$H(x)=2\sqrt{x}$ on $[0,1]$: $H$ is continuous, $H^{\mathrm{\prime }}$ is unbounded on $(0,1]$, and $H^{\mathrm{\prime }}$ is therefore not Riemann integrable

Write $\sqrt{x}:=x^{1\mathrm{/}2}$ for the unique nonnegative square root of $x\ge 0$ (Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$, Rational powers $a^r$ of a positive base) and put

$H:[0,1]\to \mathbb{R},H(x):=2\sqrt{x}\mathrm{.}$

Then:

1. $H$ is continuous on $[0,1]$;
2. $H$ is differentiable at every $x\in (0,1]$, with $H^{\mathrm{\prime }}(x)=1\mathrm{/}\sqrt{x}$, and it is not differentiable at $0$;
3. $H^{\mathrm{\prime }}$ is unbounded on $(0,1]$: $H^{\mathrm{\prime }}(1\mathrm{/}\iota (n+1{)}^2)=\iota (n+1)$ for every $n\in \mathbb{N}$;
4. consequently no function on $[0,1]$ agreeing with $H^{\mathrm{\prime }}$ on $(0,1]$ is Riemann integrable on $[0,1]$, because Darboux sums are defined only for bounded functions (For bounded $f$ on $[a,b]$ and a partition $P$: the infimum $m_i$ and supremum $M_i$ of $f$ on the $i$-th subinterval, and the lower and upper Darboux sums $L(f,P)={\sum }_i{m}_i{\mathrm{\Delta }}_i$ and $U(f,P)={\sum }_i{M}_i{\mathrm{\Delta }}_i$).

So the second fundamental theorem does not apply on $[0,1]$, even though $H$ is continuous there and differentiable on $(0,1]$: both hypotheses of The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G^{\mathrm{\prime }}=f$ and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$ fail, differentiability at $0$ and integrability of the derivative.

What is available, and what is not. On $[\eta ,1]$ with $0<\eta <1$ everything works: $H^{\mathrm{\prime }}$ is continuous there, hence integrable, and

${\int }_{\eta }^1{H}^{\mathrm{\prime }}=H(1)-H(\eta )=2-2\sqrt{\eta }\mathrm{.}$

The value that the right-hand side approaches as $\eta$ shrinks is not computed here and is not called an integral: ${\int }_0^1{H}^{\mathrm{\prime }}$ is undefined, and the object that repairs it is the improper integral, which belongs to a later page.

• Boundedness, not continuity, is what fails. $H^{\mathrm{\prime }}$ is continuous at every point of $(0,1]$; what defeats Riemann integrability on $[0,1]$ is that For bounded $f$ on $[a,b]$ and a partition $P$: the infimum $m_i$ and supremum $M_i$ of $f$ on the $i$-th subinterval, and the lower and upper Darboux sums $L(f,P)={\sum }_i{m}_i{\mathrm{\Delta }}_i$ and $U(f,P)={\sum }_i{M}_i{\mathrm{\Delta }}_i$ takes a supremum of $f$ over each subinterval and the subinterval containing $0$ has none. No choice of value at $0$ repairs this, which is why claim 4 quantifies over every extension of $H^{\mathrm{\prime }}$.

• This is the standard motivation for the improper integral, and it is left standing here. The numbers $2-2\sqrt{\eta }$ of step 8.1 approach $2$ as $\eta$ shrinks, and a later page defines an object whose value is $2$. Calling that object ${\int }_0^1{H}^{\mathrm{\prime }}$ before it has been defined would be the error this item exists to avoid; nothing above computes it.

• The square root is obtained from published material and not assumed. Existence and uniqueness of the nonnegative $n$-th root is Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$; continuity of $x\mapsto \sqrt{x}$ comes from Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f[I]$, and the inverse $g:f[I]\to I$ is continuous and strictly monotone in the same sense as $f$ applied to $x\mapsto x^2$ on $[0,\mathrm{\infty })$; and the derivative comes from Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c\in I$ with $f^{\mathrm{\prime }}(c)\mathrm{\ne }0$, then the inverse $g$ is differentiable at $f(c)$ with $g^{\mathrm{\prime }}(f(c))=1\mathrm{/}{f}^{\mathrm{\prime }}(c)$; and if $f^{\mathrm{\prime }}(c)=0$ then $g$ is not differentiable at $f(c)$, whose second clause also explains the failure at $0$: there $q^{\mathrm{\prime }}(0)=0$, so the inverse is not differentiable at $q(0)=0$.