Tonelli's Theorem from Simple Functions

Why Tonelli’s Theorem Starts with Rectangles

On a product space, a nonnegative function can depend on both coordinates, while its integral is defined with respect to the product measure. Tonelli’s theorem connects that integral to repeated integration: first integrate in one coordinate, then in the other. Its key strength is that it applies without assuming the integral is finite.

The proof follows a useful progression. A rectangle separates into one condition on each coordinate, so its indicator is easy to integrate in either order. Measure uniqueness extends that calculation from rectangles to all measurable sets. Finite linear combinations then give the result for simple functions, and monotone convergence carries it to general nonnegative functions.

Setting and Notation

Let $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ be sigma-finite measure spaces. This means that each space is a countable union of measurable sets of finite measure. Write $\mathcal A\otimes\mathcal B$ for the sigma-algebra generated by measurable rectangles $A\times B$, and write $\mu\otimes\nu$ for the product measure.

For a function $f:X\times Y\to[0,\infty]$, its section at $x\in X$ is the function $f_x:Y\to[0,\infty]$ defined by $f_x(y)=f(x,y)$. The inner integral is therefore $\int_Y f(x,y)\,\nu(dy)$, viewed as a function of $x$. For product-measurable nonnegative $f$, the sections are measurable, and the section integral is measurable in $x$, as in the earlier material on measurability of section integrals.

Definition: A nonnegative simple function is a measurable function taking only finitely many values in $[0,\infty)$. It can be written as a finite sum $s=\sum_{j=1}^m c_j\mathbf 1_{E_j}$, where the $E_j$ are disjoint measurable sets and $c_j\geq 0$.

We use the monotone convergence theorem: if measurable nonnegative functions increase pointwise to a function $g$, then their integrals increase to the integral of $g$, including when that integral is infinite.

The Rectangle Calculation

For measurable $A\in\mathcal A$ and $B\in\mathcal B$, the section of $A\times B$ at $x$ is $B$ when $x\in A$, and is empty otherwise. Thus the section measure is $\nu(B)\mathbf 1_A(x)$. Integrating this function gives the rectangle identity.

\int_X \nu\bigl((A\times B)_x\bigr)\,\mu(dx) = \mu(A)\nu(B) = (\mu\otimes\nu)(A\times B).

This calculation holds with extended nonnegative values, so neither factor needs to be finite. The reverse order gives the same value: the section at $y$ is $A$ for $y\in B$, and empty otherwise.

Worked Example: An Indicator of a Rectangle

Take $X=Y=\mathbb R$ with Lebesgue measure, and let $f=\mathbf 1_{[0,2]\times(1,4]}$. For each $x\in[0,2]$, the section is $(1,4]$, of length $3$; outside $[0,2]$, it is empty. Hence the integral in the $y$-coordinate is $3\mathbf 1_{[0,2]}(x)$, and

\int_{\mathbb R}\left(\int_{\mathbb R}f(x,y)\,dy\right)dx = \int_0^2 3\,dx = 6.

Integrating first in $x$ gives a section of length $2$ for each $y\in(1,4]$, and zero otherwise, so that order also gives $2\cdot 3=6$. The product measure of the rectangle is $2\cdot 3=6$, in agreement with both calculations.

From Rectangles to All Measurable Sets

The rectangle calculation alone is not yet enough for a general measurable set $E\subseteq X\times Y$. The bridge is uniqueness of sigma-finite measures from their values on measurable rectangles. To apply it, define a measure by integrating the sizes of the sections.

Lemma: For every $E\in\mathcal A\otimes\mathcal B$, $$ (\mu\otimes\nu)(E)=\int_X \nu(E_x)\,\mu(dx). $$

Proof. Section-integral measurability ensures that $x\mapsto\nu(E_x)$ is measurable. Define $\lambda(E)=\int_X\nu(E_x)\,\mu(dx)$ for product-measurable $E$. For pairwise disjoint measurable sets $E_1,E_2,\ldots$, their sections are pairwise disjoint and $\nu((\bigcup_n E_n)_x)=\sum_n\nu((E_n)_x)$. Applying monotone convergence to the increasing finite partial sums shows that $\lambda$ is countably additive. Also, $\lambda(\varnothing)=0$, so $\lambda$ is a measure.

On every measurable rectangle $A\times B$, the rectangle calculation gives $\lambda(A\times B)=\mu(A)\nu(B)=(\mu\otimes\nu)(A\times B)$. Both measures are sigma-finite: choose countable finite-measure covers $(X_n)$ of $X$ and $(Y_m)$ of $Y$. The rectangles $X_n\times Y_m$ cover $X\times Y$ and have finite measure for both measures. The uniqueness theorem for sigma-finite measures agreeing on a generating family of measurable rectangles therefore gives $\lambda=\mu\otimes\nu$. This proves the identity. $\square$

The same argument with the roles of $X$ and $Y$ exchanged gives the corresponding identity integrating section measures over $Y$. Sigma-finiteness matters here: it makes the uniqueness argument applicable and is part of the theorem’s hypotheses.

Simple Functions and the Main Theorem

For a nonnegative simple function $s=\sum_{j=1}^m c_j\mathbf 1_{E_j}$ with disjoint product-measurable sets $E_j$, the lemma applies to each set. Finite linearity of integration then gives the iterated-integral identity for $s$. The same reasoning applies when some $c_j=0$; those terms contribute zero.

Theorem (Tonelli): Let $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ be sigma-finite measure spaces. If $f:X\times Y\to[0,\infty]$ is $\mathcal A\otimes\mathcal B$-measurable, then its section integrals are measurable and $$ \int_{X\times Y} f\,d(\mu\otimes\nu) = \int_X\left(\int_Y f(x,y)\,\nu(dy)\right)\mu(dx) = \int_Y\left(\int_X f(x,y)\,\mu(dx)\right)\nu(dy). $$ All three values are in $[0,\infty]$.

Proof. The measurable-set lemma and finite linearity prove the asserted identity for every nonnegative simple function. Since $f$ is measurable and nonnegative, there is an increasing sequence of nonnegative simple functions $(s_n)$ with $s_n\uparrow f$ pointwise. Monotone convergence on the product space gives $$ \int_{X\times Y} f\,d(\mu\otimes\nu) = \lim_{n\to\infty}\int_{X\times Y}s_n\,d(\mu\otimes\nu). $$ For every fixed $x$, $s_n(x,\cdot)\uparrow f(x,\cdot)$, so monotone convergence on $Y$ gives $\int_Y s_n(x,y)\,\nu(dy)\uparrow\int_Y f(x,y)\,\nu(dy)$. Applying monotone convergence once more, now on $X$, yields $$ \int_X\left(\int_Y f(x,y)\,\nu(dy)\right)\mu(dx) = \lim_{n\to\infty}\int_X\left(\int_Y s_n(x,y)\,\nu(dy)\right)\mu(dx). $$ For each $n$, the simple-function identity equates the corresponding product and iterated integrals. Taking limits proves the first equality. Repeating the same argument with $X,\mu$ and $Y,\nu$ exchanged proves the second equality. The limits are allowed to be infinite, so no additional integrability hypothesis is needed. $\square$

Worked Examples Beyond Rectangles

Worked Example: A Simple Function on Two Disjoint Rectangles

On $\mathbb R^2$ with Lebesgue measure, let $E_1=[0,1]\times[0,2]$, $E_2=(1,3]\times[0,1]$, and $f=2\mathbf 1_{E_1}+3\mathbf 1_{E_2}$. The rectangles are disjoint because their $x$-coordinate sets are disjoint. Their areas are $2$ and $2$, respectively, so the product integral is $2\cdot2+3\cdot2=10$.

For $x\in[0,1]$, the section integral is $2\cdot2=4$; for $x\in(1,3]$, it is $3\cdot1=3$; and elsewhere it is zero. Integrating over $x$ gives $4\cdot1+3\cdot2=10$. In the other order, the section integral is $2\cdot1+3\cdot2=8$ for $y\in[0,1]$, is $2\cdot1=2$ for $y\in(1,2]$, and is zero otherwise. Its integral is $8\cdot1+2\cdot1=10$.

Worked Example: A Nonnegative Function on the Positive Quadrant

Let $f(x,y)=e^{-x-y}$ for $x,y\geq0$. Since $e^{-x-y}=e^{-x}e^{-y}$, integrating in $y$ first gives

\int_0^\infty e^{-x-y}\,dy = e^{-x}\int_0^\infty e^{-y}\,dy = e^{-x}.

Consequently, the outer integral is $\int_0^\infty e^{-x}\,dx=1$. Integrating first in $x$ gives $e^{-y}$ and then $\int_0^\infty e^{-y}\,dy=1$ as well. Tonelli applies because the function is measurable and nonnegative; the calculation also shows that the common integral is finite.

Worked Example: The Indicator of a Triangular Region

Let $f(x,y)=\mathbf 1_{\{(x,y):0<y<x<1\}}$ on $\mathbb R^2$. For a fixed $x\in(0,1)$, the section in $y$ is $(0,x)$, with length $x$; for $x\notin(0,1)$, it is empty. Therefore

\int_{\mathbb R}\left(\int_{\mathbb R}f(x,y)\,dy\right)dx = \int_0^1 x\,dx = \frac12.

For fixed $y\in(0,1)$, the section in $x$ is $(y,1)$, with length $1-y$. Integrating in the other order gives $\int_0^1(1-y)\,dy=1-\frac12=\frac12$. Both section descriptions include the strict inequalities defining the triangle; the endpoints have measure zero in these one-dimensional integrals.

What the Theorem Does—and Does Not—Say

Tonelli’s theorem licenses either order of integration for a nonnegative measurable function on a sigma-finite product space. It does not promise a finite answer: both iterated integrals and the product integral may equal $+\infty$. Nonnegativity is what lets the simple approximations increase to the function without cancellation. For signed or complex functions, changing integration order requires different hypotheses; that is the subject of Fubini’s theorem for integrable functions.

A common mistake is to treat a formal iterated integral as automatically interchangeable with a product integral, regardless of the integrand or measure spaces. Tonelli’s conclusion depends on its stated assumptions. In particular, its nonnegative hypothesis should not be dropped merely because the repeated integrals appear to exist.

Check Your Understanding

Use the rectangle calculation and the monotone-approximation argument to answer the following questions.

Why does the section measure of $A\times B$ equal $\nu(B)\mathbf 1_A(x)$?
Where does sigma-finiteness enter the extension from rectangles to all product-measurable sets?
Why is monotone convergence applied first on one factor and then on the other in the proof of Tonelli’s theorem?
Does Tonelli’s theorem assert that the common integral is finite? Explain.
Which hypothesis of Tonelli prevents cancellation from complicating the approximation argument?