The Darboux Integral

The upper/lower sum machinery is the entry point to essentially all of classical integration theory: it underlies the Fundamental Theorem of Calculus, numerical quadrature algorithms (trapezoidal and Simpson's rule error bounds are phrased in terms of how fast $U(f,P) - L(f,P) \to 0$), and the Riemann–Stieltjes Integral used in probability theory to unify discrete and continuous expectations. In computer science, the sandwiching idea reappears in algorithmic analysis of definite-integral approximation and in interval arithmetic, where bounding a function's range on subintervals is exactly the computation of $M_i$ and $m_i$. In physics and engineering, upper and lower sums give rigorous error bars on computed work, area, and probability quantities before any limiting or numerical-analysis argument is invoked. The equivalence between this approach and Riemann's original tagged-sum definition, established via the Equivalence of Riemann and Darboux Integrals, is what lets textbooks freely switch between "Riemann sum" intuition (useful for applications) and "upper/lower sum" rigor (useful for proofs) without any loss of correctness.

:::quiz[If $P \subseteq P^*$ are partitions of $[a,b]$ and $f$ is bounded, which inequality is always guaranteed?]

• $U(f,P) \le U(f,P^*)$
• $L(f,P^*) \le L(f,P)$
• $U(f,P^*) \le U(f,P)$ ✓
• $L(f,P) \ge U(f,P^*)$

:::quiz[Which of the following is the correct definition of Darboux integrability of a bounded $f$ on $[a,b]$?]

• There exists a partition $P$ with $U(f,P) = L(f,P)$
• $\sup_P L(f,P) = \inf_P U(f,P)$ ✓
• $U(f,P) - L(f,P) \to 0$ for the uniform partition as $n \to \infty$ only
• $f$ is continuous almost everywhere on $[a,b]$

:::quiz[For the Dirichlet function $f = \mathbf{1}_{\mathbb{Q}}$ on $[0,1]$, what are $\underline{\int_0^1} f$ and $\overline{\int_0^1} f$?]

• Both equal $1/2$
• $\underline{\int_0^1} f = 0$ and $\overline{\int_0^1} f = 1$ ✓
• Both equal $0$
• $\underline{\int_0^1} f = 1$ and $\overline{\int_0^1} f = 0$

:::faq[What goes wrong if $[a,b]$ is replaced by an unbounded interval like $[a,\infty)$?] The partition-based definitions above require finitely many points $a = x_0 < \cdots < x_n = b$ with $b$ finite, so they simply do not apply to $[a,\infty)$: there is no finite right endpoint to partition up to. Extending integration to unbounded intervals (or to unbounded functions on bounded intervals) requires the separate machinery of improper integrals, defined as a limit of proper Darboux/Riemann integrals over $[a,N]$ as $N \to \infty$. This is a genuinely different construction layered on top of the finite-interval theory developed here, not a special case of it.

Ask the tutor: How does the Cauchy Integrability Criterion turn this sandwiching inequality into a practical test for integrability?

Picture the graph of a bounded function $f$ over $[a,b]$, sitting inside a bounding box. A partition chops $[a,b]$ into thin vertical strips. On each strip, instead of trying to capture the exact area under the (possibly wild) curve, we build two rectangles: a tall one reaching up to the strip's supremum $M_i$, and a short one reaching only to the infimum $m_i$. The tall rectangles, summed up, give $U(f,P)$ — an overestimate of any reasonable notion of area. The short rectangles give $L(f,P)$ — an underestimate. No matter how $f$ oscillates inside a strip, the true "area" (if it exists) is squeezed between these two numbers, strip by strip. The crucial fact — proved via refinement — is that adding more partition points can only tighten this squeeze: upper sums can only decrease (or stay the same) and lower sums can only increase (or stay the same) as we refine. This is why it makes sense to take an infimum over all upper sums and a supremum over all lower sums: we are asking "how tight can the overestimate get?" and "how tight can the underestimate get?" independently, and the theorem below shows the overestimate side never dips below the underestimate side. Integrability is then simply the statement that these two tightening processes converge to a shared value, leaving no gap.

:::svg[A partition of $[a,b]$ with upper rectangles ($M_i$) and lower rectangles ($m_i$) sandwiching the graph of $f$]

Example 1: A step function with an asymmetric jump. Let $f:[0,4] \to \mathbb{R}$ be defined by

We compute $\overline{\int_0^4} f$ and $\underline{\int_0^4} f$ directly and show they agree. Upper sums. Take the partition $P = \{0,1,3,4\}$, which respects the jump points exactly. On $[0,1]$, $\sup f = 2$; on $[1,3]$, $\sup f = -1$; on $[3,4]$, $\sup f = 5$. So

Any refinement $P^*$ of $P$ subdivides one of these three intervals further, but $f$ is constant on each piece of $[0,1]$, $[1,3]$, $[3,4]$ that arises this way, so the supremum on each sub-piece equals the constant value; refining changes nothing. By refinement monotonicity, $U(f,P^*) \le U(f,P) = 5$ for every refinement, and one checks directly that in fact $U(f,P^*) = 5$ always. For a partition $P'$ not refining $P$ — e.g. one that splits at $x = 0.5$ instead of at $1$ — the same computation (redo the sum over the actual sub-intervals, tracking which constant value of $f$ applies on each) still yields $U(f,P') \ge 5$, since any interval straddling a jump point picks up the larger of the two neighboring values as its supremum, only increasing the sum. Hence $U(f,P) = 5$ is in fact the minimum possible upper sum, so

Lower sums. By the mirror argument, on $[0,1]$, $\inf f = 2$; on $[1,3]$, $\inf f = -1$; on $[3,4]$, $\inf f = 5$ (since $f$ is constant on each piece, sup and inf coincide there). So with the same partition,

and any other partition gives a lower sum $\le 5$ (splitting across a jump point picks up the smaller neighboring value as the infimum on that sub-piece, only decreasing the sum). Hence $L(f,P)=5$ is the maximum possible lower sum, so

Since $\overline{\int_0^4} f = \underline{\int_0^4} f = 5$, $f$ is Darboux integrable on $[0,4]$ with $\int_0^4 f = 5$. (Sanity check: this matches signed-area bookkeeping, $2\cdot 1 + (-1)\cdot 2 + 5 \cdot 1 = 2-2+5=5$.)

Example 2: Squeezing $\int_1^2 \frac{1}{x}\,dx$ with an explicit non-uniform partition. Let $f(x) = 1/x$ on $[1,2]$, which is bounded ($1/2 \le f(x) \le 1$) and strictly decreasing, so on any subinterval $[c,d] \subseteq [1,2]$, $\sup f = f(c) = 1/c$ and $\inf f = f(d) = 1/d$. Use the deliberately non-uniform partition $P = \{1,\, 1.2,\, 1.5,\, 1.8,\, 2\}$, with subintervals of lengths $0.2, 0.3, 0.3, 0.2$. Upper sum:

Numerically: $\frac{1}{1}(0.2) = 0.2$; $\frac{1}{1.2}(0.3) = 0.25$; $\frac{1}{1.5}(0.3) = 0.2$; $\frac{1}{1.8}(0.2) \approx 0.11111$. Summing,

Lower sum:

Numerically: $\frac{1}{1.2}(0.2) \approx 0.16667$; $\frac{1}{1.5}(0.3) = 0.2$; $\frac{1}{1.8}(0.3) \approx 0.16667$; $\frac{1}{2}(0.2) = 0.1$. Summing,

By the theorem just proved, we know automatically (without computing any other partition) that

so whatever the true value $\int_1^2 \frac{1}{x}\,dx = \ln 2 \approx 0.6931$ turns out to be, this single non-uniform partition already traps it inside an interval of width about $0.128$. Refining $P$ further (say, subdividing each piece into ten equal parts) shrinks this gap and squeezes both sums toward $\ln 2$ from either side — this is the mechanism by which one proves $1/x$ is integrable and evaluates the integral, though establishing integrability rigorously for all continuous functions is the job of later criteria such as the Cauchy Integrability Criterion.