Fubini on Triangles and Simplexes

Why the Geometry of the Region Matters

Changing the order of integration is not just a matter of exchanging the symbols $dx$ and $dy$. The bounds describe which points belong to the region, so they must change with the order. On a rectangle, each coordinate ranges independently over a fixed interval. On a triangle or simplex, the range of one coordinate usually depends on the others.

There is also an analytic condition to check. For a signed or complex-valued integrand, absolute integrability is the standard hypothesis that allows Fubini’s theorem to identify the integral with either iterated integral. For a Lebesgue-measurable function, the sections need only be integrable for almost every outer coordinate. An exceptional section cannot simply be assumed to have an integral.

Triangles as Sections

Consider the closed triangle

D=\{(x,y)\in\mathbb{R}^2:0\leq y\leq x\leq 1\}.

Fixing $x$ gives a vertical section: $x$ ranges from $0$ to $1$, and, for each fixed $x$, $y$ ranges from $0$ to $x$. Fixing $y$ gives a horizontal section: $y$ ranges from $0$ to $1$, and, for each fixed $y$, $x$ ranges from $y$ to $1$. Thus the same region has the two descriptions

0\leq x\leq 1,\quad 0\leq y\leq x, \qquad\text{and}\qquad 0\leq y\leq 1,\quad y\leq x\leq 1.

The bounds are obtained by fixing the outer coordinate and solving the region’s inequalities for the inner one. The diagonal $y=x$ is part of the boundary; including or excluding it does not change a Lebesgue integral, since that line has two-dimensional Lebesgue measure zero.

Definition: For a region $D\subseteq\mathbb{R}^2$, a vertical section at $x$ is the set of $y$ such that $(x,y)\in D$. A horizontal section at $y$ is the set of $x$ such that $(x,y)\in D$. Describing these sections gives the bounds for the corresponding iterated integrals.

Fubini’s Theorem and the Section Qualification

The relevant hypothesis for switching orders with a signed or complex-valued integrand is absolute integrability. For the triangle above, suppose $f$ is Lebesgue measurable on $D$ and

\int_D |f(x,y)|\,dx\,dy<\infty.

Extend $f$ by zero outside $D$. This gives a Lebesgue-measurable function on $\mathbb{R}^2$ whose absolute integral is finite. Fubini’s theorem says that the inner section integrals exist for almost every outer coordinate, and that the resulting section-integral functions have measurable versions. On any exceptional null set of outer coordinates, assign arbitrary values, such as zero, to those versions. These assignments do not affect the outer Lebesgue integrals.

Theorem: If $f$ is Lebesgue measurable on $D=\{(x,y):0\leq y\leq x\leq 1\}$ and $\int_D |f|\,dx\,dy<\infty$, then, with the section-integral functions defined almost everywhere and assigned arbitrary values on exceptional null sets, $$ \int_D f(x,y)\,dx\,dy = \int_0^1\left(\int_0^x f(x,y)\,dy\right)dx = \int_0^1\left(\int_y^1 f(x,y)\,dx\right)dy. $$

Proof. Define $g:\mathbb{R}^2\to\mathbb{C}$ to equal $f$ on $D$ and zero off $D$ (or take real-valued $f$). Since $D$ is Borel and $f$ is Lebesgue measurable on $D$, $g$ is Lebesgue measurable, and $\int_{\mathbb{R}^2}|g|\,dx\,dy=\int_D|f|\,dx\,dy<\infty$. Fubini’s theorem applied to $g$ gives both orders of integration, with the inner integrals defined for almost every outer coordinate. The zero extension means that the vertical sections integrate over $0\leq y\leq x$ when $0\leq x\leq1$, and the horizontal sections over $y\leq x\leq1$ when $0\leq y\leq1$; all other contributions vanish. The section-integral functions can be assigned arbitrary values on their exceptional null sets without changing their integrals. Since $g=f$ on $D$ and vanishes outside it, both iterated integrals equal $\int_D f\,dx\,dy$. Thus the stated identities hold.

\square

This formulation matters even when the double integral of $|f|$ is finite: measurability and integrability of the two-dimensional function do not guarantee that every individual section is measurable or integrable. Fubini’s conclusion is an almost-everywhere statement about sections, not an assertion about every fixed outer coordinate.

Worked Example: Reverse the Order on a Triangle

Evaluate $\int_D(x+2y)\,dx\,dy$ for $D=\{0\leq y\leq x\leq1\}$. The integrand is continuous and bounded on the bounded region $D$, so its absolute integral is finite and Fubini applies. Using vertical sections first gives

\int_0^1\int_0^x(x+2y)\,dy\,dx = \int_0^1\bigl[xy+y^2\bigr]_{y=0}^{y=x}\,dx = \int_0^1 2x^2\,dx = \frac{2}{3}.

For the other order, the inequalities describing the same triangle give $0\leq y\leq1$ and $y\leq x\leq1$. Therefore

\int_0^1\int_y^1(x+2y)\,dx\,dy = \int_0^1\left(\frac{1-y^2}{2}+2y(1-y)\right)dy = \int_0^1\left(\frac12+2y-\frac52y^2\right)dy = \frac12+1-\frac56 = \frac23.

Both calculations agree. The inner limits changed because fixing $y$, rather than $x$, changes which part of the triangle is traversed.

A Two-Dimensional Simplex

The standard two-dimensional simplex is the triangle

\Delta_2=\{(x,y)\in\mathbb{R}^2:x\geq0,\ y\geq0,\ x+y\leq1\}.

Its vertical sections have $0\leq x\leq1$ and $0\leq y\leq1-x$. Its horizontal sections have $0\leq y\leq1$ and $0\leq x\leq1-y$. The region is symmetric in $x$ and $y$, but the bounds still come from solving the same inequality $x+y\leq1$ for the inner coordinate.

Worked Example: Integrate over the Standard Triangle

Evaluate the integral of $e^{x+y}$ over $\Delta_2$. This integrand is continuous and bounded on the compact region, so it is absolutely integrable. With $y$ integrated first,

\int_0^1\int_0^{1-x}e^{x+y}\,dy\,dx = \int_0^1\left[e^{x+y}\right]_{y=0}^{y=1-x}dx = \int_0^1(e-e^x)\,dx = e-(e-1) = 1.

With $x$ integrated first, the bounds are $0\leq y\leq1$ and $0\leq x\leq1-y$. Direct evaluation gives

\int_0^1\int_0^{1-y}e^{x+y}\,dx\,dy = \int_0^1(e-e^y)\,dy = e-(e-1) = 1.

The agreement is a check on the bounds as well as on the calculation. In either order, the inner integral reaches the sloping boundary $x+y=1$.

Three-Dimensional Simplex Sections

The same method applies in three dimensions. Consider

\Delta_3=\{(x,y,z)\in\mathbb{R}^3:x\geq0,\ y\geq0,\ z\geq0,\ x+y+z\leq1\}.

For the order $dx\,dy\,dz$, first fix $z$, then $y$, then $x$. The inequalities imply $0\leq z\leq1$, $0\leq y\leq1-z$, and $0\leq x\leq1-y-z$. For the reverse order $dz\,dy\,dx$, they imply $0\leq x\leq1$, $0\leq y\leq1-x$, and $0\leq z\leq1-x-y$. These descriptions are obtained by isolating the innermost variable in $x+y+z\leq1$.

Worked Example: Compute the Volume of a Simplex

The constant function $1$ is integrable on the bounded simplex. Its integral is the volume. Using the first set of bounds gives

\begin{aligned} \int_0^1\int_0^{1-z}\int_0^{1-y-z}1\,dx\,dy\,dz &=\int_0^1\int_0^{1-z}(1-y-z)\,dy\,dz\\ &=\int_0^1\frac{(1-z)^2}{2}\,dz\\ &=\frac12\left[z-z^2+\frac{z^3}{3}\right]_0^1 =\frac16. \end{aligned}

Now take the reverse order. Integrating $z$ first gives

\begin{aligned} \int_0^1\int_0^{1-x}\int_0^{1-x-y}1\,dz\,dy\,dx &=\int_0^1\int_0^{1-x}(1-x-y)\,dy\,dx\\ &=\int_0^1\frac{(1-x)^2}{2}\,dx =\frac16. \end{aligned}

In both orders, the section length is the remaining distance to the sloping face $x+y+z=1$. The resulting value $1/6$ is consistent across the two calculations.

A Reliable Procedure for Setting Bounds

When changing order, begin with the region, not with the integral signs. A disciplined setup separates the geometric reasoning from the analytic justification.

Write the region as inequalities.
For example, $x+y\leq1$ with $x,y\geq0$ specifies the two-dimensional simplex.

Choose the outer coordinate.
Project the region onto that coordinate’s axis to determine its full range.

Fix the outer coordinate and solve for the inner one.
The resulting inequalities give the section bounds, which may depend on the fixed coordinate.

Check the integral hypothesis.
For signed or complex $f$, verify $\int_D|f|<\infty$ before invoking Fubini to equate the iterated integrals.

Why the Hypotheses and Bounds Matter

A common geometric error is to reverse the order while leaving the old limits unchanged. For $D=\{0\leq y\leq x\leq1\}$, for instance, the $y$-range is $0$ to $x$ when $x$ is fixed, but the $x$-range is $y$ to $1$ when $y$ is fixed. Writing $0\leq x\leq y\leq1$ instead describes the other half of the unit square.

A separate error is to treat Fubini’s theorem as if it guaranteed valid inner integrals at every outer coordinate. For Lebesgue-measurable functions, the theorem gives section integrability almost everywhere. When an exceptional null set occurs, use the measurable section-integral version supplied by Fubini, assigning arbitrary values on that null set. Do not claim that the original inner integral exists at every coordinate unless an additional condition justifies that claim.

These points make triangular and simplex integrals useful models for more general regions: the geometry determines the section bounds, while the integrability hypothesis determines whether changing the order preserves the value.

Check Your Understanding

For each question, identify the region or theorem condition involved before carrying out any computation.

For $D=\{(x,y):0\leq y\leq x\leq1\}$, what are the outer and inner bounds when $x$ is integrated first?
Describe the bounds for the same region when $y$ is integrated first.
What integrability hypothesis allows Fubini to equate the two iterated integrals for a signed or complex-valued function?
Does Fubini guarantee that every section of a Lebesgue-measurable integrable function is integrable? State the correct qualification.
For $x,y,z\geq0$ and $x+y+z\leq1$, what are the bounds in the order $dz\,dy\,dx$?