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The Fubini-Tonelli Theorems · Tutorial 403 of 798

Passing to General Nonnegative Functions

Learn how increasing simple approximations transfer the iterated-integral identity to general nonnegative measurable functions, including functions whose integrals may be infinite.

Advanced 11 min read

What You'll Learn

  • Construct an increasing sequence of finite-valued simple functions converging to any nonnegative extended-valued measurable function.
  • State the assumptions and conclusion of the product-integral identity for general nonnegative functions.
  • Apply the Monotone Convergence Theorem to the product integral and to section integrals.
  • Compute product and iterated integrals in Lebesgue and counting-measure examples.
  • Identify why a pointwise-convergent sequence that is not increasing cannot be used in this argument.

From Simple Functions to General Functions

The product-integral identity is first established for nonnegative simple functions: integrating such a function on a product space gives the same value as integrating its section integrals. To pass from simple functions to a general nonnegative measurable function, we approximate it from below by an increasing sequence of simple functions. The Monotone Convergence Theorem then transfers the identity to the limit. No finiteness of the resulting integrals is required; either side may be \(+\infty\).

Throughout, let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, not necessarily finite or complete, and equip \(X\times Y\) with \(\mathcal{M}\otimes\mathcal{N}\) and the product measure \(\mu\times\nu\). Functions in the main result are measurable and take values in \([0,\infty]\). For \(f:X\times Y\to[0,\infty]\), write \(f_x(y)=f(x,y)\). Sections of a product-measurable function are measurable, and the earlier result on nonnegative section integrals ensures that \(x\mapsto\int_Y f_x\,d\nu\) is measurable.

Simple-function identity: For a nonnegative simple \(\psi\) measurable on \((X\times Y,\mathcal{M}\otimes\mathcal{N})\), the section-integral function is measurable and \(\displaystyle \int_{X\times Y}\psi\,d(\mu\times\nu)=\int_X\left(\int_Y\psi(x,y)\,d\nu(y)\right)d\mu(x)\). The identity also holds with the order of \(X\) and \(Y\) reversed. This is the identity established for simple functions; here we extend it.

Increasing Simple Approximation

The approximation must work even when \(f\) is unbounded or takes the value \(+\infty\). For each integer \(n\ge1\), round \(f\) down to the grid of mesh \(2^{-n}\), while truncating at height \(n\). Explicitly, define \(q_n:[0,\infty]\to[0,\infty)\) and \(\varphi_n:X\times Y\to[0,\infty)\) by

$$ q_n(t)=2^{-n}\left\lfloor 2^n\min(t,n)\right\rfloor, \qquad \varphi_n(x,y)=q_n(f(x,y)). $$

Here \(\min(+\infty,n)=n\). Each \(\varphi_n\) is a finite-valued simple function: its possible values are among \(0,2^{-n},2\cdot2^{-n},\ldots,n\). It is measurable because it is obtained by applying a Borel measurable function \(q_n\) to the measurable function \(f\).

Approximation lemma: For every measurable \(f:X\times Y\to[0,\infty]\), the functions \(\varphi_n=q_n\circ f\) are nonnegative simple functions and satisfy \(\varphi_n\uparrow f\) pointwise everywhere.

Proof. First, \(0\le q_n(t)\le t\) for every \(t\in[0,\infty]\), so \(0\le\varphi_n\le f\). To check that \(q_n(t)\le q_{n+1}(t)\), consider two cases. If \(t\le n\), then \(\min(t,n)=\min(t,n+1)=t\). Since every grid point \(k2^{-n}\) is also a grid point at level \(n+1\), rounding \(t\) down to the finer grid cannot decrease its rounded value. If \(t>n\), then \(q_n(t)=n\). When \(n<t<n+1\), the finer-grid rounded value is at least \(n\); when \(t\ge n+1\), \(q_{n+1}(t)=n+1\). Thus \(q_n(t)\le q_{n+1}(t)\) in all cases.

For finite \(t\), once \(n>t\), the truncation no longer changes \(t\), and \(0\le t-q_n(t)<2^{-n}\). Hence \(q_n(t)\to t\). For \(t=+\infty\), \(q_n(t)=n\), so \(q_n(t)\to+\infty\). Applying these pointwise statements to \(t=f(x,y)\) proves \(\varphi_n(x,y)\uparrow f(x,y)\) for every \((x,y)\). \(\square\)

The Identity for General Nonnegative Functions

Theorem: Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, with product measure \(\mu\times\nu\) on \(\mathcal{M}\otimes\mathcal{N}\). If \(f:X\times Y\to[0,\infty]\) is measurable, then \(\displaystyle \int_{X\times Y}f\,d(\mu\times\nu)=\int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x)\). The inner integral defines a measurable function of \(x\), and equality is an equality in \([0,\infty]\). The identity also holds with the two factors interchanged.

Proof. Use the approximation lemma to choose nonnegative simple functions \(\varphi_n\uparrow f\) pointwise everywhere. For each \(n\), define \(A_n(x)=\int_Y\varphi_n(x,y)\,d\nu(y)\). The simple-function identity ensures that \(A_n\) is measurable and that \(\int_{X\times Y}\varphi_n\,d(\mu\times\nu)=\int_X A_n\,d\mu\).

Fix \(x\in X\). The sequence \(y\mapsto\varphi_n(x,y)\) is nonnegative, measurable, and increasing to \(y\mapsto f(x,y)\) at every \(y\). The Monotone Convergence Theorem on \((Y,\mathcal{N},\nu)\) therefore gives \(\displaystyle A_n(x)\uparrow A(x):=\int_Y f(x,y)\,d\nu(y)\). This holds for every \(x\), including when the limit is infinite. The function \(A\) is measurable by the result on nonnegative section integrals. Apply the Monotone Convergence Theorem on \((X,\mathcal{M},\mu)\) to \(A_n\uparrow A\), and on \((X\times Y,\mathcal{M}\otimes\mathcal{N},\mu\times\nu)\) to \(\varphi_n\uparrow f\). Taking limits in the simple-function identity yields \[ \int_{X\times Y} f\,d(\mu\times\nu) =\lim_{n\to\infty}\int_{X\times Y}\varphi_n\,d(\mu\times\nu) =\lim_{n\to\infty}\int_X A_n\,d\mu =\int_X A\,d\mu. \] All three limits are justified by Monotone Convergence; no interchange of a limit and an integral is being made without that theorem. The reversed-order identity follows by applying the same argument with the roles of \(X\) and \(Y\) exchanged. \(\square\)

Worked Applications

Worked Example: Integrating x + y on the Unit Square

Let \(X=Y=[0,1]\), with Lebesgue measure on each factor, and let \(f(x,y)=x+y\). This is a finite, nonnegative measurable function. The approximation lemma gives \(\varphi_n=2^{-n}\lfloor2^n\min(x+y,n)\rfloor\), with \(\varphi_n\uparrow f\); thus the theorem applies.

Compute the section integral first. For fixed \(x\in[0,1]\),

$$ \int_0^1(x+y)\,dy =\left[xy+\frac{y^2}{2}\right]_{y=0}^{y=1} =x+\frac12. $$

Integrating this function of \(x\) gives \(\displaystyle\int_0^1(x+\tfrac12)\,dx =\left[\tfrac{x^2}{2}+\tfrac{x}{2}\right]_0^1 =\tfrac12+\tfrac12=1\). The product integral is therefore \(1\). In the reverse order, the inner integral is \(y+\tfrac12\), and its integral from \(0\) to \(1\) is also \(\tfrac12+\tfrac12=1\). Check: the symmetry of \(x+y\) under exchanging \(x\) and \(y\) agrees with the equal values from the two orders.

Worked Example: An Infinite Product Region

On \([0,\infty)\times[0,\infty)\) with product Lebesgue measure, let \(f(x,y)=e^{-x-y}\). It is nonnegative and measurable, and the approximation lemma applies even though the underlying measure spaces have infinite measure.

For fixed \(x\), \(e^{-x-y}=e^{-x}e^{-y}\), so \(\displaystyle\int_0^\infty e^{-x-y}\,dy =e^{-x}\left[-e^{-y}\right]_{0}^{\infty} =e^{-x}(0-(-1))=e^{-x}\). Consequently, \(\displaystyle\int_0^\infty e^{-x}\,dx =\left[-e^{-x}\right]_{0}^{\infty}=0-(-1)=1\). The theorem gives \(\int f\,d(m\times m)=1\); reversing the roles gives the same value.

Check: on the finite square \([0,R]^2\), direct integration gives \(\displaystyle\int_0^R\int_0^R e^{-x-y}\,dy\,dx =(1-e^{-R})(1-e^{-R})=(1-e^{-R})^2\). As \(R\to\infty\), \(e^{-R}\to0\), so these values tend to \(1\), consistent with the full integral.

Worked Example: A Product of Geometric Terms on \(\mathbb{N}^2\)

Give \(\mathbb{N}=\{1,2,\ldots\}\) counting measure in each coordinate, and set \(f(m,n)=2^{-m}3^{-n}\). Every function on \(\mathbb{N}^2\) is measurable. Counting measure is sigma-finite, since \(\mathbb{N}\) is a countable union of finite sets. The product integral is the sum over all pairs.

For fixed \(m\), summing over \(n\) gives \(\displaystyle\sum_{n=1}^{\infty}2^{-m}3^{-n} =2^{-m}\frac{1/3}{1-1/3} =2^{-m}\cdot\frac12\). Summing over \(m\) then gives \(\displaystyle\sum_{m=1}^{\infty}2^{-m}\cdot\frac12 =\frac12\cdot\frac{1/2}{1-1/2} =\frac12\). In the opposite order, \(\sum_{m=1}^{\infty}2^{-m}=1\), and hence \(\displaystyle\sum_{n=1}^{\infty}3^{-n}\cdot1 =\frac{1/3}{1-1/3}=\frac12\).

Check: the sum over \(1\le m\le M,\ 1\le n\le N\) is \(\displaystyle(1-2^{-M})\frac12(1-3^{-N})\), which tends to \(\frac12\) as both \(M\) and \(N\) tend to infinity. This independently confirms the iterated sums. The spaces have infinite measure, but the integral of this particular function is finite.

Why Monotonicity Matters

The proof depends on increasing approximations: Monotone Convergence applies to both the approximating functions on the product space and their section integrals. Pointwise convergence alone is not enough. For a concrete failure, on \(\mathbb{R}^2\) with product Lebesgue measure define \(g_n(x,y)=n^2\chi_{(0,1/n)\times(0,1/n)}(x,y)\). At every fixed \((x,y)\), \(g_n(x,y)\to0\): if either coordinate is nonpositive, the point is never in the indicated open square; otherwise, eventually \(1/n\le x\) or \(1/n\le y\). But \(\displaystyle\int_{\mathbb{R}^2}g_n\,d(m\times m) =n^2\cdot\frac1n\cdot\frac1n=1\) for every \(n\), whereas the integral of the pointwise limit is \(0\). The sequence is not increasing: at \((0.4,0.2)\), \(g_2=4\), since both coordinates are strictly between \(0\) and \(1/2\), while \(g_3=0\), since \(0.4\) is not strictly less than \(1/3\). Thus the increasing-sequence hypothesis cannot be omitted.

A second common error is to infer that the iterated integral is defined as a measurable function of the outer variable merely because each section is measurable. Measurability of the section \(f_x\) for each fixed \(x\) and measurability of \(x\mapsto\int_Y f_x\,d\nu\) are different assertions. The latter is supplied here by the result on nonnegative section integrals; it is needed before the outer integral is formed. The theorem concerns nonnegative measurable functions and permits infinite values. It does not, by itself, give the corresponding identity for signed or complex-valued functions, where cancellation and integrability conditions must be addressed separately.

Check Your Understanding

For each question, identify the measure spaces and the hypotheses that justify the conclusion.

  1. For \(f(x,y)=\max(x,y)\) on \([0,1]^2\), describe an increasing simple approximation and state why it converges pointwise to \(f\).
  2. In the approximation lemma, what does \(q_n(+\infty)\) equal, and why does this still converge to the required value?
  3. For \(f(x,y)=e^{-2x-y}\) on \([0,\infty)^2\), compute the integral in each order and verify that both are finite.
  4. Explain where the proof uses the measurability of the section-integral function, in addition to measurability of each section.
  5. For the sequence \(g_n\) in the pitfall, verify directly its pointwise limit and its integral, and identify the precise missing hypothesis for Monotone Convergence.