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The Fubini-Tonelli Theorems · Tutorial 412 of 798

Fubini's Theorem: Complete Proof

See how absolute integrability makes almost every section integrable and allows both iterated integrals to equal the product-space integral.

Advanced 10 min read

What You'll Learn

  • State Fubini’s theorem for real-valued integrable functions on sigma-finite product spaces
  • Derive section integrability from Tonelli’s theorem applied to the absolute value
  • Prove both orders of iterated integration equal the product-space integral
  • Distinguish the almost-everywhere conclusion for sections from the everywhere hypothesis on the function
  • Compute signed, indicator, and counting-measure examples in both orders
  • Identify and correct a common error in integrating over rectangular sections

From Tonelli to Fubini

Tonelli’s theorem permits the order of integration to be exchanged for nonnegative measurable functions, even when the integral is infinite. For a signed function, positive and negative contributions can cancel, so applying that theorem directly to the function is not justified. Fubini’s theorem resolves this issue under absolute integrability: Tonelli applies separately to the positive and negative parts, and the resulting section integrals are finite for almost every value of the outer variable.

Throughout, let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, and give \(X\times Y\) the product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\) and product measure \(\mu\times\nu\). Neither measure space is assumed complete. The function \(f:X\times Y\to\mathbb{R}\) is real-valued and measurable, and all integrals in this tutorial are Lebesgue integrals. For fixed \(x\in X\), write \(f_x(y)=f(x,y)\); for fixed \(y\in Y\), write \(f^y(x)=f(x,y)\). The section measurability result for product-measurable functions ensures these are measurable functions on their respective spaces.

Definition: The real-valued measurable function \(f\) is integrable on \(X\times Y\) if \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\). Equivalently, its positive and negative parts \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\) both have finite integrals, since \(|f|=f^++f^-\). In that case \(\int f\,d(\mu\times\nu)=\int f^+\,d(\mu\times\nu)-\int f^-\,d(\mu\times\nu)\).

Fubini’s Theorem and Its Proof

Theorem (Fubini): Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, and let \(f:X\times Y\to\mathbb{R}\) be \(\mathcal{M}\otimes\mathcal{N}\)-measurable with \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\). Then \(f_x\) is integrable on \(Y\) for \(\mu\)-almost every \(x\), and \(f^y\) is integrable on \(X\) for \(\nu\)-almost every \(y\). The section integrals, defined arbitrarily as zero on their exceptional null sets, are integrable measurable functions, and $$ \int_{X\times Y} f\,d(\mu\times\nu) =\int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x) =\int_Y\left(\int_X f(x,y)\,d\mu(x)\right)d\nu(y). $$ The section integrals are finite almost everywhere; their values on exceptional null sets do not affect the outer integrals.

Proof. We first prove the assertion for sections indexed by \(x\). Since \(f\) is real-valued and measurable, \(f^+\), \(f^-\), and \(|f|\) are nonnegative measurable functions on the product space. Apply Tonelli’s theorem to \(|f|\), which requires nonnegativity and product measurability and is valid on the given sigma-finite spaces. It gives a measurable extended-real-valued function \(H(x)=\int_Y |f_x(y)|\,d\nu(y)\) and the identity $$ \int_X H\,d\mu=\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty. $$ The set \(A=\{x:H(x)=\infty\}\) is measurable. It is null: for every positive integer \(k\), \(H\ge k\chi_A\), so monotonicity of the integral gives \(k\mu(A)\le\int H<\infty\). If \(\mu(A)>0\), these inequalities fail for sufficiently large \(k\); hence \(\mu(A)=0\). For every \(x\notin A\), the definition of \(H\) says precisely that \(f_x\) is integrable on \(Y\).

Apply Tonelli’s theorem separately to \(f^+\) and \(f^-\). The functions $$ P(x)=\int_Y f^+_x\,d\nu,\qquad N(x)=\int_Y f^-_x\,d\nu $$ are measurable and nonnegative, and \(\int_XP\,d\mu=\int_{X\times Y}f^+\,d(\mu\times\nu)\) and \(\int_XN\,d\mu=\int_{X\times Y}f^-\,d(\mu\times\nu)\). Both integrals are finite because \(f^+\le |f|\) and \(f^-\le |f|\). Define \(P_0=P\chi_{A^c}\), \(N_0=N\chi_{A^c}\), and \(g=P_0-N_0\). On \(A^c\), both \(P\) and \(N\) are finite, and the definition of the signed Lebesgue integral gives \(g(x)=\int_Y f_x\,d\nu\). On \(A\), we have assigned the value zero. The functions \(P_0,N_0\) are integrable, and \(g\) is measurable and integrable because \(|g|\le P_0+N_0\). Since \(A\) is null, deleting it does not change the integrals of the nonnegative integrable functions \(P,N\). Linearity of the integral for integrable real-valued functions and Tonelli’s identities therefore give $$ \int_Xg\,d\mu =\int_XP\,d\mu-\int_XN\,d\mu =\int_{X\times Y}f^+\,d(\mu\times\nu)-\int_{X\times Y}f^-\,d(\mu\times\nu) =\int_{X\times Y}f\,d(\mu\times\nu). $$

For the other order, Tonelli applied to \(|f|\) gives the measurable function \(K(y)=\int_X|f^y(x)|\,d\mu(x)\) with \(\int_YK\,d\nu=\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\). The same level-set argument shows \(\{y:K(y)=\infty\}\) is \(\nu\)-null. Tonelli applied to \(f^+\) and \(f^-\) then gives measurable section integrals in \(y\), each integrable; subtracting them off that null set gives an integrable section-integral function whose integral is \(\int f^+-\int f^-=\int f\,d(\mu\times\nu)\). This proves the second equality as well. \(\square\)

The theorem’s hypotheses have distinct roles. Product measurability provides measurable sections and permits Tonelli’s theorem. Sigma-finiteness is a hypothesis for the Tonelli theorem used here. Absolute integrability makes the positive and negative product integrals finite, and it is what forces section integrability almost everywhere. The theorem does not claim that every section is integrable, nor that the section integrals are finite at every point.

Worked Example: A Signed Function on a Rectangle

On \(X=[0,1]\) and \(Y=[0,2]\) with Lebesgue measure, let \(f(x,y)=x-y\). This measurable function is bounded in absolute value by \(2\) on a rectangle of measure \(2\), so \(\int|f|\,d(m\times m)\le4<\infty\). Fubini applies.

Integrate first in \(y\). For each \(x\in[0,1]\),

$$ \int_0^2(x-y)\,dy =\left[xy-\frac{y^2}{2}\right]_0^2 =2x-2. $$

Thus the outer integral is \(\int_0^1(2x-2)\,dx=[x^2-2x]_0^1=1-2=-1\).

In the other order, for each \(y\in[0,2]\),

$$ \int_0^1(x-y)\,dx =\left[\frac{x^2}{2}-yx\right]_0^1 =\frac12-y. $$

The outer integral is \(\int_0^2(\frac12-y)\,dy=[\frac y2-\frac{y^2}{2}]_0^2=1-2=-1\). Check: integrating the two terms over the rectangle separately gives \(\int_0^1\int_0^2x\,dy\,dx=2\int_0^1x\,dx=1\) and \(\int_0^1\int_0^2y\,dy\,dx=\int_0^2y\,dy=2\). Their difference is \(1-2=-1\), agreeing with both iterated integrals.

Worked Example: An Indicator of a Triangle

Let \(X=Y=[0,1]\) with Lebesgue measure and define \(f(x,y)=\chi_E(x,y)\), where \(E=\{(x,y):0<x<y<1\}\). This is a measurable set, and \(0\le f\le1\) on a space of measure \(1\), so \(f\) is integrable. The strict inequalities mean that points with \(x=0\), \(y=1\), or \(x=y\) are not in \(E\).

For fixed \(x\in(0,1)\), the section \(E_x\) is the interval \((x,1)\), of measure \(1-x\). For \(x=0\) or \(x=1\), the section is empty. Consequently the first iterated integral is

$$ \int_0^1\left(\int_0^1\chi_E(x,y)\,dy\right)dx =\int_0^1(1-x)\,dx =\left[x-\frac{x^2}{2}\right]_0^1 =\frac12. $$

For fixed \(y\in(0,1)\), the section in \(x\) is \((0,y)\), of measure \(y\); at \(y=0\) and \(y=1\), it is empty. Hence the other order gives

$$ \int_0^1\left(\int_0^1\chi_E(x,y)\,dx\right)dy =\int_0^1y\,dy =\left[\frac{y^2}{2}\right]_0^1 =\frac12. $$

Check: the product-space integral is the area of the triangle. Its vertical sections have lengths \(1-x\), so its area is \(\int_0^1(1-x)\,dx=\frac12\), as computed from either order.

Worked Example: A Finite-Support Function with Counting Measure

Take \(X=Y=\mathbb{N}=\{1,2,\ldots\}\) with counting measure, which is sigma-finite because \(\mathbb{N}\) is a countable union of finite-measure singletons. Define \(f\) to be zero outside \(\{1,2\}\times\{1,2\}\), and on that set let its entries be \(f(1,1)=2,\ f(1,2)=-1,\ f(2,1)=-3,\ f(2,2)=4\). Every function on this countable measurable space is measurable. Its absolute integral is the finite sum \(2+1+3+4=10\), so Fubini applies.

Summing first over \(y\), the section sums are \(2+(-1)=1\) for \(x=1\) and \((-3)+4=1\) for \(x=2\); all later section sums are zero. The outer sum is therefore \(1+1=2\). Summing first over \(x\), the section sums are \(2+(-3)=-1\) for \(y=1\) and \((-1)+4=3\) for \(y=2\); all later ones are zero. The outer sum is \((-1)+3=2\). Check: the product-space integral is the sum of all four entries, \(2-1-3+4=2\), in agreement with both orders; the absolute sum \(2+1+3+4=10\) confirms integrability.

A Common Rectangular-Integral Error

When integrating \(x-y\) over \([0,1]\times[0,2]\), a common incorrect calculation is \(\int_0^1x\,dx-\int_0^2y\,dy=\frac12-2=-\frac32\). It treats the two terms as though they were integrated over one-dimensional intervals only. In the product integral, \(x\) is repeated over the full \(y\)-interval, so its contribution is \(\int_0^1\int_0^2x\,dy\,dx=2\int_0^1x\,dx=1\). The term \(y\) is repeated over the full \(x\)-interval of length \(1\), giving \(\int_0^1\int_0^2y\,dy\,dx=1\cdot2=2\). The correct difference is \(1-2=-1\). Fubini permits either order of integration, but it does not remove the cross-sectional factors.

The structural point is more important than the arithmetic: signed integration is justified here because \(\int|f|\) is finite. Tonelli controls \(f^+\) and \(f^-\) separately, with finite integrals; their section integrals are therefore finite outside a null set, and subtraction is legitimate there. Without absolute integrability, that reasoning does not establish Fubini’s conclusion.

Check Your Understanding

In each problem, identify the measure spaces, verify integrability where required, and distinguish conclusions that hold everywhere from those that hold almost everywhere.

  1. For \(f(x,y)=2x+y\) on \([0,1]\times[0,1]\) with Lebesgue measure, compute both iterated integrals and the product integral.
  2. For \(E=\{(x,y):0<y<x<1\}\), compute the two section integrals of \(\chi_E\) and their outer integrals.
  3. In the proof, why is \(\{x:\int_Y|f_x|\,d\nu=\infty\}\) a null set when \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\)?
  4. Explain why the proof applies Tonelli to \(f^+\) and \(f^-\) separately rather than directly to \(f\).
  5. For a real-valued product-measurable \(f\) with finite \(\int|f|\), what can Fubini conclude about the sections where the inner integral is finite?

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