Upper and Lower Darboux Sums

Upper and lower Darboux sums are the standard vehicle by which every rigorous real analysis course (in the style of Rudin or Abbott) defines the Riemann integral, because they avoid the awkward existential quantifier over sample-point choices that appears in Riemann's original tagged-sum definition. The Darboux criterion — that $f$ is integrable iff $\sup_P L(f,P) = \inf_P U(f,P)$ — is used to prove nearly every basic integrability theorem: continuous functions are integrable, monotone functions are integrable, and functions continuous except at finitely many points are integrable. In numerical analysis, upper and lower sums give rigorous, computable error bounds for quadrature rules — a numerically computed approximation to an integral can be certified to lie within $[L(f,P), U(f,P)]$ for a suitably fine partition. In measure theory, the sup/inf-per-piece idea generalizes directly to simple-function approximations used to build the Lebesgue integral, making Darboux sums a conceptual bridge to that later, more general theory. In computer science and interval arithmetic, the same sup/inf-per-subinterval idea underlies rigorous numerical integration with guaranteed error bounds, and in control theory and optimization, bounding a cost functional above and below via partition-based envelopes is a direct descendant of this construction.

:::quiz[If $f$ is bounded on $[a,b]$ and $P$ is any partition, which inequality must always hold?]

• $U(f,P) \le L(f,P)$
• $L(f,P) \le U(f,P)$ ✓
• $L(f,P) = U(f,P)$
• $U(f,P) < 0$

:::quiz[Why must $M_i = \sup_{[x_{i-1},x_i]} f$ exist as a finite real number?]

• Because $f$ is continuous
• Because the subinterval has positive length
• Because $f$ is bounded, so the set of values on each subinterval is a nonempty bounded set of reals, hence has a supremum ✓
• Because the partition has finitely many points

:::quiz[For the Dirichlet-type function equal to 1 on rationals and 0 on irrationals, on any subinterval $[c,d]$ with $c<d$, what are $M_i$ and $m_i$?]

• $M_i = 1, m_i = 0$, always, since both rationals and irrationals are dense in every interval ✓
• $M_i = m_i = 1/2$
• They depend on whether $c$ is rational
• Undefined, since the function has no limit

:::quiz[Why does the article's counterexample function $f(x)=1/x$ on $(0,1]$, $f(0)=0$, fail to have well-defined Darboux sums?]

• Because $f$ is discontinuous at every point
• Because $f$ is negative somewhere
• Because $f$ is unbounded near $x=0$, so $\sup f$ on any subinterval touching $0$ is $+\infty$, not a real number ✓
• Because $[0,1]$ is not compact

:::faq[Do upper and lower sums make sense on unions of intervals or only on a single $[a,b]$?] As defined here, they are built for a single closed bounded interval $[a,b]$ with a bounded function on it, since a partition itself is defined as a finite increasing sequence of points spanning one such interval. Extensions to finite unions of intervals are typically handled by additivity of the integral over adjacent subintervals once the theory is developed further, rather than by redefining the partition or sum concept itself.

Ask the tutor: How does refining a partition rigorously push the lower sum up and the upper sum down, and why does that guarantee $\sup_P L(f,P)$ and $\inf_P U(f,P)$ both exist?

Picture the graph of $f$ over $[a,b]$. A partition slices the domain into narrow strips. On each strip, instead of trying to track the exact shape of the graph — which may be erratic — we build one rectangle that is guaranteed to lie entirely above the graph on that strip (using height $M_i$, the least upper bound of $f$ there) and one rectangle guaranteed to lie entirely below (using height $m_i$, the greatest lower bound). Summing the areas of the "roof" rectangles gives $U(f,P)$; summing the "floor" rectangles gives $L(f,P)$. Because every value of $f$ on a subinterval is squeezed between $m_i$ and $M_i$ by definition of sup and inf, the true (yet-to-be-defined) area under the graph — whatever it should mean — is trapped between $L(f,P)$ and $U(f,P)$ for this particular $P$. The deeper fact, proved below, is that this trapping is far more robust: even comparing a lower sum from one partition to an upper sum from a completely different, unrelated partition, the lower sum never exceeds the upper sum. This is what makes it sensible to later define the integral as the common value obtained by squeezing the gap between the best possible lower sums and the best possible upper sums to zero.

:::svg[A partition of $[0,4]$ with upper (outer) and lower (inner) rectangles for a bounded, non-monotonic $f$]

This corollary is exactly the ingredient needed to define, in The Darboux Integral, the lower integral $\underline{\int_a^b} f = \sup_P L(f,P)$ and the upper integral $\overline{\int_a^b} f = \inf_P U(f,P)$, and to know in advance that $\underline{\int_a^b} f \le \overline{\int_a^b} f$ always holds.

Example 1: A non-monotonic function, uneven partition. Let $f(x) = x(4-x)$ on $[0,4]$ (an upward-opening-down parabola, maximum at $x=2$), and take the uneven partition

So the four subintervals are $[0,1]$, $[1,1.5]$, $[1.5,3]$, $[3,4]$, with lengths $\Delta x_1 = 1$, $\Delta x_2 = 0.5$, $\Delta x_3 = 1.5$, $\Delta x_4 = 1$. Since $f$ is continuous and $f'(x) = 4-2x$, $f$ is increasing on $[0,2]$ and decreasing on $[2,4]$. We compute sup and inf on each piece directly (no need for continuity/extreme-value-theorem machinery beyond evaluating this simple quadratic):

• $[0,1]$: $f$ increasing here (since $1 < 2$), so $m_1 = f(0) = 0$, $M_1 = f(1) = 1\cdot 3 = 3$.
• $[1,1.5]$: still increasing ($1.5<2$), so $m_2 = f(1) = 3$, $M_2 = f(1.5) = 1.5 \cdot 2.5 = 3.75$.
• $[1.5,3]$: contains the peak $x=2$, and $f$ increases then decreases, so $M_3 = f(2) = 2\cdot 2 = 4$; the inf is at an endpoint, whichever is smaller: $f(1.5)=3.75$, $f(3) = 3\cdot1 = 3$, so $m_3 = 3$.
• $[3,4]$: decreasing here (since $3>2$), so $m_4 = f(4) = 4\cdot 0 = 0$, $M_4 = f(3) = 3$.

Now assemble:

Indeed $L(f,P) = 6 \le 13.875 = U(f,P)$, consistent with the theorem, and both lie inside $[m(b-a), M(b-a)] = [0, 16]$ since $0 \le f \le 4$ on $[0,4]$. (For reference, since $f$ is continuous, the true area is $\int_0^4 x(4-x)\,dx = \left[2x^2 - x^3/3\right]_0^4 = 32 - 64/3 = 32/3 \approx 10.667$, comfortably trapped between $6$ and $13.875$.)

Example 2: A bounded, discontinuous function — sums do NOT converge to a common value. Let

(the Dirichlet-type function on $[0,2]$, bounded with $m=0, M=1$). Take the partition $P = \{0, 0.5, 1.3, 2\}$, subintervals of lengths $0.5, 0.8, 0.7$. On every subinterval $[x_{i-1},x_i]$ with $x_{i-1}<x_i$, both rationals and irrationals are present (density of $\mathbb{Q}$ and of $\mathbb{R}\setminus\mathbb{Q}$ in every nondegenerate interval), so $f$ attains values arbitrarily close to (in fact exactly) $0$ and $1$ on each piece. Hence for every $i$: $M_i = \sup f = 1$ and $m_i = \inf f = 0$. Then

This holds for every partition $P$ of $[0,2]$, not just this one: any nondegenerate subinterval contains both a rational and an irrational, forcing $M_i=1, m_i=0$ always. So

The upper and lower sums are consistent with the theorem ($L \le U$ always), but they never squeeze together no matter how fine the partition — foreshadowing that this $f$ fails to be Darboux integrable, a phenomenon studied fully once the integral itself is defined.