From One Integral to Two
The Lebesgue integral has been developed so far on a single measure space. On a product space, a function depends on two variables, and it is natural to ask whether its integral can be computed by integrating one variable at a time. Fubini’s Theorem answers this question for absolutely integrable functions: almost every section is integrable, the resulting section integrals are integrable functions, and either order of integration gives the product-space integral.
The condition of absolute integrability is essential. It is not enough that the two iterated integrals happen to exist: without absolute integrability, they may have different values. We will first set up product spaces and establish the nonnegative integration result that Fubini’s Theorem uses.
Here, sigma-finiteness means that each space is a countable union of measurable sets of finite measure. It ensures that the product measure is well-defined and unique. For a measurable set \(E\subseteq X\times Y\), its sections are $$ E_y=\{x\in X:(x,y)\in E\}, \qquad E^x=\{y\in Y:(x,y)\in E\}. $$ For a measurable function \(f:X\times Y\to[0,\infty]\), write \(f_y(x)=f(x,y)\) and \(f^x(y)=f(x,y)\). Sections of sets in \(\mathcal{F}\otimes\mathcal{G}\) are measurable, so sections of product-measurable functions are measurable as well.
The Nonnegative Foundation
The nonnegative result says that product integration agrees with iterated integration, even when the common value is infinite. It is the key step: once we know how to integrate a nonnegative function in either order, we can apply the result to the positive and negative parts of an absolutely integrable function.
Proof. First suppose both measures are finite. For a measurable set \(E\subseteq X\times Y\), define $$ \phi_E(y)=\mu(E_y). $$ Consider the class of product-measurable sets \(E\) for which \(\phi_E\) is measurable and $$ \int_Y\phi_E\,d\nu=(\mu\times\nu)(E). $$ This class contains every measurable rectangle \(A\times B\): its section is \(A\) when \(y\in B\) and empty otherwise, so \(\phi_{A\times B}=\mu(A)\mathbf{1}_B\), and both sides of the equality are \(\mu(A)\nu(B)\).
The class contains \(X\times Y\). It is closed under complements: for a product-measurable \(E\), \(\phi_{(X\times Y)\setminus E}(y)=\mu(X)-\phi_E(y)\), and finiteness gives $$ \int_Y\bigl(\mu(X)-\phi_E(y)\bigr)\,d\nu(y) =\mu(X)\nu(Y)-(\mu\times\nu)(E) =(\mu\times\nu)((X\times Y)\setminus E). $$ It is also closed under countable disjoint unions. For pairwise disjoint \(E_k\), the sections \((E_k)_y\) are pairwise disjoint, so countable additivity gives \(\phi_{\cup_k E_k}(y)=\sum_k\phi_{E_k}(y)\). Measurability of this sum and the Monotone Convergence Theorem give the required integral identity. Thus the class is a lambda-system containing the measurable rectangles, which form a pi-system generating \(\mathcal{F}\otimes\mathcal{G}\). The pi-lambda theorem shows that the identity holds for every product-measurable \(E\).
For sigma-finite spaces, cover \(X\) and \(Y\) by increasing sequences of measurable sets of finite measure. Their product rectangles increase to \(X\times Y\). Apply the finite-measure result to the intersections of \(E\) with these rectangles, and use the Monotone Convergence Theorem to pass to the limit. This proves the section measurability and integral identity for indicators of all product-measurable sets. By linearity, the identity follows for nonnegative simple functions. Finally, approximate \(f\) from below by an increasing sequence of nonnegative measurable simple functions, using the Increasing Simple Approximation theorem. Apply the identity to each simple function and use the Monotone Convergence Theorem for the product integral and both iterated integrals. This proves the theorem in the stated generality. Interchanging the roles of \(X\) and \(Y\) gives the equality in the other order.
Worked Example: Integrating an Indicator of a Rectangle
Let \(E=[1,3]\times[0,2]\) in \(\mathbb{R}^2\), with Lebesgue measure in each coordinate, and take \(f=\mathbf{1}_E\). For a fixed \(y\), the \(x\)-section is \([1,3]\) if \(0\leq y\leq2\), and empty otherwise. Consequently, $$ \int_{\mathbb{R}}\mathbf{1}_E(x,y)\,dx = \begin{cases} 2,&0\leq y\leq2,\\ 0,&\text{otherwise}. \end{cases} $$ Integrating this section integral gives $$ \int_{\mathbb{R}}\left(\int_{\mathbb{R}}\mathbf{1}_E(x,y)\,dx\right)dy =\int_0^2 2\,dy=4. $$ In the other order, the \(y\)-section has length \(2\) for \(1\leq x\leq3\) and length \(0\) otherwise, so the iterated integral is \(\int_1^3 2\,dx=4\). This agrees with \((\lambda\times\lambda)(E)=(3-1)(2-0)=4\), as required.
Fubini’s Theorem for Integrable Functions
For a signed function, the positive and negative parts are \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\), so \(f=f^+-f^-\) and \(|f|=f^++f^-\). Absolute integrability means that \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\). Applying the nonnegative theorem to \(|f|\) shows that almost every section has finite absolute integral. The same argument controls the iterated integrals of \(f^+\) and \(f^-\), allowing their difference to be taken without an undefined subtraction of infinities.
Proof. Apply the Nonnegative Iterated-Integration Theorem to \(|f|\). It gives $$ \int_Y\left(\int_X|f(x,y)|\,d\mu(x)\right)d\nu(y) =\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty. $$ A nonnegative measurable function with finite integral is finite almost everywhere. Thus \(\int_X|f(x,y)|\,d\mu(x)<\infty\) for \(\nu\)-almost every \(y\), which means \(f_y\) is integrable there. Reversing the roles of the variables proves that \(f^x\) is integrable for \(\mu\)-almost every \(x\).
The nonnegative theorem also shows that the section integrals of \(f^+\) and \(f^-\) are measurable. Their difference gives the section integral of \(f\) wherever both are finite; assign any value, such as zero, on the exceptional measurable null set. Since the integral of each of these nonnegative section-integral functions is bounded above by the corresponding integral of \(|f|\), both are integrable. In particular, their difference is integrable. Applying the nonnegative theorem separately to \(f^+\) and \(f^-\), and subtracting the resulting finite equalities, yields $$ \int_Y\left(\int_X f(x,y)\,d\mu(x)\right)d\nu(y) =\int_{X\times Y}(f^+-f^-)\,d(\mu\times\nu) =\int_{X\times Y}f\,d(\mu\times\nu). $$ The same argument with \(X\) and \(Y\) exchanged proves the other equality. This completes the proof.
Worked Example: A Polynomial on a Rectangle
On \([0,1]\times[0,2]\), let \(f(x,y)=xy+y\). The function is bounded and measurable on a set of finite measure, so it is integrable. Integrate first with respect to \(x\): $$ \int_0^1(xy+y)\,dx =y\left[\frac{x^2}{2}+x\right]_0^1 =\frac{3y}{2}. $$ Therefore, $$ \int_0^2\left(\int_0^1(xy+y)\,dx\right)dy =\int_0^2\frac{3y}{2}\,dy =\frac{3}{2}\left[\frac{y^2}{2}\right]_0^2=3. $$ In the opposite order, $$ \int_0^2(xy+y)\,dy =2x+2, \qquad \int_0^1(2x+2)\,dx =\left[x^2+2x\right]_0^1=3. $$ Both orders give the product integral, as Fubini’s Theorem guarantees.
Worked Example: The Area Below the Diagonal
Let \(E=\{(x,y)\in[0,1]^2:0\leq y\leq x\}\) and \(f=\mathbf{1}_E\). This is a measurable set of finite measure, so its indicator is integrable. For fixed \(x\), the \(y\)-section is \([0,x]\), with length \(x\). Thus $$ \int_0^1\left(\int_0^1\mathbf{1}_E(x,y)\,dy\right)dx =\int_0^1 x\,dx=\frac12. $$ For fixed \(y\), the \(x\)-section is \([y,1]\), with length \(1-y\). Integrating in the reverse order gives $$ \int_0^1\left(\int_0^1\mathbf{1}_E(x,y)\,dx\right)dy =\int_0^1(1-y)\,dy =\left[y-\frac{y^2}{2}\right]_0^1=\frac12. $$ The two computations agree and show that the triangle has area \(1/2\).
Why Absolute Integrability Matters
The integrability hypothesis is not a technical formality. Here is a measurable function for which both orders of integration give finite values, but the values disagree. On \((0,1)^2\), define $$ f(x,y)=\frac{x-y}{(x+y)^3}. $$ For fixed \(x>0\), direct integration in \(y\) gives $$ \int_0^1\frac{x-y}{(x+y)^3}\,dy =\left[-\frac{x}{(x+y)^2}+\frac{1}{x+y}\right]_{y=0}^{y=1} =\frac{1}{(x+1)^2}. $$ The terms at \(y=0\) cancel, since \(-x/x^2+1/x=0\). Hence integrating in \(x\) gives \(\int_0^1(1+x)^{-2}\,dx=1/2\). The function satisfies \(f(y,x)=-f(x,y)\), so reversing the order gives \(-1/2\).
There is no contradiction with Fubini’s Theorem: \(f\) is not absolutely integrable. On the region \(0<x<1\) and \(0<y<x/2\), we have \(x-y>x/2\) and \(x+y<3x/2\). It follows that $$ |f(x,y)|>\frac{x/2}{(3x/2)^3}=\frac{4}{27x^2}. $$ Consequently, the integral of \(|f|\) over this region is at least $$ \int_0^1\int_0^{x/2}\frac{4}{27x^2}\,dy\,dx =\int_0^1\frac{2}{27x}\,dx =\infty. $$ The iterated integrals exist here, but absolute integrability fails, and the orders disagree.
A common pitfall is to exchange integration order merely because both iterated integrals can be calculated. Fubini’s Theorem licenses the exchange when the function is product-measurable and integrable. For nonnegative functions, the Nonnegative Iterated-Integration Theorem applies without requiring a finite integral; its common value may be infinite. The next tutorial develops that nonnegative result further.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What set generates the product sigma-algebra \(\mathcal{F}\otimes\mathcal{G}\)?
- Why does applying the nonnegative result to \(|f|\) show that almost every section of an integrable \(f\) is integrable?
- For which hypothesis of Fubini’s Theorem does the function \((x-y)/(x+y)^3\) fail, and how does the disagreement of iterated integrals illustrate that failure?
- Compute both orders of integration for \(\mathbf{1}_{[0,1]\times[0,3]}\) on \(\mathbb{R}^2\).
- Does the nonnegative iterated-integration result require the common integral to be finite?