Prove Riemann Integrability of a Continuous Function

From Continuity to Riemann Integrability

Taylor’s Theorem, the subject of the previous tutorial, depends on continuity and differentiability across an interval. Here we turn to another central question about functions on intervals: when can their values be integrated in the Riemann sense? The key link is that continuity on a closed bounded interval gives uniform control of how much the function can vary over short subintervals.

We will use that control to make the difference between upper and lower sums as small as desired. First, we set up those sums and prove the criterion that makes this strategy work. Then uniform continuity will provide the needed estimate for every continuous function on the interval.

Definition: Let $a<b$, and let $f:[a,b]\to\mathbb{R}$ be bounded. A partition of $[a,b]$ is a finite ordered set $P=\{x_0,x_1,\ldots,x_n\}$ with $a=x_0<x_1<\cdots<x_n=b$. Write $\Delta x_i=x_i-x_{i-1}$. For each subinterval $[x_{i-1},x_i]$, let $M_i=\sup\{f(x):x\in[x_{i-1},x_i]\}$ and $m_i=\inf\{f(x):x\in[x_{i-1},x_i]\}$. The upper sum and lower sum of $f$ for $P$ are $$ U(f,P)=\sum_{i=1}^{n}M_i\Delta x_i \qquad\text{and}\qquad L(f,P)=\sum_{i=1}^{n}m_i\Delta x_i. $$ The upper and lower integrals are, respectively, $\inf_P U(f,P)$ and $\sup_P L(f,P)$, where $P$ ranges over all partitions. The bounded function $f$ is Riemann integrable on $[a,b]$ if these two numbers are equal; their common value is its Riemann integral.

The supremum and infimum in each subinterval exist because $f$ is bounded. Also $m_i\leq M_i$, so $L(f,P)\leq U(f,P)$. A refinement of a partition adds division points without removing any of the original ones. Refinement can only decrease the upper sum and increase the lower sum: splitting a subinterval replaces its supremum by suprema no larger than the original one, and its infimum by infima no smaller than the original one. The lengths of the pieces add back to the length of that subinterval, which gives the claimed inequalities for the sums.

Any two partitions have a common refinement, obtained by taking the union of their division points. In particular, for arbitrary partitions $P$ and $Q$, a common refinement $R$ gives $L(f,P)\leq L(f,R)\leq U(f,R)\leq U(f,Q)$. Thus every lower sum is at most every upper sum, and the lower integral cannot exceed the upper integral.

The Darboux Criterion

The definition by upper and lower integrals has a useful equivalent form. It says that it is enough to find, for each prescribed error, a partition whose upper and lower sums differ by less than that error. We prove this criterion so that the later continuity argument has a precise target.

Theorem (Darboux Criterion for Riemann Integrability): A bounded function $f:[a,b]\to\mathbb{R}$ is Riemann integrable if and only if, for every $\varepsilon>0$, there is a partition $P$ of $[a,b]$ such that $$ U(f,P)-L(f,P)<\varepsilon. $$

Proof. Denote the upper integral by $\overline I=\inf_P U(f,P)$ and the lower integral by $\underline I=\sup_P L(f,P)$. We have already seen that $\underline I\leq\overline I$.

Suppose first that $f$ is Riemann integrable, so $\overline I=\underline I=I$. Given $\varepsilon>0$, the definition of infimum supplies a partition $P$ with $U(f,P)<I+\varepsilon/2$. The definition of supremum supplies a partition $Q$ with $L(f,Q)>I-\varepsilon/2$. Let $R$ be a common refinement. By the refinement inequalities,

U(f,R)\leq U(f,P)<I+\frac{\varepsilon}{2}, \qquad L(f,R)\geq L(f,Q)>I-\frac{\varepsilon}{2}.

Subtracting the second strict bound from the first gives $U(f,R)-L(f,R)<\varepsilon$, as required.

Conversely, suppose that for every $\varepsilon>0$ there is a partition $P$ with $U(f,P)-L(f,P)<\varepsilon$. Since $\overline I\leq U(f,P)$ and $L(f,P)\leq\underline I$, we have

0\leq \overline I-\underline I \leq U(f,P)-L(f,P)<\varepsilon.

This holds for every positive $\varepsilon$, so $\overline I-\underline I=0$. Therefore the upper and lower integrals are equal, and $f$ is Riemann integrable. $\square$

Continuous Functions on Closed Intervals

Now let $f$ be continuous on $[a,b]$, where $a<b$. By the Extreme Value Theorem, $f$ is bounded there, so its upper and lower sums are defined. By the theorem on uniform continuity on a compact set, $f$ is uniformly continuous on $[a,b]$. That uniform control is what lets us choose one partition that works across the entire interval.

Theorem (Continuous Functions on Closed Intervals Are Riemann Integrable): If $f:[a,b]\to\mathbb{R}$ is continuous and $a<b$, then $f$ is Riemann integrable on $[a,b]$.

Proof. Let $\varepsilon>0$. Uniform continuity of $f$ on $[a,b]$ gives a $\delta>0$ such that, for all $x,y\in[a,b]$,

|x-y|<\delta \quad\Longrightarrow\quad |f(x)-f(y)|<\frac{\varepsilon}{2(b-a)}.

Choose a positive integer $n$ large enough that $(b-a)/n<\delta$, and use the equally spaced partition $x_i=a+i(b-a)/n$, for $i=0,\ldots,n$. Its subintervals all have length $\Delta x=(b-a)/n<\delta$. If $x$ and $y$ belong to the same subinterval, then $|x-y|\leq\Delta x<\delta$. Hence the difference between any two values of $f$ on that subinterval is less than $\varepsilon/(2(b-a))$. Taking the supremum and infimum shows that its oscillation satisfies

M_i-m_i\leq\frac{\varepsilon}{2(b-a)}.

(The bound is non-strict because a supremum or infimum need not be attained.) Consequently, for this partition,

U(f,P)-L(f,P) =\sum_{i=1}^{n}(M_i-m_i)\Delta x_i \leq \frac{\varepsilon}{2(b-a)}\sum_{i=1}^{n}\Delta x_i =\frac{\varepsilon}{2}<\varepsilon.

The Darboux Criterion now implies that $f$ is Riemann integrable on $[a,b]$. $\square$

Worked Estimates with Upper and Lower Sums

Worked Example: The Square Function on a Unit Interval

Consider $f(x)=x^2$ on $[0,1]$, with the equally spaced partition $x_i=i/n$. Since $x^2$ is increasing on this interval, the supremum on $[(i-1)/n,i/n]$ is $i^2/n^2$, and the infimum is $(i-1)^2/n^2$. Each subinterval has length $1/n$. Therefore

U(f,P)-L(f,P) =\sum_{i=1}^{n}\left(\frac{i^2-(i-1)^2}{n^2}\right)\frac1n =\frac{1}{n^3}\sum_{i=1}^{n}(2i-1) =\frac{n^2}{n^3} =\frac1n.

Here we used $\sum_{i=1}^{n}(2i-1)=n^2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$; then $U(f,P)-L(f,P)=1/n<\varepsilon$. This verifies integrability directly by the Darboux Criterion.

Worked Example: An Absolute-Value Function

Let $f(x)=|x-1|$ on $[0,2]$, and divide the interval into $n$ equal subintervals of length $2/n$. The inequality $\bigl||x-1|-|y-1|\bigr|\leq|x-y|$ follows from the triangle inequality: $|x-1|\leq|x-y|+|y-1|$, and interchanging $x$ and $y$ gives the reverse difference bound. Thus, on each subinterval, any two function values differ by at most $2/n$, so $M_i-m_i\leq2/n$. It follows that

U(f,P)-L(f,P) =\sum_{i=1}^{n}(M_i-m_i)\frac{2}{n} \leq n\left(\frac{2}{n}\right)\left(\frac{2}{n}\right) =\frac{4}{n}.

For any $\varepsilon>0$, take an integer $n>4/\varepsilon$. Then the upper-minus-lower sum is less than $\varepsilon$, proving integrability. The estimate remains valid for the subinterval that contains the corner at $x=1$; no differentiability at that point is needed.

Worked Example: The Square-Root Function Near an Endpoint

Consider $f(x)=\sqrt{x}$ on $[0,1]$. For $x\geq y\geq0$, we have $(\sqrt{x}-\sqrt{y})^2\leq x-y$, because $\sqrt{x}-\sqrt{y}\leq\sqrt{x}+\sqrt{y}$ and $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=x-y$. Therefore $|\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$ for all $x,y\in[0,1]$. On the equally spaced partition into $n$ subintervals, any two points in a subinterval are at distance at most $1/n$, so $M_i-m_i\leq1/\sqrt n$. Hence

U(f,P)-L(f,P) =\sum_{i=1}^{n}(M_i-m_i)\frac1n \leq n\left(\frac1{\sqrt n}\right)\frac1n =\frac1{\sqrt n}.

Choose an integer $n>1/\varepsilon^2$. Then $1/\sqrt n<\varepsilon$, so the Darboux Criterion proves that $\sqrt{x}$ is Riemann integrable. The estimate works even on the first subinterval adjoining zero, where the derivative of $\sqrt{x}$ is not defined at the endpoint.

Why the Closed Interval Matters

The proof has two distinct ingredients. Boundedness ensures that the upper and lower sums are finite. Uniform continuity ensures that the oscillation on every short subinterval is small, allowing the total gap between the sums to be made small. On a closed bounded interval, continuity supplies both properties: boundedness by the Extreme Value Theorem and uniform continuity by the compactness theorem for continuous functions.

It is not enough to know that a function is continuous at each point without checking the domain and the uniform estimate. For instance, $f(x)=1/x$ is continuous on $(0,1]$, but it is unbounded there and so does not meet the bounded-function definition used here on a closed interval. The theorem is specifically about continuity on the whole closed interval $[a,b]$.

A useful proof habit is to keep the roles of the quantities separate. The function’s oscillation $M_i-m_i$ controls the height of each contribution to the gap, while $\Delta x_i$ is its width. Uniform continuity makes the heights small; summing the widths gives exactly $b-a$. That is why the resulting estimate is a total bound, rather than merely a pointwise observation.

Ensure the sums exist.
Use boundedness of the function on the interval so each subinterval has finite supremum and infimum.

Control oscillation uniformly.
Use uniform continuity to make the difference between any two function values on a sufficiently short subinterval small.

Sum the gaps.
Bound each $M_i-m_i$, multiply by its subinterval length, and use $\sum_i\Delta x_i=b-a$.

Apply the criterion.
Make the upper-minus-lower sum smaller than an arbitrary positive error and invoke the Darboux Criterion.

Check Your Understanding

Use the definitions, criterion, and proof above to answer these questions.

Why does boundedness matter when defining the upper and lower sums?
What refinement inequalities relate the upper and lower sums of a partition and its refinement?
In the proof of the Darboux Criterion, why is a common refinement useful when integrability is assumed?
How does uniform continuity bound $M_i-m_i$ on a sufficiently short subinterval?
Why is the total length $b-a$ essential to the final estimate for $U(f,P)-L(f,P)$?