Exact Tonelli Integrals on Rectangular and Triangular Regions

From Double Integrals to Section Lengths

A nonnegative double integral can often be evaluated by replacing each two-dimensional region with a one-dimensional measurement: at each base point, find the length of the vertical section. For an indicator function, this computes area directly. For a general nonnegative function, the same idea can be applied to its superlevel regions, then combined with the layer-cake identity.

Throughout, the underlying one-dimensional measure space is $(\mathbb R,\mathcal B_{\mathbb R},m)$, where $m$ is Lebesgue measure restricted to the Borel sigma-algebra. The product is $(\mathbb R^2,\mathcal B_{\mathbb R}\otimes\mathcal B_{\mathbb R},m\times m)$, with $m\times m$ the usual planar Lebesgue measure on Borel sets. Both spaces are sigma-finite: the intervals $[-n,n]$ have finite measure and cover $\mathbb R$. Unless stated otherwise, functions are Borel measurable and take values in $[0,\infty]$; all displayed integral identities are equalities of extended nonnegative numbers, so $+\infty$ is allowed.

For a Borel set $E\subseteq\mathbb R^2$, its vertical section at $x$ is $E_x=\{y\in\mathbb R:(x,y)\in E\}$. The section length is $m(E_x)$, which may be infinite. Tonelli’s theorem says that this section length is measurable as a function of $x$, and that its integral equals the product measure of $E$. More generally, for a nonnegative Borel function $f$, Tonelli identifies its double integral with either order of iterated integration. The layer-cake identity, established earlier in the course, expresses $\int f$ as the integral over $t\ge0$ of the measures of $\{f>t\}$. These results apply here because the measures are sigma-finite and the functions and sets used below are nonnegative and measurable.

Section formula: For every Borel set $E\subseteq\mathbb R^2$, $(m\times m)(E)=\int_{\mathbb R}m(E_x)\,dm(x)$. For every nonnegative Borel function $f$, Tonelli gives $\int_{\mathbb R^2}f\,d(m\times m)=\int_{\mathbb R}\left(\int_{\mathbb R}f(x,y)\,dm(y)\right)dm(x)$. The identities also hold with the order of integration reversed.

In geometric problems, $m(E_x)$ is the length of the vertical slice at $x$. In function integrals, the inner integral $\int f(x,y)\,dm(y)$ is the corresponding weighted slice. The formula is not restricted to finite values: a section length or section integral may be $+\infty$.

Rectangles and Triangles by Slicing

A rectangle with horizontal interval $I$ and vertical interval $J$ has section $J$ for each $x\in I$, and an empty section for $x\notin I$. Thus its area is the length of $I$ times the length of $J$, when both lengths are finite. A triangle can be handled just as directly, except that the section length changes with $x$. For example, the region $0\le y\le c-x$, $0\le x\le c$, has vertical section length $c-x$, and its area is the integral of that length.

Superlevel-section formula: If $f:\mathbb R^2\to[0,\infty]$ is Borel measurable, then $$ \int_{\mathbb R^2} f\,d(m\times m) =\int_0^\infty (m\times m)(\{(x,y):f(x,y)>t\})\,dm(t) =\int_0^\infty\int_{\mathbb R}m(\{y:f(x,y)>t\})\,dm(x)\,dm(t). $$ Each equality is valid in $[0,\infty]$.

Proof. Apply the layer-cake identity to $f$ on the sigma-finite product measure space to obtain the first equality. For each $t\ge0$, the set $\{f>t\}$ is Borel. Apply the section formula, which is Tonelli’s theorem applied to its nonnegative indicator, to write its product measure as $\int_{\mathbb R}m(\{y:f(x,y)>t\})\,dm(x)$. The integrand as a function of $(x,t)$ is nonnegative and measurable, so Tonelli permits the order of integration in $x$ and $t$ to be written as displayed. No finiteness assumption is needed. $\square$

Worked Example: A Weighted Rectangle

Let $R=[0,2]\times[1,4]$, and define $f(x,y)=x$ on $R$ and $f(x,y)=0$ outside $R$. Compute $\int_{\mathbb R^2} f\,d(m\times m)$ using superlevel sections.

For $0\le t<2$, the set $\{f>t\}$ consists of points in $R$ with $x>t$. At each $x$ with $t<x\le2$, its vertical section is $[1,4]$, of length $4-1=3$; outside that $x$-range the section is empty. Consequently, its area is $3(2-t)$. For $t\ge2$, the superlevel set is empty. The superlevel-section formula therefore gives

\int_{\mathbb R^2}f\,d(m\times m) =\int_0^2 3(2-t)\,dt =3\left[2t-\frac{t^2}{2}\right]_0^2 =3(4-2)=6.

Check by direct sections. For $0\le x\le2$, the integral over the vertical section is $\int_1^4 x\,dy=3x$. Hence the iterated integral is $\int_0^2 3x\,dx=3[x^2/2]_0^2=3(2)=6$, agreeing with the superlevel calculation.

Worked Example: A Function on a Triangular Region

Let $T=\{(x,y):x\ge0,\ y\ge0,\ x+y\le1\}$, and let $f(x,y)=1-x-y$ on $T$ and $0$ elsewhere. Find its integral exactly.

On $T$, $f$ is nonnegative. For $0\le t<1$, the strict superlevel condition $f>t$ is equivalent to $x+y<1-t$, with $x,y\ge0$. For a fixed $x$ between $0$ and $1-t$, the vertical section has length $1-t-x$; for other $x$, it is empty. Thus

(m\times m)(\{f>t\}) =\int_0^{1-t}(1-t-x)\,dx =\left[(1-t)x-\frac{x^2}{2}\right]_0^{1-t} =\frac{(1-t)^2}{2}.

For $t\ge1$, the strict superlevel set is empty. Applying the layer-cake identity and evaluating the integral gives

\int_{\mathbb R^2}f\,d(m\times m) =\int_0^1\frac{(1-t)^2}{2}\,dt =\frac12\left[t-t^2+\frac{t^3}{3}\right]_0^1 =\frac16.

Check by direct integration. At fixed $x\in[0,1]$, the vertical section of $T$ runs from $y=0$ to $y=1-x$, so

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

Integrating this section value gives $\int_0^1(1-x)^2/2\,dx=\frac12[x-x^2+x^3/3]_0^1=\frac16$. Both methods give the same exact value.

Infinite Section Integrals

Tonelli’s theorem allows the inner section integral to be infinite. What matters for the outer integral is how the section values behave across the base. An infinite value at one base point does not by itself force the double integral to be infinite, because a singleton has Lebesgue measure zero. By contrast, if section integrals are infinite throughout a set of positive measure, then their outer integral is infinite.

Worked Example: One Infinite Section on a Null Fiber

On $Q=[0,1]\times[0,1]$, define $f(x,y)=+\infty$ when $x=0$, and $f(x,y)=1$ when $0<x\le1$; set $f=0$ outside $Q$. This is an extended-real-valued Borel function. Compute its double integral, including its section integrals.

At $x=0$, the section integral is $\int_{[0,1]}+\infty\,dm(y)=+\infty$, because the section has length $1$. At each $x\in(0,1]$, the section integral is $\int_{[0,1]}1\,dm(y)=1$. For $x\notin[0,1]$, it is $0$. Thus the section-integral function is infinite at the single point $0$, equals $1$ on $(0,1]$, and equals $0$ elsewhere. Its integral is $1$, since changing a nonnegative measurable function at a null set does not change its Lebesgue integral.

Check by superlevel areas. For each $t$ with $0\le t<1$, the superlevel set is $Q$, whose area is $(1-0)(1-0)=1$. For $t\ge1$, the superlevel set is $\{0\}\times[0,1]$, whose area is zero because its horizontal base is a singleton of measure zero. Therefore the layer-cake integral is $\int_0^1 1\,dt=1$, as before.

Worked Example: Infinite Sections on a Positive-Measure Base

Let $E=[0,1]\times[0,\infty)$, and take $f=\chi_E$ on $\mathbb R^2$. For each $x\in[0,1]$, the vertical section is $[0,\infty)$, of infinite length. For $x\notin[0,1]$, it is empty. Hence the section integral is $+\infty$ on $[0,1]$ and $0$ elsewhere, so its outer integral is $+\infty\cdot m([0,1])=+\infty$, since $m([0,1])=1>0$.

Check by superlevel sets. For every $t$ with $0\le t<1$, $\{f>t\}=E$, whose area is infinite: its section length is infinite at every $x\in[0,1]$, a set of measure $1$. For $t\ge1$, the strict superlevel set is empty. Thus the layer-cake integral is $\int_0^1+\infty\,dt=+\infty$, confirming the section calculation.

Common Pitfall: Multiplying Zero by Infinity

In the null-fiber example, it would be incorrect to write “the section at $x=0$ contributes $+\infty\cdot m(\{0\})=+\infty\cdot0=0$” as though this were an ordinary product. The product $0\cdot\infty$ is undefined, and the section formula is an integral of a function, not a sum of pointwise products. The correct calculation is to integrate the section-integral function: it is $1$ almost everywhere on $(0,1]$, infinite at $0$, and zero elsewhere, so its Lebesgue integral is $1$. Equivalently, the superlevel sets have area $1$ for $0\le t<1$ and area zero for $t\ge1$.

The contrast between the last two examples is the essential diagnostic. An infinite inner integral on a null set of base points can coexist with a finite double integral. Infinite inner integrals on a positive-measure set force the outer integral to be infinite. Tonelli’s identity applies in both situations; it does not assert that either side must be finite.

Check Your Understanding

In each problem, identify the vertical sections and state whether the relevant integrals are finite or infinite.

Compute the area of $[1,4]\times[2,5]$ by integrating vertical section lengths.
For $T=\{(x,y):x\ge0,\ y\ge0,\ x+y\le2\}$, find the vertical section length at each $x$ and compute $m^2(T)$.
On $[0,1]^2$, set $f(x,y)=x$. Find $\int f\,d(m\times m)$ by calculating the areas of its strict superlevel sets.
Suppose a nonnegative Borel function has section integral $+\infty$ at exactly one point $x$, and has section integral $1$ on $(0,1]$ and $0$ elsewhere. What is its outer section integral, and which measure property determines the answer?
For $E=[0,1]\times[0,\infty)$, explain why the section formula gives an infinite product-measure integral even though the base interval has finite measure.