Why Continuous Functions Are Integrable on Rectangles

Why Continuity Is Enough

A double integral on a rectangle is built from rectangular Riemann sums. Each sum uses a function value from each cell, multiplied by that cell’s area. For such sums to approach one definite value as the cells become small, the function cannot vary too much within the cells.

Continuity prevents abrupt changes near each individual point. On a closed rectangle, it gives something stronger: uniform continuity, which controls the function’s variation everywhere with one common distance bound. This lets us make the difference between upper and lower sums as small as we wish.

Rectangular Partitions and Upper and Lower Sums

Let $R=[a,b]\times[c,d]$, where $a<b$ and $c<d$. A rectangular partition $P$ is formed by choosing division points $a=x_0<x_1<\cdots<x_m=b$ and $c=y_0<y_1<\cdots<y_n=d$. Its cells are the rectangles $C_{ij}=[x_{i-1},x_i]\times[y_{j-1},y_j]$. Their interiors do not overlap, and their areas add to the area of $R$.

For a bounded function $f:R\to\mathbb{R}$, define the supremum and infimum of $f$ on each cell. These are used even when the function does not attain its extreme values.

Definition: For each cell $C_{ij}$, let $M_{ij}=\sup_{(x,y)\in C_{ij}}f(x,y)$ and $m_{ij}=\inf_{(x,y)\in C_{ij}}f(x,y)$. The upper sum $U(f,P)$ and lower sum $L(f,P)$ are $$ U(f,P)=\sum_{i=1}^{m}\sum_{j=1}^{n}M_{ij}\,\Delta A_{ij}, \qquad L(f,P)=\sum_{i=1}^{m}\sum_{j=1}^{n}m_{ij}\,\Delta A_{ij}, $$ where $\Delta A_{ij}=(x_i-x_{i-1})(y_j-y_{j-1})$. Since $f$ is bounded, all these suprema and infima are finite.

For every choice of sample point $(x_{ij}^*,y_{ij}^*)$ in each cell, its Riemann sum lies between the lower and upper sums: each sampled function value is at least $m_{ij}$ and at most $M_{ij}$, and each cell area is positive. Thus $L(f,P)\leq S(f,P)\leq U(f,P)$.

Theorem (Darboux criterion): A bounded function on a rectangle is Riemann integrable if and only if, for every $\eta>0$, there is a rectangular partition $P$ such that $U(f,P)-L(f,P)<\eta$. The upper and lower sums squeeze all sampled sums; when their difference can be made arbitrarily small, those sums have a common limiting value as the partition mesh tends to zero.

The difference between the upper and lower sums has a useful cell-by-cell form:

U(f,P)-L(f,P) = \sum_{i=1}^{m}\sum_{j=1}^{n} (M_{ij}-m_{ij})\,\Delta A_{ij}.

The quantity $M_{ij}-m_{ij}$ is the oscillation of $f$ on that cell: it measures how much the function can vary there. The plan is to make every cell’s oscillation small while the total cell area remains the fixed area of $R$.

Uniform Continuity Controls Every Cell

A function $f:R\to\mathbb{R}$ is uniformly continuous if, for every $\varepsilon>0$, there is a $\delta>0$ such that for all points $p,q\in R$, distance less than $\delta$ implies $|f(p)-f(q)|<\varepsilon$. The same $\delta$ works throughout the rectangle; it does not depend on the location of $p$ or $q$.

Every continuous function on a closed, bounded rectangle is uniformly continuous. This is the Heine–Cantor theorem. The compactness of the rectangle is essential to this conclusion. A short proof explains the role it plays.

Assume uniform continuity fails.
Then there is some $\varepsilon_0>0$ such that, for every positive integer $k$, points $p_k,q_k\in R$ can be chosen with $\lVert p_k-q_k\rVert<1/k$ but $|f(p_k)-f(q_k)|\geq\varepsilon_0$.

Use compactness.
Because $R$ is closed and bounded, the sequence $(p_k)$ has a convergent subsequence $(p_{k_\ell})$ with limit $p\in R$. Also, $\lVert q_{k_\ell}-p\rVert\leq\lVert q_{k_\ell}-p_{k_\ell}\rVert+\lVert p_{k_\ell}-p\rVert\to0$, so $q_{k_\ell}\to p$ as well.

Apply continuity and reach a contradiction.
Continuity at $p$ gives $f(p_{k_\ell})\to f(p)$ and $f(q_{k_\ell})\to f(p)$. Therefore $|f(p_{k_\ell})-f(q_{k_\ell})|\to0$, contradicting the fact that each difference is at least $\varepsilon_0$. Hence $f$ is uniformly continuous.

Proof That Continuous Functions Are Integrable

We can now combine uniform continuity with the upper-minus-lower sum formula. The diameter of a cell is the greatest distance between two points in it. If a cell has side lengths $\Delta x_i$ and $\Delta y_j$, its diameter is $\sqrt{(\Delta x_i)^2+(\Delta y_j)^2}$.

Theorem: If $f$ is continuous on a closed, bounded rectangle $R=[a,b]\times[c,d]$, then $f$ is Riemann integrable on $R$.

Proof. Write $A=(b-a)(d-c)>0$ for the area of $R$, and let $\eta>0$ be any desired upper bound for the difference between the upper and lower sums. By uniform continuity, there is a $\delta>0$ such that points less than $\delta$ apart have function values differing by less than $\eta/(2A)$.

Divide each side of $R$ into $N$ equal parts. The width of each cell is $(b-a)/N$, and its height is $(d-c)/N$. Therefore its diameter is

\frac{\sqrt{(b-a)^2+(d-c)^2}}{N}.

Choose $N$ large enough that this diameter is less than $\delta$. Any two points in the same cell are then less than $\delta$ apart, so their function values differ by less than $\eta/(2A)$. Consequently, the supremum minus the infimum on each cell is at most $\eta/(2A)$. Multiplying these oscillation bounds by the cell areas and adding gives

U(f,P)-L(f,P) \leq \frac{\eta}{2A}\sum_{i,j}\Delta A_{ij} = \frac{\eta}{2A}A = \frac{\eta}{2} < \eta.

The sum of the cell areas equals $A$, because the cells partition the rectangle. We have shown that for every $\eta>0$, there is a partition with upper-minus-lower sum less than $\eta$. The Darboux criterion therefore implies that $f$ is Riemann integrable. $\square$

Worked Examples: Bounding the Upper-minus-Lower Sum

Worked Example: A Linear Function on the Unit Square

Let $f(x,y)=x+y$ on $R=[0,1]\times[0,1]$. Divide each side into $n$ equal intervals. The cell indexed by $i,j$, for $i,j=0,\ldots,n-1$, is

C_{ij}= \left[\frac{i}{n},\frac{i+1}{n}\right] \times \left[\frac{j}{n},\frac{j+1}{n}\right].

Since $x+y$ increases with either coordinate, the infimum is $m_{ij}=(i+j)/n$, and the supremum is $M_{ij}=(i+j+2)/n$. Thus $M_{ij}-m_{ij}=2/n$. Each cell has area $1/n^2$, and there are $n^2$ cells. It follows that

U(f,P)-L(f,P) = n^2\left(\frac{2}{n}\right)\left(\frac{1}{n^2}\right) = \frac{2}{n}.

For example, with $n=4$, the difference is $2/4=1/2$. In general, if a target $\eta>0$ is given, choosing an integer $n>2/\eta$ makes $2/n<\eta$. This agrees with the theorem’s conclusion.

Worked Example: A Quadratic and Linear Function

Let $f(x,y)=x^2+y$ on $R=[0,1]\times[0,2]$. Use $n$ equal intervals in each coordinate. For $i,j=0,\ldots,n-1$, the cell has $x$-interval $[i/n,(i+1)/n]$ and $y$-interval $[2j/n,2(j+1)/n]$. The function increases in both coordinates on this rectangle, so

m_{ij}=\frac{i^2}{n^2}+\frac{2j}{n}, \qquad M_{ij}=\frac{(i+1)^2}{n^2}+\frac{2(j+1)}{n}.

Subtracting gives the oscillation on this cell:

M_{ij}-m_{ij} = \frac{(i+1)^2-i^2}{n^2}+\frac{2}{n} = \frac{2i+1}{n^2}+\frac{2}{n}.

Each cell has area $(1/n)(2/n)=2/n^2$. Summing the oscillations first over $i$, for one fixed $j$, gives

\sum_{i=0}^{n-1}\left(\frac{2i+1}{n^2}+\frac{2}{n}\right) = \frac{1+3+\cdots+(2n-1)}{n^2}+2 = \frac{n^2}{n^2}+2 = 3.

There are $n$ rows of cells, so the sum of all cell oscillations is $3n$. Multiplying by the common cell area yields

U(f,P)-L(f,P) = 3n\left(\frac{2}{n^2}\right) = \frac{6}{n}.

For $n=6$, the difference is $6/6=1$; as $n$ increases, $6/n$ tends to zero. More generally, choosing an integer $n>6/\eta$ makes this difference less than any prescribed $\eta>0$.

Worked Example: A Distance Function on the Unit Square

Let $f(x,y)=\sqrt{x^2+y^2}$ on $R=[0,1]\times[0,1]$. Divide both sides into $n$ equal intervals. The diameter of each cell is $\sqrt{(1/n)^2+(1/n)^2}=\sqrt{2}/n$.

For points $p,q$ in one cell, the reverse triangle inequality gives $\big|\lVert p\rVert-\lVert q\rVert\big|\leq\lVert p-q\rVert$. Hence the function values on that cell differ by at most $\sqrt{2}/n$, and its supremum minus infimum is at most $\sqrt{2}/n$. Each cell has area $1/n^2$, and there are $n^2$ cells, so

U(f,P)-L(f,P) \leq n^2\left(\frac{\sqrt{2}}{n}\right)\left(\frac{1}{n^2}\right) = \frac{\sqrt{2}}{n}.

For $n=4$, this bound is $\sqrt{2}/4$. To make the upper-minus-lower sum less than $\eta>0$, choose $n>\sqrt{2}/\eta$. This example uses a direct distance bound on the oscillation; it does not require finding the exact supremum and infimum on every cell.

A Common Pitfall: Maximum Is Not Always Supremum

For a bounded function, it is not always correct to define $M_{ij}$ as the largest value the function takes on a cell. Consider the bounded function on the unit square given by

f(x,y)= \begin{cases} x, & x<1,\\ 0, & x=1. \end{cases}

On the entire square, the values approach $1$ as $x$ approaches $1$ from below, but the function never takes the value $1$. There is no largest value. The statement “the maximum is $1$” is therefore incorrect: $1$ is not attained. The correct upper bound for the cell is its supremum, $\sup_R f=1$. Its infimum is $0$.

For continuous functions on closed cells, the extreme value theorem ensures that the supremum and infimum are attained, so they are also a maximum and minimum. But the upper- and lower-sum definitions must work for all bounded functions, including ones that do not attain their bounds.

Key takeaway: Uniform continuity makes the oscillation on every sufficiently small cell small. The cell areas add to the fixed area of the rectangle, so the total upper-minus-lower sum can be made arbitrarily small. The Darboux criterion then gives Riemann integrability.

Check Your Understanding

Use the definitions and estimates in this tutorial to answer the following.

Why are suprema and infima used to define upper and lower sums for a general bounded function?
What does uniform continuity provide that pointwise continuity alone does not state directly?
In the integrability proof, why does the sum of the cell areas equal the area of the whole rectangle?
For $f(x,y)=x+y$ on the unit square with $n$ equal subdivisions per side, what is $U(f,P)-L(f,P)$?
Why does the compactness of a closed rectangle matter in the proof that a continuous function there is uniformly continuous?