Undergraduate counterexamples

(A series of continuous functions that converges non-uniformly to a continuous function)

For \(x>0\) let \[ g(x) \coloneqq \sum_{n\geq1} {1 \over 1 + n^2 x}. \] On \([\delta,\infty)\) each term is at most \(1/(n^2\delta)\), so the series converges uniformly there by the Weierstrass \(M\)-test, and \(g\) is continuous on \((0,\infty)\). The convergence is not uniform on \((0,\infty)\): at \(x = 1/N^2\) the \(N\)th term equals \(1/2\), so the tails \(\sum_{n\geq N} (1+n^2x)^{-1}\) are not uniformly small.

(The Dirichlet function)

\[ f(x) \coloneqq \chi_\QQ(x) = \begin{cases} 1, & x\in \QQ, \\ 0, & x\notin\QQ. \end{cases} \] Every interval contains rationals and irrationals, so \(f\) is continuous nowhere and differentiable nowhere. On \([0,1]\) every upper Darboux sum is \(1\) and every lower Darboux sum is \(0\), so \(f\) is not Riemann integrable. Since \(f=0\) almost everywhere, \(f\) is Lebesgue integrable with \(\int_\RR f = 0\).

(A Dirichlet-type function continuous at exactly one point)

\[ f(x) \coloneqq x\chi_\QQ(x) = \begin{cases} x, & x\in \QQ, \\ 0, & x\notin\QQ. \end{cases} \] Since \(\abs{f(x)}\leq\abs x\), \(f\) is continuous at \(0\); at \(x_0\neq0\), rationals and irrationals approaching \(x_0\) give limits \(x_0\) and \(0\), so \(D_f = \RR\smz\). At \(0\) the difference quotient \(f(h)/h = \chi_\QQ(h)\) has no limit, so \(f\) is differentiable nowhere. On \([0,1]\) the upper Darboux integral is \(\int_0^1 x\dx = 1/2\) and the lower Darboux integral is \(0\), so \(f\) is not Riemann integrable; \(f=0\) almost everywhere, so \(f\) is Lebesgue integrable with \(\int_\RR \abs f = 0\).

(A Dirichlet-type function differentiable at exactly one point)

\[ f(x) \coloneqq x^2\chi_\QQ(x) = \begin{cases} x^2, & x\in \QQ, \\ 0, & x\notin\QQ. \end{cases} \] As before \(D_f = \RR\smz\). At \(0\), \(\abs{f(h)/h}\leq\abs h\to0\), so \(f'(0)=0\), and \(f\) is not differentiable at any point of discontinuity, so \(D_f'=\RR\smz\). The Darboux integrals on \([0,1]\) are \(1/3\) and \(0\), so \(f\) is not Riemann integrable; \(f=0\) almost everywhere, so \(\int_\RR\abs f=0\).

(A Dirichlet-type function taking two values of opposite sign)

\[ f(x) \coloneqq x\qty{\chi_\QQ(x) - \chi_{\RR\setminus\QQ}(x)} = \begin{cases} x, & x\in \QQ, \\ -x, & x\notin\QQ. \end{cases} \] Since \(\abs{f(x)} = \abs x\), \(f\) is continuous at \(0\), and \(D_f = \RR\smz\). At \(0\) the difference quotient \(f(h)/h = \pm1\) according to whether \(h\) is rational, so \(f\) is differentiable nowhere. Since \(\abs f = \abs x\) everywhere, \(f\) is not Lebesgue integrable on \(\RR\).

(The Thomae function)

\[ f(x) \coloneqq \begin{cases} \frac 1 q, & x = \frac p q \in \QQ \text{ in lowest terms with } q\geq1, \\ 0, & x\notin\QQ. \end{cases} \] For \(\varepsilon>0\), only finitely many rationals in a bounded interval have denominator at most \(1/\varepsilon\), so \(f(x)\to0\) as \(x\to x_0\) for every \(x_0\). Hence \(f\) is continuous exactly on \(\RR\setminus\QQ\) and \(D_f=\QQ\). \(f\) is differentiable nowhere. Since \(D_f\) is countable, hence null, and \(f\) is bounded, \(f\) is Riemann integrable on every \([a,b]\) by the Lebesgue criterion, with \(\int_a^b f = 0\); it is also Lebesgue integrable on \(\RR\) with integral \(0\).

(The Weierstrass function)

For \(a \in (0, 1)\) and an odd integer \(b\geq1\) with \(ab > 1 + {3\pi \over 2}\), let \[ f(x)\coloneqq\sum_{n=0}^{\infty} a^{n} \cos \left(b^{n} \pi x\right). \] The \(n\)th term is bounded by \(a^n\), so the series converges uniformly by the Weierstrass \(M\)-test, \(f\) is continuous, and \(\abs f\leq(1-a)^{-1}\). Weierstrass’s theorem states that \(f\) is differentiable nowhere. \(f\) is Riemann integrable on every compact interval, and not Lebesgue integrable on \(\RR\), since it is continuous, \(2\)-periodic, and not almost everywhere zero.