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

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

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

Then:

1. $H$ is continuous on $[0,1]$;
2. $H$ is differentiable at every $x \in (0,1]$, with $H'(x) = 1/\sqrt{x}$, and it is not differentiable at $0$;
3. $H'$ is unbounded on $(0,1]$: $H'\bigl(1/\iota(n+1)^{2}\bigr) = \iota(n+1)$ for every $n \in \mathbb{N}$;
4. consequently no function on $[0,1]$ agreeing with $H'$ 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 $L(f,P) = \sum_i m_i \Delta_i$ and $U(f,P) = \sum_i M_i \Delta_i$).

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 $G' = f$ 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: $H'$ is continuous there, hence integrable, and

$\int_{\eta}^{1} H' \;=\; H(1) - H(\eta) \;=\; 2 - 2\sqrt{\eta} .$

The value that the right-hand side approaches as $\eta$ shrinks is not computed here and is not called an integral: $\int_0^1 H'$ is undefined, and the object that repairs it is the improper integral, which belongs to a later page.

• Boundedness, not continuity, is what fails. $H'$ 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 $L(f,P) = \sum_i m_i \Delta_i$ and $U(f,P) = \sum_i M_i \Delta_i$ 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 $H'$.

• 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 $\int_0^1 H'$ 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 $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$; 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 $[0,\infty)$; 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 $f'(c) \ne 0$, then the inverse $g$ is differentiable at $f(c)$ with $g'(f(c)) = 1/f'(c)$; and if $f'(c) = 0$ then $g$ is not differentiable at $f(c)$, whose second clause also explains the failure at $0$: there $q'(0) = 0$, so the inverse is not differentiable at $q(0) = 0$.