Real Analysis 52 | Riemann Integral - Examples [dark version]

Q: What integral value is obtained for f(x)=x and why is it consistent?

The lower and upper bounds converge to 1/2 as n→∞, since 1/2 ± 1/(2n) both approach 1/2. The integral value is therefore 1/2. This matches the geometric expectation of the area under y=x from 0 to 1 (a triangle with base and height 1), but the key point is that the rectangle-based approximation proves it within the Riemann framework.

TL;DR

Riemann integrability can be checked by forcing the gap between upper and lower step-function integrals below any epsilon.

Briefing Cornell Notes

Briefing

Riemann integrability hinges on whether the “gap” between step-function approximations from below and from above can be forced arbitrarily small. The lesson is made concrete through two contrasting examples: the Dirichlet (rational/irrational) function on [0,1] fails the test, while the identity function f(x)=x succeeds and yields the familiar integral value 1/2.

The counterexample is the function f(x)=1 for rational x and f(x)=0 for irrational x on [0,1]. Because every subinterval contains both rationals and irrationals, any step function that lies above f must effectively sit at height 1 throughout each interval: there is always a rational point where f equals 1. That forces the upper approximation to have integral at least 1. Symmetrically, any step function lying below f must be at height 0 throughout each interval, since every subinterval also contains irrationals where f=0; thus the lower approximation’s integral is at most 0. Since the difference between the upper and lower integrals is always at least 1, no matter how fine the partition gets, the “epsilon gap” condition cannot be satisfied. This blocks Riemann integrability outright.

The second example demonstrates how integrability can be verified constructively. For f(x)=x on [0,1], the area under the curve is known to be 1/2, but the goal is to reproduce that value using rectangle (step-function) approximations. For each integer n, a step function F_n is built from below using n equal subintervals of width 1/n. On the k-th subinterval, the height is (k−1)/n, so each rectangle has width 1/n and height varying linearly across the partition. The integral of this step function becomes a sum of rectangle areas: (1/n^2) times the sum of integers from 0 to n−1. Using the arithmetic-series formula, the lower integral simplifies to (n(n−1)/2)/n^2 = 1/2 − 1/(2n).

To approximate from above, a second step function S_n is shifted upward so that it stays above x on each subinterval while using the same partition. The rectangle heights now run through 1/n, 2/n, …, n/n, producing an upper integral of (1/n^2) times the sum of integers from 1 to n. The same series formula yields 1/2 + 1/(2n). The crucial quantity is the gap between these bounds: (1/2 + 1/(2n)) − (1/2 − 1/(2n)) = 1/n. As n grows, 1/n can be made smaller than any chosen epsilon, satisfying the Riemann criterion. With that, f(x)=x is confirmed Riemann integrable, and the integral value is pinned down as the common limit 1/2.

Cornell Notes

Riemann integrability can be tested by comparing integrals of step-function approximations from below and from above. The Dirichlet function on [0,1] (1 on rationals, 0 on irrationals) is not Riemann integrable because every interval contains both types of points, forcing any upper step approximation to have integral at least 1 and any lower step approximation to have integral at most 0. For f(x)=x, step functions built on n equal subintervals produce lower and upper integrals of 1/2 − 1/(2n) and 1/2 + 1/(2n). Their difference is exactly 1/n, which can be made smaller than any epsilon by choosing n large. The integral therefore exists and equals 1/2.

Why does the rational/irrational (Dirichlet) function fail the Riemann integrability test on [0,1]?

Every subinterval of [0,1] contains both rational and irrational numbers. If a step function lies above f, then on each subinterval it must be at least 1 because there is always a rational point where f=1. That makes the upper integral ≥ 1. If a step function lies below f, it must be at most 0 on each subinterval because there is always an irrational point where f=0, making the lower integral ≤ 0. The gap between upper and lower integrals is therefore always ≥ 1, so it cannot be reduced below an arbitrary epsilon.

How are the lower step-function approximations F_n for f(x)=x constructed?

Divide [0,1] into n equal subintervals of width 1/n. On the k-th subinterval (k=1,...,n), set the step height to (k−1)/n, so the step function stays below x. Each rectangle has width 1/n and height (k−1)/n, so the lower integral equals (1/n^2) times the sum of integers 0 through n−1. Using the arithmetic-series formula gives the lower bound 1/2 − 1/(2n).

What changes in the upper step-function approximations S_n for f(x)=x?

The same partition into n equal subintervals is used, but the step heights are shifted upward so the step function stays above x. Instead of heights (k−1)/n, the k-th rectangle uses height k/n. The upper integral becomes (1/n^2) times the sum of integers 1 through n, which simplifies to 1/2 + 1/(2n).

How does the epsilon condition get verified for f(x)=x?

The difference between the upper and lower integrals is (1/2 + 1/(2n)) − (1/2 − 1/(2n)) = 1/n. Given any epsilon > 0, choosing n > 1/epsilon makes 1/n < epsilon. That means the upper and lower step-function integrals can be forced arbitrarily close, which is exactly the Riemann integrability criterion used in the alternative epsilon formulation.

What integral value is obtained for f(x)=x and why is it consistent?