From Finite Pieces to Sigma-Finite Spaces
On a product space, a measurable function can be integrated by first fixing one coordinate and integrating in the other. Tonelli’s theorem makes this precise for nonnegative functions: the section integrals are measurable, and both iterated integrals agree with the product integral. No finiteness of the resulting integrals is required; all three values may be \(+\infty\).
Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces. Here sigma-finite means that \(X=\bigcup_{n=1}^{\infty}X_n\) and \(Y=\bigcup_{n=1}^{\infty}Y_n\) for measurable sets with \(\mu(X_n)<\infty\) and \(\nu(Y_n)<\infty\). The product measure on \((X\times Y,\mathcal{M}\otimes\mathcal{N})\) is denoted by \(\mu\times\nu\). Functions in this tutorial are extended nonnegative real-valued, with values in \([0,\infty]\), and measurable with respect to the indicated sigma-algebra.
For \(E\in\mathcal{M}\otimes\mathcal{N}\), its sections are \(E_x=\{y\in Y:(x,y)\in E\}\) and \(E^y=\{x\in X:(x,y)\in E\}\). For a function \(f\), write \(f_x(y)=f(x,y)\) and \(f^y(x)=f(x,y)\). Sections of product-measurable sets are measurable: for fixed \(x\), the class of \(E\in\mathcal{M}\otimes\mathcal{N}\) for which \(E_x\in\mathcal{N}\) is a sigma-algebra containing every measurable rectangle \(A\times B\), and those rectangles generate \(\mathcal{M}\otimes\mathcal{N}\). The same argument applies to \(E^y\). Consequently, sections of product-measurable functions are measurable.
The Finite-Measure Identity
We first prove the result when both measures are finite. This is the step that uses the product sigma-algebra and the uniqueness of product measure. We use the pi-lambda theorem: if a lambda-system contains a pi-system, then it contains the sigma-algebra generated by that pi-system. This standard set-class theorem is established in earlier measure-theory material.
Proof. Let \(\mathcal{D}\) be the collection of product-measurable sets \(E\) for which \(\int_X\nu(E_x)\,d\mu(x)=(\mu\times\nu)(E)\). Sections are measurable by the section argument above, so the displayed integral is defined. The whole space belongs to \(\mathcal{D}\), since both sides equal \(\mu(X)\nu(Y)\). If \(E\in\mathcal{D}\), then \(\nu((E^c)_x)=\nu(Y)-\nu(E_x)\); finiteness of \(\nu(Y)\) gives measurability, and finiteness of the measures gives \(\int_X\nu((E^c)_x)\,d\mu=\mu(X)\nu(Y)-(\mu\times\nu)(E)=(\mu\times\nu)(E^c)\). Thus \(\mathcal{D}\) is closed under complements. If \(E_1,E_2,\ldots\in\mathcal{D}\) are pairwise disjoint, their sections are pairwise disjoint, so \(\nu((\bigcup_jE_j)_x)=\sum_j\nu((E_j)_x)\). The Monotone Convergence Theorem makes the section-measure function measurable and gives \(\int_X\sum_j\nu((E_j)_x)\,d\mu=\sum_j(\mu\times\nu)(E_j)=(\mu\times\nu)(\bigcup_jE_j)\). Hence \(\mathcal{D}\) is a lambda-system. Measurable rectangles form a pi-system; for \(A\times B\), the section-measure function is \(\nu(B)\chi_A\), and its integral is \(\mu(A)\nu(B)=(\mu\times\nu)(A\times B)\). The pi-lambda theorem now gives \(\mathcal{D}=\mathcal{M}\otimes\mathcal{N}\).
For a nonnegative simple function \(\varphi=\sum_{j=1}^k a_j\chi_{E_j}\), with disjoint product-measurable \(E_j\) and \(a_j\geq0\), the section integral is \(\sum_{j=1}^k a_j\nu((E_j)_x)\). It is measurable by the set result, and linearity for nonnegative simple functions gives \(\int_X\int_Y\varphi_x\,d\nu\,d\mu=\int\varphi\,d(\mu\times\nu)\). For general measurable \(f\geq0\), choose nonnegative simple functions \(\varphi_n\uparrow f\), as supplied by the standard simple-function approximation for the Lebesgue integral. For each \(x\), the Monotone Convergence Theorem on \(Y\) gives \(\int_Y\varphi_n(x,y)\,d\nu(y)\uparrow\int_Yf(x,y)\,d\nu(y)\). Thus the section integral is a limit of measurable functions and is measurable. Applying Monotone Convergence on \(X\) and on the product space to the simple-function identities proves \(\int_X\int_Y f_x\,d\nu\,d\mu=\int f\,d(\mu\times\nu)\). Reversing the roles of the two spaces proves the other order. \(\square\)
Proof by Sigma-Finite Exhaustion
Choose increasing measurable exhaustions \(X_n\uparrow X\) and \(Y_n\uparrow Y\) with \(\mu(X_n)<\infty\) and \(\nu(Y_n)<\infty\). Such increasing exhaustions are obtained from any sigma-finite covers by taking finite unions. Set \(R_n=X_n\times Y_n\), so \(R_n\uparrow X\times Y\), and define \(f_n=f\chi_{R_n}\). Then \(f_n\uparrow f\) pointwise.
Proof of Tonelli’s theorem. On each finite-measure pair \((X_n,Y_n)\), apply the lemma to the restriction of \(f\) to \(R_n\). The product measure restricted to \(R_n\) agrees with the product of the restricted measures: both are finite measures on \(\mathcal{M}|_{X_n}\otimes\mathcal{N}|_{Y_n}\) and agree on measurable rectangles, so uniqueness of finite product measures applies. Therefore, for each \(n\), $$ \int_{R_n} f\,d(\mu\times\nu) =\int_{X_n}\left(\int_{Y_n}f(x,y)\,d\nu(y)\right)d\mu(x) =\int_X\left(\int_Y f_n(x,y)\,d\nu(y)\right)d\mu(x). $$ The last equality holds because \(f_n\) vanishes outside \(X_n\times Y_n\).
For every fixed \(x\), \(f_n(x,\cdot)\uparrow f(x,\cdot)\): if \(x\notin X_n\), the value at stage \(n\) is zero, and eventually \(x\in X_n\); also \(Y_n\uparrow Y\). The sequence is increasing because the rectangles \(R_n\) increase. By the Monotone Convergence Theorem on \(Y\), the measurable functions \(g_n(x)=\int_Y f_n(x,y)\,d\nu(y)\) increase pointwise to \(g(x)=\int_Y f(x,y)\,d\nu(y)\). Each \(g_n\) is measurable by the finite-measure lemma, extended by zero outside \(X_n\); hence \(g\), their pointwise limit, is measurable.
Now apply the Monotone Convergence Theorem on \(X\) to \(g_n\uparrow g\), and on \(X\times Y\) to \(f_n\uparrow f\). Since \(\bigcup_nR_n=X\times Y\), these give $$ \int_X g\,d\mu =\lim_{n\to\infty}\int_Xg_n\,d\mu =\lim_{n\to\infty}\int_{R_n}f\,d(\mu\times\nu) =\int_{X\times Y}f\,d(\mu\times\nu). $$ Every limit is justified by monotone convergence; no finiteness of these integrals is assumed. Repeating the argument with \(X,\mu\) and \(Y,\nu\) interchanged proves measurability of \(y\mapsto\int_Xf(x,y)\,d\mu(x)\) and the identity in the other order. Thus all three quantities agree, including when their common value is infinite. \(\square\)
Worked Applications
Worked Example: A Product of Exponential Functions
On \(\mathbb{R}^2\) with Lebesgue measure \(m\times m\), let \(f(x,y)=e^{-|x|}e^{-|y|}\). This function is nonnegative and measurable, and Lebesgue measure on \(\mathbb{R}\) is sigma-finite. For fixed \(x\), $$ \int_{\mathbb{R}}f(x,y)\,dy=e^{-|x|}\int_{\mathbb{R}}e^{-|y|}\,dy =e^{-|x|}\left(\int_{-\infty}^{0}e^y\,dy+\int_0^\infty e^{-y}\,dy\right) =2e^{-|x|}. $$ Then $$ \int_{\mathbb{R}}2e^{-|x|}\,dx =2\left(\int_{-\infty}^0e^x\,dx+\int_0^\infty e^{-x}\,dx\right) =2(1+1)=4. $$ Interchanging the coordinates gives the same value, so Tonelli gives \(\int_{\mathbb{R}^2}f\,d(m\times m)=4\). Check: on \([-R,R]^2\), the product integral is \([2(1-e^{-R})]^2\), which tends to \(4\) as \(R\to\infty\). The rectangles increase to \(\mathbb{R}^2\), so Monotone Convergence confirms the value.
Worked Example: Integrating the Indicator of a Rectangle
Let \(E=[0,2]\times[1,4]\subset\mathbb{R}^2\), with both intervals including their endpoints. For every \(x\in[0,2]\), the section \(E_x=[1,4]\) has Lebesgue measure \(4-1=3\); for \(x\notin[0,2]\), its measure is zero. Thus \(x\mapsto m(E_x)=3\chi_{[0,2]}(x)\), and $$ \int_{\mathbb{R}}m(E_x)\,dx=3\cdot(2-0)=6. $$ In the other direction, \(E^y=[0,2]\) for \(y\in[1,4]\), and is empty otherwise. Hence $$ \int_{\mathbb{R}}m(E^y)\,dy=2\cdot(4-1)=6. $$ Tonelli applied to \(\chi_E\) gives \((m\times m)(E)=6\). Check: the rectangle’s side lengths are \(2-0=2\) and \(4-1=3\), whose product is \(2\cdot3=6\), in agreement with both section calculations.
Worked Example: Finite Section Integrals but Infinite Total Integral
Take \(X=[0,\infty)\) and \(Y=[0,1]\), each with the restriction of Lebesgue measure, and let \(f(x,y)=1\). Both spaces are sigma-finite. For every \(x\in X\), $$ \int_Y f(x,y)\,dy=\int_0^1 1\,dy=1, \qquad \int_X\left(\int_Y f(x,y)\,dy\right)dx=\int_0^\infty1\,dx=\infty. $$ For each \(y\in Y\), \(\int_X f(x,y)\,dx=\int_0^\infty1\,dx=\infty\), so the reverse iterated integral is also infinite. Therefore Tonelli gives \(\int_{X\times Y}f\,d(\mu\times\nu)=\infty\). Check: on \([0,N]\times[0,1]\), the product integral is \(N\cdot1=N\), and these values increase without bound as \(N\to\infty\).
What Tonelli Does—and Does Not—Say
Tonelli applies to nonnegative product-measurable functions on sigma-finite measure spaces. It establishes measurability of the section integrals as well as equality of the integrals. It does not require the product integral to be finite, nor does it assert that each section integral is finite. The equalities are identities of extended nonnegative integrals, not statements about conditional cancellation of positive and negative parts.
Common pitfall. In the last example, every integral over a \(Y\)-section is finite: it equals \(1\). It would be incorrect to infer that the integral over \(X\times Y\) is therefore finite. The outer integral is \(\int_0^\infty1\,dx=\infty\), and the product integral is infinite as well. Sectionwise finiteness at each individual point does not imply integrability over the other coordinate.
Check Your Understanding
Use the hypotheses and proof structure of Tonelli’s theorem to organize each argument.
- For \(f(x,y)=\chi_{[0,1]}(x)e^{-y}\) on \(\mathbb{R}\times[0,\infty)\), compute both section integrals and the product integral.
- In the finite-measure lemma, where is the assumption \(\nu(Y)<\infty\) used when passing from \(E\) to \(E^c\)?
- Explain why \(f_n=f\chi_{X_n\times Y_n}\) increases pointwise to \(f\) when both exhaustions are increasing.
- Give a nonnegative function whose section integrals are finite at every point of the first space but whose product integral is infinite.
- Which theorem justifies passing from \(\int_X g_n\,d\mu\) to \(\int_X\lim_n g_n\,d\mu\) in the exhaustion proof, and what property of \(g_n\) is required?