A function differentiable on $[0,1]$ whose derivative is unbounded, hence not Riemann integrable

By [L1], $\varphi$ is differentiable everywhere with ${\varphi }^{\mathrm{\prime }}(t)=2t(1-t{)}^2-2t^2(1-t)=2t(1-t)(1-2t)$; in particular $\varphi (0)=\varphi (1)=0$ and ${\varphi }^{\mathrm{\prime }}(0)={\varphi }^{\mathrm{\prime }}(1)=0$, and ${\varphi }^{\mathrm{\prime }}(1\mathrm{/}4)=2\cdot \frac{1}{4}\cdot \frac{3}{4}\cdot \frac{1}{2}=\frac{3}{16}$.

For $t\in [0,1]$, $0\le \varphi (t)=(t(1-t){)}^2\le (1\mathrm{/}4{)}^2=1\mathrm{/}16$, by [L9].

Each ${\alpha }_n,h_n,c_n$ is a positive real, $h_n\le {\alpha }_n$ and $c_n\le {\alpha }_n$ because $\iota (n+2)>1$, and ${\alpha }_n\le 1\mathrm{/}2$; so $I_n\subseteq (0,1]$, since ${\beta }_n={\alpha }_n+h_n\le 2{\alpha }_n\le 1$.

The $I_n$ are pairwise disjoint. For $n\ge 1$, ${\alpha }_{n-1}-{\alpha }_n=1\mathrm{/}(\iota (n+1)\iota (n+2))$ while $h_n=1\mathrm{/}\iota (n+2{)}^4$, and $\iota (n+2{)}^4>\iota (n+1)\iota (n+2)$ because $\iota (n+2{)}^3\ge 2\iota (n+2)>\iota (n+1)$ by [L3]; so ${\beta }_n<{\alpha }_{n-1}$. Since $({\alpha }_n)$ is strictly decreasing, $m<n$ gives ${\beta }_n<{\alpha }_{n-1}\le {\alpha }_m$, and $I_n$ lies strictly below $I_m$.

For each $n$ define the polynomial ${\mathrm{\Phi }}_n(y):=c_n\varphi ((y-{\alpha }_n)\mathrm{/}{h}_n)$ on $\mathbb{R}$. By [L1] and [L2] it is differentiable everywhere with ${\mathrm{\Phi }}_n^{\mathrm{\prime }}(y)=(c_n\mathrm{/}{h}_n){\varphi }^{\mathrm{\prime }}((y-{\alpha }_n)\mathrm{/}{h}_n)$, and ${\mathrm{\Phi }}_n({\alpha }_n)={\mathrm{\Phi }}_n({\beta }_n)=0$, ${\mathrm{\Phi }}_n^{\mathrm{\prime }}({\alpha }_n)={\mathrm{\Phi }}_n^{\mathrm{\prime }}({\beta }_n)=0$ by step 1.1.

Differentiability at a point of $(0,1]$ outside every $I_n$. Let $x\in (0,1]$ with $x\mathrm{\notin }{I}_m$ for every $m$. By [L3] fix $N$ with $2\mathrm{/}x<\iota (N+2)$, so $2\mathrm{/}\iota (N+2)<x$; for $m\ge N$, ${\beta }_m\le 2{\alpha }_m=2\mathrm{/}\iota (m+2)\le 2\mathrm{/}\iota (N+2)<x$ by step 1.3 and [L3].

So $G$ is a well-defined function on $[0,1]$, no $x$ lying in two of the $I_n$.

Gaps around the endpoints. For each $n$, $({\beta }_{n+1},{\alpha }_n)$ meets no $I_m$: for $m\le n$ one has ${\alpha }_m\ge {\alpha }_n$ by step 2.1, and for $m\ge n+1$ one has ${\beta }_m\le {\beta }_{n+1}$, again by step 2.1. Likewise $({\beta }_n,{\alpha }_{n-1})$ meets no $I_m$ for $n\ge 1$, and $({\beta }_0,1]$ meets none.

The finitely many closed intervals $I_0,\dots ,I_{N-1}$ do not contain $x$, so each of the positive reals $\mathrm{\mid }x-{\alpha }_m\mathrm{\mid }$ and $\mathrm{\mid }x-{\beta }_m\mathrm{\mid }$ for $m<N$, together with $x-2\mathrm{/}\iota (N+2)$, forms a nonempty finite set of positive reals; let $\rho$ be its least element, which is positive by [L8].

$G$ agrees with ${\mathrm{\Phi }}_n$ on $I_n$ and with the zero function off ${\bigcup }_m{I}_m$; in particular $G({\alpha }_n)=G({\beta }_n)=0$ and $G\ge 0$ everywhere.

Differentiability at an interior point of a bump. Let $x\in ({\alpha }_n,{\beta }_n)$. The difference quotients of $G$ and of ${\mathrm{\Phi }}_n$ at $x$ agree on the punctured neighbourhood $0<\mathrm{\mid }y-x\mathrm{\mid }<\min \{x-{\alpha }_n,{\beta }_n-x\}$ inside $[0,1]$, so by [L5] and step 2.2, $G$ is differentiable at $x$ with $G^{\mathrm{\prime }}(x)={\mathrm{\Phi }}_n^{\mathrm{\prime }}(x)$.

Differentiability at a left endpoint ${\alpha }_n$. On the right, $G$ agrees with ${\mathrm{\Phi }}_n$ on $[{\alpha }_n,{\beta }_n]$ and $G({\alpha }_n)={\mathrm{\Phi }}_n({\alpha }_n)$, so the right-hand limit of the difference quotient is ${\mathrm{\Phi }}_n^{\mathrm{\prime }}({\alpha }_n)=0$ by [L5] and step 2.2. On the left, $G$ vanishes on $({\beta }_{n+1},{\alpha }_n]$ by step 3.2 and step 4.1, so the quotient is identically $0$ there and its left-hand limit is $0$. By [L4], $G^{\mathrm{\prime }}({\alpha }_n)=0$.

Differentiability at a right endpoint ${\beta }_n$. Symmetrically: on the left $G$ agrees with ${\mathrm{\Phi }}_n$, giving limit ${\mathrm{\Phi }}_n^{\mathrm{\prime }}({\beta }_n)=0$; on the right $G$ vanishes on $[{\beta }_n,{\alpha }_{n-1})$ when $n\ge 1$ and on $[{\beta }_0,1]$ when $n=0$, by step 3.2, giving limit $0$. By [L4], $G^{\mathrm{\prime }}({\beta }_n)=0$.

Then $G$ vanishes on $(x-\rho ,x+\rho )\cap [0,1]$, so its difference quotient at $x$ is identically $0$ there and $G^{\mathrm{\prime }}(x)=0$ by [L5] and [L6].

Differentiability at $0$. For $y\in (0,1]$: if $y\mathrm{\notin }{\bigcup }_m{I}_m$ then $G(y)\mathrm{/}y=0$; and if $y\in I_m$ then $0\le G(y)\le c_m\mathrm{/}16$ by step 1.2 and step 4.1 and $y\ge {\alpha }_m$, so $0\le G(y)\mathrm{/}y\le c_m\mathrm{/}(16{\alpha }_m)=1\mathrm{/}(16\iota (m+2))$.

Given $\epsilon >0$, fix by [L3] a natural $N$ with $1\mathrm{/}\epsilon <\iota (N+2)$ and put $\delta :={\alpha }_N>0$. If $0<y<\delta$ and $y\in I_m$ then ${\alpha }_m\le y<{\alpha }_N$, so $\iota (N+2)<\iota (m+2)$ by [L3], and step 5.5 gives $\mathrm{\mid }G(y)\mathrm{/}y\mathrm{\mid }\le 1\mathrm{/}(16\iota (m+2))<\epsilon$; otherwise $G(y)\mathrm{/}y=0$.

$G^{\mathrm{\prime }}$ is unbounded. Put $u_n:={\alpha }_n+h_n\mathrm{/}4$, an interior point of $I_n$; by step 5.1 and step 2.2, $G^{\mathrm{\prime }}(u_n)=(c_n\mathrm{/}{h}_n){\varphi }^{\mathrm{\prime }}(1\mathrm{/}4)=\iota (n+2{)}^2\cdot \frac{3}{16}$, using $c_n\mathrm{/}{h}_n=\iota (n+2{)}^4\mathrm{/}\iota (n+2{)}^2=\iota (n+2{)}^2$. Given a real $M\ge 0$, [L3] supplies $n$ with $\frac{16}{3}M<\iota (n+2)\le \iota (n+2{)}^2$, so $G^{\mathrm{\prime }}(u_n)>M$.

So the difference quotient of $G$ at $0$, which is $y\mapsto G(y)\mathrm{/}y$ on $(0,1]$, tends to $0$; hence $G$ is differentiable at $0$ with $G^{\mathrm{\prime }}(0)=0$ by [L6].

By steps 5.1, 5.2, 5.3, 5.4 and 7.1, $G$ is differentiable at every point of $[0,1]$: every $x$ is either $0$, or an interior point of some $I_n$, or an endpoint of some $I_n$, or a point of $(0,1]$ outside every $I_n$.

Hence $G^{\mathrm{\prime }}$ is a function on $[0,1]$ that is not bounded, so it has no Darboux sums and is not Riemann integrable on $[0,1]$ by [L7]; the claim is false, and ${\int }_0^1{G}^{\mathrm{\prime }}$ is an undefined symbol, so [L10] gives nothing here.