Introduction
A discontinuity does not automatically make an integral improper. The important question is not simply whether the function is continuous everywhere, but whether it is bounded and how many points of discontinuity it has.
In particular, a bounded function that is continuous except at one point on a finite closed interval is Riemann integrable. This includes a removable discontinuity and even an endpoint oscillation with no limiting value. By contrast, an unbounded function at an endpoint is not an ordinary Riemann-integrable function on the full interval; its integral must be treated using a limit, and that limit may fail to be finite.
Ordinary and Improper Integrals
A bounded function \(f:[a,b]\to\mathbb{R}\) is Riemann integrable if its lower and upper sums can be made arbitrarily close by choosing a sufficiently fine partition of \([a,b]\). The integral is the common value approached by those sums.
An improper integral is different: it is defined through a limit when the integrand is unbounded or is not defined at an endpoint or an interior point. For example, if the only problematic point is the left endpoint, its value, when it exists finitely, is defined by \(\lim_{\varepsilon\to 0^+}\int_{a+\varepsilon}^{b} f(x)\,dx\). For an interior singularity, the two one-sided integrals must each converge; cancellation between them does not establish ordinary convergence.
The result in this tutorial concerns a bounded function that is defined on the entire closed interval. If such a function has just one discontinuity, no improper-limit definition is needed: the ordinary Riemann integral exists.
The Bounded Single-Discontinuity Theorem
The endpoint cases matter. A theorem stated only for an interior discontinuity does not automatically cover a function whose only possible discontinuity is at \(a\) or \(b\). We include all three cases in the proof.
Why the Theorem Holds
We use the Darboux criterion: a bounded function on \([a,b]\) is Riemann integrable if, for every \(\varepsilon>0\), there is a partition \(P\) for which the upper sum \(U(f,P)\) and lower sum \(L(f,P)\) satisfy \(U(f,P)-L(f,P)<\varepsilon\). On a subinterval, the upper-minus-lower contribution is its length multiplied by the oscillation of \(f\) there, where oscillation means the supremum of the function minus its infimum.
Because \(f\) is bounded, choose \(M\geq 0\) such that \(|f(x)|\leq M\) throughout \([a,b]\). On any interval, the oscillation is at most \(2M\). We make the interval around the possible discontinuity short, so its contribution to the difference of the sums is small. On the rest of \([a,b]\), continuity on a compact interval gives uniform continuity, which makes the remaining contributions small.
For an interior point \(a<c<b\), choose \(\delta>0\) with \(c-\delta>a\) and \(c+\delta<b\). Include \(c-\delta\) and \(c+\delta\) as partition points. The contribution from \([c-\delta,c+\delta]\) is at most \(2M(2\delta)=4M\delta\).
The function is continuous on each of the compact intervals \([a,c-\delta]\) and \([c+\delta,b]\), so it is uniformly continuous on each. Choose the mesh of the partition there small enough that the oscillation on every subinterval is less than \(\varepsilon/(2(b-a))\). The total contribution away from \(c\) is then at most \(\varepsilon/2\).
Choose \(\delta\) so that \(4M\delta<\varepsilon/2\) when \(M>0\). The full upper-minus-lower sum is then less than \(\varepsilon\). If \(M=0\), the function is identically zero and is already integrable.
If the exceptional point is the endpoint \(c=a\), use the short interval \([a,a+\delta]\). Its upper-minus-lower contribution is at most \(2M\delta\). The function is continuous, hence uniformly continuous, on the remaining compact interval \([a+\delta,b]\). Choose \(\delta\) so that \(2M\delta<\varepsilon/2\), and choose the partition on the remaining interval so that its total contribution is less than \(\varepsilon/2\). Thus the Darboux difference is less than \(\varepsilon\). If \(c=b\), use \([b-\delta,b]\) and the same argument. This proves the theorem for an endpoint as well as for an interior point.
Worked Example: A Removable Discontinuity
Worked Example: A Removable Discontinuity
Define \(f:[0,2]\to\mathbb{R}\) by
For \(x\ne 1\), factor the numerator and cancel the common factor:
Therefore, as \(x\to 1\), \(f(x)\to 2\), whereas \(f(1)=7\). The discontinuity is removable: the finite limit exists, but the assigned value differs from that limit. The function is bounded and continuous except at \(x=1\), so the theorem guarantees an ordinary Riemann integral.
Changing the value at a single point does not change the integral. Here \(f\) agrees with \(g(x)=x+1\) except at \(x=1\), so their integrals are equal. To see why a single-point change has no effect, enclose that point in an interval of length as small as desired. The functions agree outside it, and the contribution of that short interval is bounded in absolute value by its length times a fixed bound for their difference. That contribution tends to zero as the interval shrinks.
Consequently, compute the integral using \(g\):
Check by differentiating the antiderivative: \(\frac{d}{dx}\left(\frac{x^2}{2}+x\right)=x+1\), as required away from the single changed point. The integral is \(4\); the value \(f(1)=7\) does not add a separate contribution.
Worked Example: Oscillation at an Endpoint
Worked Example: Oscillation at an Endpoint
Define \(h:[0,1]\to\mathbb{R}\) by
For every \(x\in[0,1]\), \(|h(x)|\leq 1\). For every \(x>0\), \(h\) is continuous. It is not continuous at \(0\): the sequences \(x_n=1/(\pi/2+2\pi n)\) and \(y_n=1/(3\pi/2+2\pi n)\) both tend to \(0\), while \(h(x_n)=1\) and \(h(y_n)=-1\).
The theorem applies to this endpoint case. In particular, isolate \([0,\delta]\). Since \(-1\leq h(x)\leq 1\), its upper-minus-lower contribution is at most \(2\delta\). On \([\delta,1]\), \(h\) is continuous on a compact interval, hence uniformly continuous, so a fine enough partition makes the contribution there as small as desired. Thus \(h\) has an ordinary Riemann integral on \([0,1]\), despite having no limit at \(0\).
This example does not require evaluating the integral in closed form. It establishes that the ordinary integral exists. After that existence is known, the truncated integrals approach it: the difference between the integral on \([0,1]\) and on \([\delta,1]\) is \(\int_0^\delta h(x)\,dx\), whose absolute value is at most \(\delta\), since \(|h(x)|\leq 1\). This also shows why a limit of truncated integrals can agree with the ordinary integral without being needed to define it.
Worked Example: A Bounded Jump
Worked Example: A Bounded Jump
Let \(q:[0,2]\to\mathbb{R}\) be given by \(q(x)=1\) for \(0\leq x<1\) and \(q(x)=3\) for \(1\leq x\leq 2\). This function is bounded and continuous except for a jump at \(x=1\), so it is Riemann integrable by the theorem.
Using additivity over the two subintervals gives
As a check, each constant value multiplied by its interval width gives the area of the corresponding rectangle: the widths are \(1\) and \(1\), and the heights are \(1\) and \(3\). Their areas sum to \(1+3=4\). The jump changes the function at one point but does not require an improper limit.
Contrast: An Unbounded Endpoint Singularity
Boundedness is essential to this theorem. Consider \(r(x)=1/x\) on \(0<x\leq 1\). This function is unbounded near \(0\), so it does not meet the theorem’s assumptions and is not an ordinary Riemann-integrable function on \([0,1]\). Its integral must be considered as an improper integral:
The limit is not finite, so the improper integral diverges. The contrast is precise: the oscillatory function in the previous example stays bounded, allowing the short interval near its discontinuity to contribute arbitrarily little; \(1/x\) becomes arbitrarily large there, and the corresponding improper limit does not converge finitely.
What the Result Does—and Does Not—Say
The theorem is a sufficient condition for integrability, not a claim that every bounded function is integrable. Its hypothesis is specific: there is at most one point where continuity may fail, and the function is continuous everywhere else on the interval. A function with many discontinuities requires a separate integrability argument.
Also distinguish an assigned value from a missing value. The theorem concerns a function defined on all of \([a,b]\). If a formula is undefined at one point, assigning it a finite value can produce a function on the whole interval. For a removable discontinuity, the value at that point may be chosen to equal the limit to make the extension continuous; a different finite assignment still gives a bounded function with at most one discontinuity and the same integral.
Check Your Understanding
Use the hypotheses of the theorem and the distinction between ordinary and improper integration to answer these questions.
- Why does boundedness let the upper-minus-lower contribution from a short interval around the exceptional point be made small?
- Why must the proof separately address a discontinuity at an endpoint?
- For the removable-discontinuity example, what is the limit of \(f(x)\) as \(x\to 1\), and why does the assigned value \(f(1)=7\) not change the integral?
- Why is \(\sin(1/x)\), with a finite value assigned at \(x=0\), Riemann integrable on \([0,1]\) even though it has no limit there?
- What fails when applying the bounded-single-discontinuity theorem to \(1/x\) on \((0,1]\)?
