Fubini’s Theorem on Rectangular Boxes

Why the Order of Integration Matters

A triple integral over a box can be written as an iterated integral: integrate with respect to one variable, then another, then the third. The order can affect how much work the calculation requires, but under the conditions of Fubini’s theorem it does not affect the value.

The theorem makes a precise connection between the triple Riemann integral and its iterated integrals. Its hypotheses matter: here we consider a function continuous on a closed, bounded rectangular box. Continuity ensures that the integral exists and that integrating in any order gives the same result.

Boxes and Iterated Integrals

Let \(a\leq b\), \(c\leq d\), and \(e\leq f\). The rectangular box is

B=[a,b]\times[c,d]\times[e,f].

A function \(g:B\to\mathbb{R}\) is continuous on \(B\) if it is continuous at every point of the box, including its boundary. Since \(B\) is closed and bounded, a continuous function on \(B\) is bounded and Riemann integrable.

In an iterated integral, the innermost integral is evaluated first. For example,

\int_a^b\int_c^d\int_e^f g(x,y,z)\,dz\,dy\,dx

means: first integrate in \(z\), holding \(x\) and \(y\) fixed; next integrate the result in \(y\); and finally integrate in \(x\). Each variable keeps its own bounds, wherever it appears in the order.

Definition: An iterated integral over a box is the result of performing successive one-variable integrals, from the innermost differential outward. Its bounds specify the interval for the variable next to each differential.

Fubini’s Theorem on a Rectangular Box

Theorem (Fubini for continuous functions): If \(g\) is continuous on \(B=[a,b]\times[c,d]\times[e,f]\), then its triple Riemann integral exists. It equals each of the six iterated integrals obtained by integrating once with respect to each variable, with the bounds for each variable unchanged:

\begin{aligned} \iiint_B g\,dV &=\int_a^b\int_c^d\int_e^f g(x,y,z)\,dz\,dy\,dx\\ &=\int_a^b\int_e^f\int_c^d g(x,y,z)\,dy\,dz\,dx\\ &=\int_c^d\int_a^b\int_e^f g(x,y,z)\,dz\,dx\,dy\\ &=\int_c^d\int_e^f\int_a^b g(x,y,z)\,dx\,dz\,dy\\ &=\int_e^f\int_a^b\int_c^d g(x,y,z)\,dy\,dx\,dz\\ &=\int_e^f\int_c^d\int_a^b g(x,y,z)\,dx\,dy\,dz. \end{aligned}

In each expression the outermost integral is performed last. For instance, in the second expression \(y\) is integrated first, then \(z\), then \(x\). The theorem does not say that the symbols \(dx\), \(dy\), and \(dz\) can be shuffled while their numerical limits stay in their original positions. A variable’s bounds travel with that variable.

Why the Order Can Be Changed

The reason behind the theorem is that every order can be approximated by the same rectangular Riemann sums. Partition each side of \(B\) into small intervals and choose a sample point in every resulting small box. A Riemann sum adds the value of \(g\) at each sample point multiplied by that small box’s volume.

Because \(g\) is continuous on the closed, bounded box, it is uniformly continuous. This means that, for any \(\varepsilon>0\), points close enough to one another have function values differing by less than \(\varepsilon\). As the partitions become finer, the variation of \(g\) within each small box therefore becomes uniformly small. Consequently, the Riemann sums converge to the triple integral.

The same idea applies to an iterated integral. For example, start with the order \(dz\,dy\,dx\). Approximate each one-dimensional integration by midpoint sums over its partition. The resulting nested sum uses the same grid points and the same small-box volumes as the triple Riemann sum. Repeating this for any other order changes the sequence of additions, not the values being added.

Partition the box.
Divide each coordinate interval into small subintervals. Their products form a grid of small boxes.

Form the common sum.
Sample \(g\) once in each small box and multiply by that box’s volume. This gives a triple Riemann sum.

Approximate each iterated integral.
Use midpoint sums for its successive one-variable integrals. The result approaches the same sum, with the same grid and sample values.

Pass to the limit.
Uniform continuity makes the approximation error tend to zero as the grid is refined. Every order therefore has the same limiting value.

More quantitatively, let \(\omega(\delta)\) be the largest possible change in \(g\) between points in \(B\) whose distance is at most \(\delta\). Uniform continuity says \(\omega(\delta)\to0\) as \(\delta\to0\). If the largest diameter of a small box is \(\delta\), the error between its sample value and the values of \(g\) throughout that box is at most \(\omega(\delta)\). Summed over the box, the corresponding error is at most its volume times \(\omega(\delta)\); summed over the whole partition, it is at most \(\operatorname{vol}(B)\omega(\delta)\). The same vanishing-error reasoning applies to the nested midpoint approximations in each order. Thus each iterated integral has the same limit as the triple Riemann sums, proving the theorem.

This argument relies on continuity on the whole box: it supplies uniform control of the function’s variation. Fubini’s theorem has broader versions with different hypotheses, but this continuous-function result does not claim that every function, or every improper or unbounded integral, permits arbitrary changes of order.

Worked Examples

Worked Example: A Constant Function

Let \(g(x,y,z)=1\) on \(B=[0,1]\times[0,2]\times[0,3]\). Compute the iterated integral in two orders.

For the order \(dz\,dy\,dx\), integrate from the inside outward:

\begin{aligned} \int_0^1\int_0^2\int_0^3 1\,dz\,dy\,dx &=\int_0^1\int_0^2 [z]_0^3\,dy\,dx\\ &=\int_0^1\int_0^2 3\,dy\,dx\\ &=\int_0^1 [3y]_0^2\,dx\\ &=\int_0^1 6\,dx\\ &=[6x]_0^1=6. \end{aligned}

For the order \(dx\,dz\,dy\), the bounds remain attached to their variables:

\begin{aligned} \int_0^2\int_0^3\int_0^1 1\,dx\,dz\,dy &=\int_0^2\int_0^3 [x]_0^1\,dz\,dy\\ &=\int_0^2\int_0^3 1\,dz\,dy\\ &=\int_0^2 [z]_0^3\,dy\\ &=\int_0^2 3\,dy=6. \end{aligned}

Both orders give \(6\), the volume \(1\cdot2\cdot3\) of the box. Substituting \(g=1\) confirms that the integrand is unchanged; the agreement of the two computations illustrates the theorem.

Worked Example: An Affine Function

Let \(g(x,y,z)=x+2y+3z\) on \([0,1]\times[0,2]\times[0,1]\). First use the order \(dz\,dy\,dx\):

\begin{aligned} \int_0^1\int_0^2\int_0^1 (x+2y+3z)\,dz\,dy\,dx &=\int_0^1\int_0^2 [xz+2yz+\tfrac32z^2]_0^1\,dy\,dx\\ &=\int_0^1\int_0^2 (x+2y+\tfrac32)\,dy\,dx\\ &=\int_0^1 [xy+y^2+\tfrac32y]_0^2\,dx\\ &=\int_0^1 (2x+4+3)\,dx\\ &=[x^2+7x]_0^1=8. \end{aligned}

Now integrate in the order \(dy\,dx\,dz\):

\begin{aligned} \int_0^1\int_0^1\int_0^2 (x+2y+3z)\,dy\,dx\,dz &=\int_0^1\int_0^1 [xy+y^2+3zy]_0^2\,dx\,dz\\ &=\int_0^1\int_0^1 (2x+4+6z)\,dx\,dz\\ &=\int_0^1 [x^2+4x+6zx]_0^1\,dz\\ &=\int_0^1 (5+6z)\,dz\\ &=[5z+3z^2]_0^1=8. \end{aligned}

The two results agree. In particular, the \(y\)-bounds are \(0\) to \(2\) in both expressions, and the \(x\)- and \(z\)-bounds are each \(0\) to \(1\), even though their positions in the iterated integrals differ.

Worked Example: A Signed Integral

Let \(g(x,y,z)=xz-y\) on \([-1,1]\times[0,2]\times[0,1]\). The function takes both positive and negative values, so the integral is signed. In the order \(dz\,dy\,dx\):

\begin{aligned} \int_{-1}^1\int_0^2\int_0^1 (xz-y)\,dz\,dy\,dx &=\int_{-1}^1\int_0^2 [\tfrac12xz^2-yz]_0^1\,dy\,dx\\ &=\int_{-1}^1\int_0^2 (\tfrac12x-y)\,dy\,dx\\ &=\int_{-1}^1 [\tfrac12xy-\tfrac12y^2]_0^2\,dx\\ &=\int_{-1}^1 (x-2)\,dx\\ &=[\tfrac12x^2-2x]_{-1}^1\\ &=(\tfrac12-2)-(\tfrac12+2)=-4. \end{aligned}

For a different order, \(dy\,dz\,dx\), integrate first with respect to \(y\):

\begin{aligned} \int_{-1}^1\int_0^1\int_0^2 (xz-y)\,dy\,dz\,dx &=\int_{-1}^1\int_0^1 [xyz-\tfrac12y^2]_0^2\,dz\,dx\\ &=\int_{-1}^1\int_0^1 (2xz-2)\,dz\,dx\\ &=\int_{-1}^1 [xz^2-2z]_0^1\,dx\\ &=\int_{-1}^1 (x-2)\,dx=-4. \end{aligned}

The repeated value \(-4\) is consistent with Fubini’s theorem. The negative result is not a failure of the theorem: a triple integral adds signed contributions, not just unsigned volume.

A Common Bounds Mistake

A frequent error is to change the order of the differentials but leave the bounds in their old positions. For the affine function in the previous example, the correct order \(dz\,dx\,dy\) is

\int_0^2\int_0^1\int_0^1 (x+2y+3z)\,dz\,dx\,dy=8.

If one instead writes \(\int_0^1\int_0^2\int_0^1 (x+2y+3z)\,dz\,dx\,dy\), the first, innermost bounds now assign \(x\) the interval \([0,1]\), the middle bounds assign \(z\) the interval \([0,2]\), and the outer bounds assign \(y\) the interval \([0,1]\). That is a different box. Indeed, this incorrect expression evaluates to

\begin{aligned} \int_0^1\int_0^2\int_0^1 (x+2y+3z)\,dz\,dx\,dy &=\int_0^1\int_0^2 (x+2y+\tfrac32)\,dx\,dy\\ &=\int_0^1 (5+4y)\,dy=7, \end{aligned}

not \(8\). The correction is to attach each interval to its variable: \(x\) gets \([0,1]\), \(y\) gets \([0,2]\), and \(z\) gets \([0,1]\), regardless of their order of integration.

Why the Theorem Is Useful

On a rectangular box, Fubini’s theorem gives a choice: the same integral can be approached using any of the six orders. One order may make an inner integral especially simple, while another may reveal a symmetry or make the bounds easier to track. The theorem guarantees that this choice does not change the value, provided the function is continuous on the box and each variable retains its correct interval.

The theorem also gives a conceptual foundation for interpreting triple Riemann sums. A sum over small boxes has a single limiting value; performing the corresponding one-variable integrations successively is another route to that same limit. This is why the order of integration is a computational choice here, not a change in the quantity being integrated.

Check Your Understanding

Use the definitions and theorem to check your understanding of integration over rectangular boxes.

State the continuity condition and the shape of the domain in the version of Fubini’s theorem presented here.
In \(\int_a^b\int_c^d\int_e^f g(x,y,z)\,dz\,dy\,dx\), which variable is integrated first?
When changing from \(dz\,dy\,dx\) to \(dy\,dz\,dx\), which bounds belong to each variable?
In a Riemann-sum justification, why does uniform continuity matter as the partition is refined?
Does Fubini’s theorem require a continuous integrand to be nonnegative? Explain briefly.