Page 292 Problem 13
In the statement of Theorem 31.5 we require that f 0 be integrable on [a, b]. Show that this
requirement is necessary. That is, find a differentiable function f such that f 0 is not integrable.
Solution
(
x2 sin
Let f : [0, 1] → R be defined piecewise: f (x) =
0
1
x2
if x 6= 0,
if x = 0.
It is not too difficult to show that
(
2x sin x12 − x2 cos x12
if x 6= 0,
0
f (x) =
0
if x = 0.
On the interval [0, 1], then, f 0 exists. But f 0 is not Riemann integrable on that interval because it
is not bounded there (see exercise 25.8).
Remarks
The intellectual history surrounding the fundamental theorem of calculus is impressive. It began
with works published by Gottfried Leibniz (in 1684: Acta eruditorum his Nova methodus pro
maximis et minimis, itemque tangentibus, quae nec fractas nec irrationales quantitates moratur,
et singulare pro illis calculi genus) and Isaac Newton (in 1687: Methodus fluxionum et serierum
infinitarum and the De quadratura curvarum). These initial treatises did not have proofs up to the
standards of modern mathematics. The necessary rigor was developed over a period of more than
two centuries. It came from a variety of mathematicians including Augustin Cauchy (1789–1857),
Bernard Riemann (1826–1866, from whom we get the term Riemann integral), Gaston Darboux
(1842–1917), Vito Volterra (1860–1940), and Henri Lebesgue (1875–1941).
Cauchy first gave a rigorous proof of the fundamental theorem in 1823. His proof, however, required
that f 0 be continuous on [a, b]. It was not until 1875 that Darboux proved a stronger version of the
fundamental theorem, stated above, requiring only that f 0 be Riemann integrable. The question
arose, then, what minimal assumptions might be required of the derivative to ensure that the
conclusion of the fundamental theorem still holds true. Might it be enough, for example, just to
require that f 0 be bounded?
Unfortunately, no. In 1881 Volterra found an example of a Riemann integrableR function whose
b
derivative, although bounded, is not Riemann integrable. Thus, the expression a f 0 is not even
defined, so it could hardly equal f (b) − f (a). This sad state of affairs, however, ultimately came to
a happy conclusion. In his doctoral dissertation of 1901 Lebesgue defined a generalization of the
Riemann integral,
he proved that, for this
R b 0 now called the Lebesgue integral. Among other results
0
new integral, a f = f (b) − f (a) with the requirement only that f be bounded.