Multiple Riemann integration and Fubini

An iterated integral is a sequence of one-variable integrals. A multiple integral is a limit of sums over small cells. Their equality requires a theorem. We establish it first for continuous functions on rectangles, where the hypotheses are transparent, and then use the result to compute and interpret region integrals.

1 Upper and lower sums on rectangles

Let $R=[a,b]\times[c,d]$ with $a<b$, $c<d$. A rectangular grid partitions $R$ into cells $R_{ij}$. For bounded $f$, let $M_{ij}$ and $m_{ij}$ be the supremum and infimum on each cell. Its upper and lower sums are

$$U(f,P)=\sum M_{ij}|R_{ij}|,\qquad L(f,P)=\sum m_{ij}|R_{ij}|.$$

The upper integral is the infimum of upper sums, and the lower integral is the supremum of lower sums. Refinement increases lower sums and decreases upper sums; common refinements show every lower sum is at most every upper sum. If the two integrals coincide, $f$ is Riemann integrable and their common value is $\iint_Rf$.

The equivalent criterion is: for every $\varepsilon>0$ there is a partition with $U-L<\varepsilon$. To prove sufficiency, put the upper and lower integrals between these two sums. Necessity follows by choosing nearly optimal upper and lower partitions and taking a common refinement.

Theorem. A continuous function on $R$ is integrable.

Proof. Compactness makes it uniformly continuous. Choose a grid whose cells have diameter small enough that the oscillation of $f$ on each cell is below $\varepsilon/|R|$. Then $U-L=\sum(M_{ij}-m_{ij})|R_{ij}|<\varepsilon$. $\square$

Boundedness alone is not enough: the indicator of points whose two coordinates are rational has infimum 0 and supremum 1 in every positive-area cell, so every lower sum is 0 and every upper sum is $|R|$.

2 Fubini for continuous functions

Theorem. If $f$ is continuous on $R$, then

$$\iint_R f(x,y)\,dA=\int_a^b\left(\int_c^d f(x,y)\,dy\right)dx =\int_c^d\left(\int_a^b f(x,y)\,dx\right)dy.$$

Proof. Every slice is continuous and hence integrable. Let $F(x)=\int_c^d f(x,y)dy$. Uniform continuity gives

$$|F(x)-F(x')|\le(d-c)\sup_y|f(x,y)-f(x',y)|,$$

so $F$ is continuous. For a fixed grid, on each $x$ interval the inner integral lies between the sum of $m_{ij}\Delta y_j$ and the sum of $M_{ij}\Delta y_j$. Integrate these bounds over that interval and add: $L(f,P)\le\int_a^bF\le U(f,P)$. Choose grids with $U-L\to0$ to identify this value with the double integral. Interchange $x$ and $y$ for the second equality. $\square$

Continuity is a convenient sufficient hypothesis, not a necessary one. More general Riemann or Lebesgue Fubini theorems have different hypotheses; neither is being silently invoked for arbitrary unbounded or improper integrals here.

2 A precise noncontinuous Fubini theorem

Theorem. Let bounded $f$ be Riemann integrable on $[a,b]\times[c,d]$. Suppose for every $x\in[a,b]$ its slice $y\mapsto f(x,y)$ is Riemann integrable. Then

$$F(x)=\int_c^df(x,y)dy$$

is Riemann integrable and $\int_a^bF(x)dx=\iint_Rf$. If every horizontal slice is also integrable, the two iterated integrals agree.

Proof. Choose a rectangular grid. On an $x$ interval $I_i$ and each $y$ interval $J_j$, let $m_{ij},M_{ij}$ be the infimum and supremum of $f$ on the product cell. For every $x\in I_i$, monotonicity of one-variable integrals gives

$$\sum_jm_{ij}|J_j|\le F(x)\le\sum_jM_{ij}|J_j|.$$

Therefore the oscillation of $F$ on $I_i$ is at most $\sum_j(M_{ij}-m_{ij})|J_j|$. Multiplying by $|I_i|$ and summing shows

$$U(F,P_x)-L(F,P_x)\le U(f,P)-L(f,P).$$

The right side can be made arbitrarily small, proving integrability of $F$. Integrating the same lower and upper inequalities over $I_i$ and adding puts $\int F$ between $L(f,P)$ and $U(f,P)$. Taking partitions with vanishing Darboux difference identifies the value. The horizontal result follows by the same argument. $\square$

The slice hypothesis is real. Define $f(x,y)=1$ when $x=0$ and $y$ is rational, and zero otherwise on the unit square. Its double integral is zero because only cells touching the vertical edge can contribute. But the slice at $x=0$ is the rational indicator, not Riemann integrable. Therefore the displayed inner integral is not defined for every $x$, despite existence of the double integral. In this particular example, assigning any finite value to $F(0)$ and zero elsewhere still gives outer integral zero; that is a separately justified finite-point modification, not automatic slice existence.

3 Worked examples and nonrectangular regions

The zero extensions in the region examples below are not continuous on the rectangle, so the continuous theorem of Section 2 alone is insufficient. Use the bounded-Riemann slicing theorem proved in the noncontinuous-Fubini chapter: double Riemann integrability and integrability of every relevant one-variable slice are separate requirements. For these triangle and parabolic-region examples, boundary covers establish the double integral, while each fixed-coordinate slice is a continuous polynomial on a finite interval, extended by zero with at most endpoint jumps. Thus every slice is Riemann integrable, and that broader theorem applies.

On $[0,1]\times[0,2]$,

$$\iint(x+2y)\,dA=\int_0^1[x y+y^2]_{y=0}^{2}dx=\int_0^1(2x+4)dx=5.$$

For the triangle $D=\{(x,y):0\le x\le1,\ 0\le y\le x\}$, integrate over a rectangle after extending by zero outside $D$. For continuous integrands, the only new discontinuities lie on the triangle's boundary. This boundary can be covered by rectangles of arbitrarily small total area; away from that cover uniform continuity controls upper-minus-lower sums. Thus the extension is Riemann integrable and slicing gives

$$\iint_D(x+y)\,dA=\int_0^1\int_0^x(x+y)\,dy\,dx=\frac12.$$

Reversing the order uses $0\le y\le1$ and $y\le x\le1$, not the original bounds with symbols exchanged. It yields $\int_0^1\int_y^1(x+y)dx\,dy=1/2$.

For a nonnegative density $\rho$, mass is $M=\iint_D\rho$ and the centroid is $(M^{-1}\iint_Dx\rho,M^{-1}\iint_Dy\rho)$ when $M>0$. Unit density $\rho=1$ on the triangle gives mass $1/2$ and centroid $(2/3,1/3)$, reflecting that most horizontal extent lies toward the right.

Exercises with complete solutions

E1 Foundation. Integrate $xy$ over $[0,2]\times[0,3]$ in both orders.

S1. Separation or direct integration gives $(\int_0^2x dx)(\int_0^3y dy)=2\cdot9/2=9$. Integrating $y$ first or $x$ first gives the same product because the continuous-rectangle theorem applies.

E2 Region bounds. Reverse the order and evaluate $\int_0^1\int_{x^2}^{1}x\,dy\,dx$.

S2. The region is $0\le y\le1$, $0\le x\le\sqrt y$. The reversed integral is $\int_0^1\int_0^{\sqrt y}x dx\,dy=\int_0^1y/2\,dy=1/4$. Original order gives $\int_0^1(x-x^3)dx=1/4$.

E3 Density. On the unit square, let $\rho(x,y)=1+x$. Find the mass and centroid.

S3. $M=3/2$. The $x$ moment is $\int_0^1x(1+x)dx=5/6$, so $\bar x=5/9$. The $y$ moment is $(3/2)(1/2)=3/4$, so $\bar y=1/2$.

E4 Proof. Show that the function equal to 1 on the line $y=x$ in the unit square and 0 elsewhere is Riemann integrable with integral zero.

S4. Every cell has infimum zero, so lower sums vanish. A uniform $N\times N$ grid has at most $3N$ cells meeting the diagonal, including boundary-touching neighbors. Its upper sum is at most $3/N$, tending to zero. The Darboux criterion proves integral zero despite discontinuity on the diagonal.

Summary

Multiple integration begins with cell sums, not iterated notation. Uniform continuity establishes integrability for continuous functions, and a squeezing proof establishes Fubini on rectangles. Region geometry determines the bounds, while improper and more general measurable settings require their own theorems.