y: ∫(∫f dx) dy.

Fubini says: same volume, whether you slice with vertical or horizontal planes. The volume is intrinsic to the solid; the slicing direction is a computational choice.

Worked example — a simple double integral

Compute ∫₀¹ ∫₀² (x + y²) dy dx. Fubini lets us choose either order.

Order 1 — integrate y first:
  ∫₀¹ (∫₀² (x + y²) dy) dx
= ∫₀¹ [xy + y³/3]_{y=0}^{y=2} dx
= ∫₀¹ (2x + 8/3) dx
= [x² + 8x/3]_0^1
= 1 + 8/3 = 11/3

Order 2 — integrate x first:
  ∫₀² (∫₀¹ (x + y²) dx) dy
= ∫₀² [x²/2 + xy²]_{x=0}^{x=1} dy
= ∫₀² (1/2 + y²) dy
= [y/2 + y³/3]_0^2
= 1 + 8/3 = 11/3   ✓

Same answer. The function is bounded and the domain is bounded, so ∫|f| < ∞ trivially — Fubini applies.

Counterexample — when Fubini fails

The classical counterexample on [0, 1] × [0, 1]:

f(x, y) = (x² − y²) / (x² + y²)²    for (x, y) ≠ (0, 0)

Iterated one way:
  ∫₀¹ (∫₀¹ f dy) dx = ∫₀¹ (1/(1 + x²)) ... = π/4

Iterated the other way:
  ∫₀¹ (∫₀¹ f dx) dy = -π/4

Two different answers — π/2 apart!

What went wrong? Compute the absolute integral:

∫∫ |f| dA = ∞   (divergent near origin)

Fubini's hypothesis ∫|f| < ∞ fails. The positive and negative parts both have infinite mass, and the order of summation rearranges the cancellation. This is exactly analogous to conditionally convergent series — Σ (−1)^k/k can be rearranged to converge to any value. Absolute integrability is the protection against this.

Proof idea — extension from indicators

The standard proof builds up in stages:

1. Indicator of rectangle. For f = 1_{A × B} with A ⊂ X, B ⊂ Y measurable, (μ × ν)(A × B) = μ(A) ν(B). Iterated integrals each give μ(A) ν(B). Trivial agreement.
2. Indicator of general measurable set. Approximate by countable unions of rectangles. The σ-finite hypothesis ensures the product σ-algebra contains them.
3. Simple non-negative functions. Linear combinations of indicators. Linearity of integration.
4. Non-negative measurable f (Tonelli). Monotone limits of simple functions; pass through monotone convergence. Both iterated integrals well-defined in [0, ∞] and equal.
5. Signed f (Fubini). Write f = f⁺ − f⁻ where f⁺ = max(f, 0), f⁻ = max(−f, 0). The hypothesis ∫|f| < ∞ means ∫f⁺ < ∞ and ∫f⁻ < ∞ separately. Apply Tonelli to each, subtract.

This "machine" — prove for indicators, extend by linearity to simple, by monotone convergence to non-negative, by decomposition to signed — is the workhorse of measure-theoretic proofs.

Numerical verification

// Numerical Fubini check on f(x, y) = x · sin(y) on [0, 1] × [0, π]
// Analytical answer: ∫₀^π sin(y) dy = 2, ∫₀^1 x dx = 1/2, product = 1
// Both iterated orders should give 1.

function trapezoidalDouble(f, x0, x1, y0, y1, nx, ny) {
  const dx = (x1 - x0) / nx, dy = (y1 - y0) / ny;
  let sum = 0;
  for (let i = 0; i <= nx; i++) {
    const x = x0 + i * dx;
    const wx = (i === 0 || i === nx) ? 0.5 : 1;
    for (let j = 0; j <= ny; j++) {
      const y = y0 + j * dy;
      const wy = (j === 0 || j === ny) ? 0.5 : 1;
      sum += wx * wy * f(x, y);
    }
  }
  return sum * dx * dy;
}

const f = (x, y) => x * Math.sin(y);
trapezoidalDouble(f, 0, 1, 0, Math.PI, 200, 200);  // ~1.0000 — matches analytical

// Counterexample where Fubini fails:
const fBad = (x, y) => {
  if (x === 0 && y === 0) return 0;
  return (x * x - y * y) / Math.pow(x * x + y * y, 2);
};

// Iterated y-first then x-first
function iteratedYX(f, x0, x1, y0, y1, nx, ny) {
  const dx = (x1 - x0) / nx;
  let sum = 0;
  for (let i = 0; i <= nx; i++) {
    const x = x0 + i * dx;
    const w = (i === 0 || i === nx) ? 0.5 : 1;
    let innerY = 0;
    const dy = (y1 - y0) / ny;
    for (let j = 0; j <= ny; j++) {
      const y = y0 + j * dy;
      const wy = (j === 0 || j === ny) ? 0.5 : 1;
      innerY += wy * f(x, y);
    }
    sum += w * innerY * dy;
  }
  return sum * dx;
}

iteratedYX(fBad, 1e-6, 1, 1e-6, 1, 400, 400);     // ≈  0.785 ≈ π/4
iteratedYX((x, y) => fBad(y, x), 1e-6, 1, 1e-6, 1, 400, 400);  // ≈ -0.785 ≈ -π/4

Variants and comparison

Theorem	Setting	Hypothesis	Conclusion	Year
Fubini	σ-finite product measures, signed f	∫|f| < ∞	Double = iterated (both orders)	1907
Tonelli	σ-finite product measures, f ≥ 0	None (just measurable)	Double = iterated in [0, ∞]	1909
Fubini-Tonelli	Combined workflow	Use Tonelli on |f|; if finite, apply Fubini	Same as Fubini	—
Classical Fubini (Riemann)	Bounded rectangle, continuous f	Continuity	Same equality	Pre-1907
Fubini for distributions	Tempered distributions, test functions	One factor is a distribution	Pairing factors	20th c.
Fubini on σ-finite vs general measure	Non-σ-finite needs care	Without σ-finiteness, can fail	Counterexamples exist	—
Sard's theorem (related slice idea)	Smooth maps, slicing	Smoothness	Critical values measure zero	1942

Common pitfalls

• Forgetting the ∫|f| < ∞ hypothesis. The signed Fubini absolutely requires it. Without absolute integrability, the iterated integrals can disagree (counterexample above). Always check ∫|f| first — that's exactly what Tonelli is for.
• Confusing σ-finite with finite. ℝⁿ with Lebesgue measure has infinite measure but is σ-finite (cover by balls). Counting measure on an uncountable set is not σ-finite — Fubini can fail there.
• Treating "Fubini fails ⇒ the integrand isn't integrable." If both iterated integrals exist and agree, the conclusion holds — but Fubini's clean statement requires the absolute hypothesis. In practice, when iterated integrals disagree, you've confirmed ∫|f| = ∞.
• Swapping limits and integrals without justification. Limits inside integrals also need a theorem (Dominated Convergence, Monotone Convergence, Fatou); Fubini is for swapping integrals with each other.
• Applying it on the wrong product space. The measure on X × Y must be the product μ × ν, not some other measure on the rectangle. For non-product measures (Borel measures with non-product structure), Fubini doesn't apply directly.
• Treating Riemann iterated and Lebesgue double as automatically equal. Continuous f on a bounded rectangle — yes, trivially. For unbounded domains or singular f, you need the Lebesgue version explicitly.

Applications

• Probability — independence. If X, Y are independent random variables, their joint distribution is the product measure. E[g(X)h(Y)] = E[g(X)] E[h(Y)] is Fubini on the product. Convolutions of independent X + Y come from Fubini on the joint density.
• Fourier analysis. Plancherel's theorem ‖f‖₂ = ‖f̂‖₂, Parseval, Fourier inversion all require swapping integrals; Fubini-Tonelli is the engine.
• Convolutions. The integral defining (f * g)(x) = ∫f(y)g(x − y) dy and identities like ‖f * g‖₁ ≤ ‖f‖₁ ‖g‖₁ all use Fubini on a double integral.
• Integral transforms. Laplace, Mellin, Hankel transforms all involve inversion formulas that exchange orders of integration — Fubini justifies them when absolute integrability holds.
• PDE — Green's functions. Solving Δu = f via u(x) = ∫G(x, y)f(y) dy and integrating over a domain requires swapping the order of integration when integrating against a test function.
• Statistics — joint distributions and marginals. Marginal density f_X(x) = ∫f_{X,Y}(x, y) dy is a partial Fubini integration; computing E[X] from joint distribution uses Fubini to integrate y out before x.
• Physics — multiple integrals. Volume integrals in cylindrical / spherical coordinates, computations of moments of inertia, mass integrals over 3D solids — all use Fubini to convert triple integrals into iterated single integrals.

History and significance

Guido Fubini proved the signed-function version in 1907 as part of his work on integration on product spaces. Leonida Tonelli followed in 1909 with the non-negative version that doesn't need absolute integrability. The two are often quoted together as Fubini-Tonelli because the working pattern is unified: use Tonelli on |f|, then Fubini on f.

The theorem is foundational because before Fubini, switching the order of integration was a delicate calculation, justified case by case. After Fubini, it became a one-line check: does ∫|f| converge? The result enabled clean development of probability theory (Kolmogorov 1933), Fourier analysis on L^p spaces, and the modern theory of PDE.