Integrability of Continuous Functions

The integrability of continuous functions is the load-bearing beam beneath the entire calculus curriculum: the First Fundamental Theorem of Calculus and the Mean Value Theorem for Integrals both presuppose that $\int_a^b f$ is a well-defined real number before differentiating or estimating it. In numerical analysis, quadrature methods (Simpson's rule, trapezoidal rule) are justified by exactly this kind of uniform-continuity error control — the same $\delta$-vs-mesh argument bounds numerical quadrature error. In probability and statistics, continuity of density functions on bounded supports guarantees that expectations $\mathbb{E}[g(X)] = \int g(x) f_X(x)\,dx$ are well-posed integrals rather than formal expressions. In physics and engineering, continuity of physical fields (temperature, potential, stress) over bounded regions is routinely invoked to justify that line and volume integrals used in energy or work computations actually exist. In theoretical computer science, uniform continuity estimates underlie rigorous error bounds for algorithms that approximate integrals or continuous functions by discretized samples, connecting this 19th-century theorem directly to modern computational guarantees.

:::quiz[Which property of $[a,b]$ is essential to upgrade continuity to uniform continuity?]

• $[a,b]$ is an interval of real numbers
• $[a,b]$ is compact (closed and bounded) ✓
• $[a,b]$ contains rational numbers
• $[a,b]$ has positive length

:::quiz[What breaks in the theorem if $f$ is continuous on $(0,1]$ but domain is not compact, as with $f(x) = \sin(1/x)$?]

• Nothing breaks; the theorem still applies
• $f$ fails to be bounded near $0$
• Continuity no longer guarantees uniform continuity, so the oscillation on subintervals near $0$ cannot be uniformly controlled ✓
• The Cauchy criterion becomes false in general

:::faq[Does uniform continuity alone guarantee integrability, or is compactness of $[a,b]$ still doing work elsewhere in the proof?] Compactness of $[a,b]$ is used twice: first, to upgrade continuity to uniform continuity (Step 2), and second, to guarantee that $f$ attains its supremum and infimum on each compact subinterval $[x_{i-1},x_i]$ via the Extreme Value Theorem (Step 5), which is what lets us write $M_i - m_i$ as an actual difference of function values $f(u_i)-f(v_i)$ rather than merely a difference of suprema/infima that might not be attained. Both uses trace back to the same underlying compactness of closed bounded intervals in $\mathbb{R}$.

Ask the tutor: How does this integrability proof change if $f$ is allowed finitely many jump discontinuities on $[a,b]$?

Picture the graph of a continuous function over $[a,b]$, and imagine slicing $[a,b]$ into thin vertical strips. Over each strip $[x_{i-1}, x_i]$, the function's values range between a minimum $m_i$ and maximum $M_i$. The upper sum $U(f,P)$ stacks up rectangles that overestimate the area (using the tallest value on each strip); the lower sum $L(f,P)$ underestimates (using the shortest value). The gap $U(f,P) - L(f,P)$ measures how much "slop" the partition leaves. Now here is the crucial fact that continuity supplies: because $f$ is continuous on a closed, bounded interval, it is uniformly continuous. This means that if two points $x,y$ are close together (within some $\delta$, chosen once and for all, not depending on where $x,y$ sit in $[a,b]$), then $f(x)$ and $f(y)$ are guaranteed to be close together (within $\varepsilon$). If every strip of the partition has width less than this $\delta$, then $f$ cannot vary by more than $\varepsilon$ within any single strip — so $M_i - m_i \le \varepsilon$ for every $i$ simultaneously. This is exactly the uniform control needed to make $U(f,P)-L(f,P)$ small, strip by strip, all at once. Without uniform continuity — if $\delta$ were allowed to depend on the location — some strip somewhere might harbor huge oscillation no matter how fine the partition, and the gap between upper and lower sums could refuse to shrink. Uniform continuity is what forecloses that possibility.

:::theorem[Continuous Functions on $[a,b]$ are Riemann Integrable] Let $f:[a,b] \to \mathbb{R}$ be continuous on the closed, bounded interval $[a,b]$, where $a < b$. Then $f$ is Riemann integrable on $[a,b]$.

Example 1. Let $f(x) = x^3 - 4x$ on $[-1, 3]$. This is a polynomial, hence continuous on $\mathbb{R}$, hence continuous on the compact interval $[-1,3]$; by the theorem, $f$ is Riemann integrable on $[-1,3]$. Let us carry out the $\varepsilon$-construction explicitly for $\varepsilon = 0.6$, to see the machine work on an asymmetric, non-monotonic function. First bound the derivative to get an explicit modulus of continuity: for $x \in [-1,3]$,

since $3x^2 - 4$ is increasing in $|x|$ on this interval and $|{-1}| < |3|$. So by the Mean Value Theorem, for any $x,y \in [-1,3]$,

(This is a Lipschitz bound, a stronger and more explicit version of uniform continuity available here because $f$ is differentiable with bounded derivative; it is not needed in general but illustrates the mechanism concretely.) To force $|f(x)-f(y)| < \varepsilon' = \frac{0.6}{3-(-1)} = 0.15$, it suffices to take $\delta = \frac{0.15}{23} \approx 0.006522$. The interval has length $b - a = 4$. Choose $n$ so that $\frac{4}{n} < \delta$, i.e. $n > \frac{4}{0.006522} \approx 613.3$, so take $n = 614$. Partitioning $[-1,3]$ into $614$ equal pieces of width $\frac{4}{614} \approx 0.006515 < \delta$ guarantees, by Step 5 of the proof,

and hence

This confirms concretely that a fine enough uniform partition squeezes the upper and lower sums together, exactly as the general theorem promises — no cleverness about where the partition points sit is required, only that the mesh is small enough.

Example 2. Let $g(x) = \sqrt{x}$ on $[0, 5]$. This function is continuous on $[0,5]$ (in particular, right-continuous at $0$, since $\lim_{x\to 0^+}\sqrt x = 0 = \sqrt 0$), so the theorem applies and $g$ is Riemann integrable on $[0,5]$, even though $g$ is not differentiable at $x=0$ and has unbounded derivative near $0$ (so the Lipschitz shortcut of Example 1 fails here). This shows why the theorem's proof must genuinely use uniform continuity rather than a derivative bound. To see uniform continuity directly: given $\varepsilon > 0$, we want $\delta > 0$ with $|\sqrt x - \sqrt y| < \varepsilon$ whenever $x, y \in [0,5]$, $|x-y| < \delta$. Using the identity $\sqrt{x} - \sqrt{y} = \frac{x - y}{\sqrt x + \sqrt y}$ for $x, y \ge 0$ not both zero, and separately checking the case where one of them is $0$:

• If $x, y \in [0,5]$ and (say) $y = 0$: $|\sqrt x - \sqrt 0| = \sqrt x < \varepsilon \iff x < \varepsilon^2$.
• If $x, y > 0$: one shows directly $|\sqrt x - \sqrt y| \le \sqrt{|x-y|}$ (squaring both sides of $\sqrt x \le \sqrt y + \sqrt{x-y}$ when $x \ge y$, which holds since $(\sqrt y + \sqrt{x-y})^2 = y + (x-y) + 2\sqrt{y(x-y)} \ge x$).

So in all cases $|\sqrt x - \sqrt y| \le \sqrt{|x - y|}$ for $x, y \ge 0$. Given $\varepsilon = 0.4$, choose $\delta = \varepsilon^2 = 0.16$; then $|x - y| < 0.16 \implies |\sqrt x - \sqrt y| \le \sqrt{0.16} = 0.4$. Take $\varepsilon' = \frac{0.4}{5} = 0.08$, needing $\delta = 0.08^2 = 0.0064$. Choosing $n$ with $\frac{5}{n} < 0.0064$, i.e. $n > 781.25$, so $n = 782$ subintervals of width $\frac{5}{782} \approx 0.006394 < \delta$, forces

Again the Cauchy criterion is met, confirming integrability at the level of an explicit partition — and demonstrating that the square-root modulus of continuity (rather than a linear Lipschitz bound) still suffices, because the theorem only ever needs some uniform continuity estimate, of whatever shape.