Definition of the Riemann Integral

You already know how to find areas of simple shapes — rectangles, triangles, circles. But what is the area under the curve $y = x^2$ from $0$ to $1$? No geometric formula applies directly. The Riemann integral is the answer: it defines “area under a curve” as a limit of increasingly fine rectangular approximations, and does so in a way that is precise enough to prove theorems.

Partitions and Darboux sums

Let $f: [a, b] \to \mathbb{R}$ be a bounded function. A partition of $[a, b]$ is a finite sequence

$$P = \{a = x_0 < x_1 < x_2 < \cdots < x_n = b\}.$$

On each subinterval $[x_{i-1}, x_i]$, $f$ is bounded, so its supremum and infimum exist:

$$M_i \coloneqq \sup_{x \in [x_{i-1},\, x_i]} f(x), \qquad m_i \coloneqq \inf_{x \in [x_{i-1},\, x_i]} f(x).$$

The upper Darboux sum and lower Darboux sum of $f$ over $P$ are:

$$U(f, P) \coloneqq \sum_{i=1}^{n} M_i\,(x_i - x_{i-1}), \qquad L(f, P) \coloneqq \sum_{i=1}^{n} m_i\,(x_i - x_{i-1}).$$

Geometrically, $U(f,P)$ is the total area of rectangles that lie above the graph, and $L(f,P)$ is the total area of rectangles that lie below it. Since $m_i \le M_i$ for every $i$, you always have $L(f,P) \le U(f,P)$.

Refining a partition

A partition $Q$ is a refinement of $P$ if $P \subseteq Q$ (every boundary point of $P$ is also in $Q$). Adding points can only lower the upper sum and raise the lower sum:

$$P \subseteq Q \implies L(f, P) \le L(f, Q) \le U(f, Q) \le U(f, P).$$

A consequence is that every lower sum is at most every upper sum, even for different partitions: if $P_1$ and $P_2$ are any two partitions, then $L(f, P_1) \le U(f, P_2)$.

Upper and lower integrals

Because the set of all lower sums is bounded above (by any upper sum), its supremum exists in $\mathbb{R}$. Likewise, the infimum of all upper sums exists. These define the lower integral and upper integral of $f$:

$$\underline{\int_a^b} f \;\coloneqq\; \sup_P L(f, P), \qquad \overline{\int_a^b} f \;\coloneqq\; \inf_P U(f, P).$$

Since every lower sum is $\le$ every upper sum, you always have $\underline{\int} f \le \overline{\int} f$.

Riemann integrability

A bounded function $f: [a, b] \to \mathbb{R}$ is Riemann integrable if the upper and lower integrals agree:

$$\underline{\int_a^b} f \;=\; \overline{\int_a^b} f.$$

When they do, their common value is called the Riemann integral of $f$ on $[a, b]$, written

$$\int_a^b f(x)\,dx \;=\; \int_a^b f.$$

The Darboux criterion

A practical reformulation: $f$ is Riemann integrable if and only if for every $\varepsilon > 0$ there exists a partition $P$ with

$$U(f, P) - L(f, P) \;<\; \varepsilon. \tag{1}$$

This is the working tool for most integrability proofs.

Equivalent Riemann sum definition

You can also define the integral via Riemann sums directly. For a partition $P$ and a choice of sample points $\xi_i \in [x_{i-1}, x_i]$, the Riemann sum is

$$S(f, P, \boldsymbol\xi) \;\coloneqq\; \sum_{i=1}^n f(\xi_i)\,(x_i - x_{i-1}).$$

The mesh of $P$ is $\|P\| \coloneqq \max_i(x_i - x_{i-1})$. Then $f$ is Riemann integrable with value $I$ if and only if: for every $\varepsilon > 0$ there exists $\delta > 0$ such that $\|P\| < \delta$ implies $|S(f,P,\boldsymbol\xi) - I| < \varepsilon$ for every choice of sample points.

The Darboux and Riemann-sum formulations are equivalent; the Darboux approach tends to be cleaner for proofs.

Every continuous function is integrable

Theorem. If $f: [a, b] \to \mathbb{R}$ is continuous, then $f$ is Riemann integrable.

Proof. A continuous function on a closed bounded interval is uniformly continuous: for every $\varepsilon > 0$ there exists $\delta > 0$ such that $|x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon/(b-a)$.

Given $\varepsilon > 0$, choose $\delta$ as above and let $P$ be any partition with mesh $\|P\| < \delta$. On each subinterval $[x_{i-1}, x_i]$, the oscillation satisfies

$$M_i - m_i \;\le\; \frac{\varepsilon}{b-a},$$

because any two points in $[x_{i-1}, x_i]$ are within distance $\delta$ of each other. Therefore

$$U(f,P) - L(f,P) \;=\; \sum_{i=1}^n (M_i - m_i)(x_i - x_{i-1}) \;\le\; \frac{\varepsilon}{b-a} \cdot (b-a) \;=\; \varepsilon.$$

By the Darboux criterion $(1)$, $f$ is integrable. $\square$

Worked example: $\int_0^1 x^2\,dx$

Take $f(x) = x^2$ and the uniform partition $P_n = \{0, \tfrac{1}{n}, \tfrac{2}{n}, \ldots, 1\}$. Since $f$ is increasing on $[0,1]$:

$$M_i = \left(\frac{i}{n}\right)^2, \qquad m_i = \left(\frac{i-1}{n}\right)^2, \qquad \Delta x_i = \frac{1}{n}.$$

The upper and lower sums are:

$$U(f, P_n) = \frac{1}{n} \sum_{i=1}^n \frac{i^2}{n^2} = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2},$$ $$L(f, P_n) = \frac{1}{n} \sum_{i=0}^{n-1} \frac{i^2}{n^2} = \frac{(n-1)(2n-1)}{6n^2}.$$

Their difference is $U - L = \frac{(n+1)(2n+1) - (n-1)(2n-1)}{6n^2} = \frac{6n}{6n^2} = \frac{1}{n} \to 0$. So the Darboux criterion is satisfied. Both sums converge to $\frac{1}{3}$, confirming

$$\int_0^1 x^2\,dx \;=\; \frac{1}{3}.$$

Summary

• A partition of $[a,b]$ divides it into subintervals; Darboux sums $L(f,P)$ and $U(f,P)$ bound the “area under $f$” from below and above.
• The lower integral $\underline{\int} f$ and upper integral $\overline{\int} f$ are the supremum and infimum of all lower and upper sums, respectively.
• $f$ is Riemann integrable when $\underline{\int} f = \overline{\int} f$; the common value is $\int_a^b f$.
• Darboux criterion: $f$ is integrable iff for every $\varepsilon > 0$ some partition achieves $U(f,P) - L(f,P) < \varepsilon$.
• Every continuous function on $[a,b]$ is integrable, by uniform continuity.