The Fundamental Theorems of Calculus: Examples and Counterexamples

Since $\mathrm{\mid }G(x)\mathrm{\mid }\le x^2$, the quotient $(G(x)-G(0))\mathrm{/}x$ has absolute value at most $x$ and tends to $0$; hence $G^{\mathrm{\prime }}(0)=0$.

For $x>0$, [L1] gives $G^{\mathrm{\prime }}(x)=2x\sin (1\mathrm{/}x)-\cos (1\mathrm{/}x)$.

By [L2], $\mathrm{\mid }{G}^{\mathrm{\prime }}(x)\mathrm{\mid }\le 2x+1\le 3$ on $(0,1]$, so $G^{\mathrm{\prime }}$ is bounded. The displayed formula is continuous away from $0$.

By [L3], sine vanishes and cosine equals $1$ at $2\pi k$, while sine vanishes and cosine equals $-1$ at $(2k+1)\pi$; substituting gives $G^{\mathrm{\prime }}(x_k)=-1$ and $G^{\mathrm{\prime }}(y_k)=1$. Both sequences tend to $0$ by [L4], so $G^{\mathrm{\prime }}$ is discontinuous there.

Thus the discontinuity set is exactly $\{0\}$, and [L5] makes $G^{\mathrm{\prime }}$ Riemann integrable.

Applying [L6] to $G$ and $G^{\mathrm{\prime }}$ yields ${\int }_0^1{G}^{\mathrm{\prime }}=G(1)-G(0)=\sin 1$.

A concrete cutoff is obtained by taking $\chi (s)=1$ for $s\le 1\mathrm{/}2$, $\chi (s)=1-3(2s-1{)}^2+2(2s-1{)}^3$ for $1\mathrm{/}2\le s\le 1$, and $\chi (s)=0$ for $s\ge 1$; the values and first derivatives agree at both joins.

On a gap $(u,v)$ the two supports lie in $(u,u+r]$ and $[v-r,v)$ and are disjoint because $2r<v-u$. Each summand and its derivative vanish at its cutoff join, so $W$ is differentiable throughout every gap.

By [L4], $\mathrm{\mid }g(t)\mathrm{\mid }\le t^2$. If $x$ lies in a gap, then $\mathrm{\mid }W(x)\mathrm{\mid }\le \mathrm{dist}(x,S{)}^2$; if $s\in S$, this is at most $\mathrm{\mid }x-s{\mathrm{\mid }}^2$. Hence $W^{\mathrm{\prime }}(s)=0$ for every $s\in S$, including $0$ and $1$.

By [L3]--[L4], $\mathrm{\mid }{g}^{\mathrm{\prime }}(t)\mathrm{\mid }\le 2t+1\le 3$ on $0<t\le 1$. Differentiating inside a support gives a sum of $g^{\mathrm{\prime }}\chi$ and $g{\chi }^{\mathrm{\prime }}\mathrm{/}r$, and $\mathrm{\mid }g(t)\mathrm{\mid }\mathrm{/}r\le t^2\mathrm{/}r\le r\le 1$ on $0\le t\le r$. Since ${\chi }^{\mathrm{\prime }}$ is bounded on its two polynomial pieces, one constant bounds $W^{\mathrm{\prime }}$ on all gaps and on $S$.

Let $I$ be a nondegenerate closed interval meeting $S$. If $S$ meets the interior of $I$, choose such a point and then, using the shrinking bound in [L1], a retained interval around it contained in $I$; its next-stage middle gap lies in $I$. If $S$ does not meet the interior, an endpoint of $I$ lies in $S$ and the interior lies in one removed gap, so gap points approach that endpoint from inside $I$. In either case the endpoint recursion in [L1] keeps the relevant gap endpoint in every later retained stage, hence in $S$. On its adjacent half-support $\chi =1$. By [L3]--[L5], the points at distances $1\mathrm{/}(2\pi k)$ and $1\mathrm{/}((2k+1)\pi )$ from that endpoint eventually lie in the half-support and give derivative values $-1$ and $1$, whereas step 3.1 gives value $0$ at the endpoint. Thus ${\omega }_{W^{\mathrm{\prime }}}(I)\ge 1\mathrm{/}2$.

Fix any partition and retain its subintervals that meet $S$. These finitely many closed intervals cover $S$, so their total lengths are at least $1\mathrm{/}2$ by [L2]. Step 4.2 therefore gives $U(W^{\mathrm{\prime }},P)-L(W^{\mathrm{\prime }},P)\ge (1\mathrm{/}2)(1\mathrm{/}2)=1\mathrm{/}4$.

The derivative is bounded by step 4.1, but the fixed lower bound in step 5.1 contradicts [L7]. Hence $W^{\mathrm{\prime }}$ is not Riemann integrable.

Apply [L2] to $E=\mathbb{Q}$, which is permitted by [L1], and call the resulting function $f$.

Its restriction to $[a,b]$ is nondecreasing and therefore integrable by [L3].

At every irrational point of $[a,b]$, $f$ is continuous, so [L4] gives $F^{\mathrm{\prime }}=f$.

At every rational $c\in (a,b)$, [L2] gives unequal left and right limits. By [L4] these are the left and right derivatives of $F$, so the two-sided derivative does not exist.

The rational points in $(a,b)$ are countable by [L1] and dense in every nondegenerate real interval by [L5], so the asserted nondifferentiability set is dense and countable.

The function satisfies $0\le t\le 1$, and by [L1] its discontinuity set is $\mathbb{Q}\cap [0,1]$, which is countable by [L2]. Hence [L3] makes $t$ integrable.

Fix $\epsilon >0$ and choose an integer $N\ge 1$ with $1\mathrm{/}N<\epsilon \mathrm{/}2$. The rationals in $[0,1]$ with least denominator at most $N$ form a finite set, since each has a representation $p\mathrm{/}q$ with $0\le p\le q\le N$.

Choose finitely many intervals around that finite set with total length below $\epsilon \mathrm{/}2$. A partition containing their endpoints has upper contribution below $\epsilon \mathrm{/}2$ on those intervals because $t\le 1$, and below $1\mathrm{/}N<\epsilon \mathrm{/}2$ on their complement because every positive value there has denominator greater than $N$. Thus it has upper sum below $\epsilon$.

Since $t\ge 0$, [L4] and the arbitrarily small upper sums in step 2.1 force ${\int }_0^{1}t=0$. The same argument on every subinterval gives ${\int }_x^{y}t=0$.

By [L5], $T(y)-T(x)=0$ for all $x,y$, so $T$ is identically zero and $T^{\mathrm{\prime }}=0$ everywhere, with relative derivatives at $0$ and $1$.

By [L6], the rationals and irrationals are both dense. Combining the definition of $t$, [L1], and step 4.1 gives the claimed equality at irrationals and failure at rationals.

The function is bounded between $0$ and $1$, and every nondegenerate interval contains a point outside the countable spike set, so every lower Darboux sum is $0$.

Given $\epsilon >0$, choose $N$ with $2^{-N}<\epsilon \mathrm{/}2$ by [L1]. Put the finitely many spikes $1,2^{-1},\dots ,2^{-(N-1)}$ in partition intervals of total length below $\epsilon \mathrm{/}2$, and put all remaining spikes in $[0,2^{-N}]$. The resulting upper sum is below $\epsilon$.

By [L2], $f$ is integrable, and steps 1.1--1.2 force its integral to be $0$. The same construction after restriction gives integral $0$ on every subinterval.

By [L3] and step 2.1, $F(x)=0$ for every $x$, so its relative derivative at $0$ is $0$.

Along the spike sequence $2^{-n}\to 0$, the values are $1$, while $f(0)=0$; thus $f$ is discontinuous at $0$ and $F^{\mathrm{\prime }}(0)=0=f(0)$.

If $x\mathrm{\notin }C$, [L2] makes $c$ constant near $x$, so every sufficiently local difference quotient is $0$ and [L4] gives $c^{\mathrm{\prime }}(x)=0$.

The exceptional set is contained in $C$, which has measure zero by [L3]; therefore $c^{\mathrm{\prime }}=0$ almost everywhere.

By [L1], $c(1)-c(0)=1-0=1$, despite step 2.1.

Hence the implication from an almost-everywhere zero derivative to zero endpoint change is false without further regularity.