Theorem 6.21, sandwich form — $F(b)-F(a)$ between the lower and upper integrals of $F'$

Let $a \le b$, let $f, F : \mathbb{R} \to \mathbb{R}$, and suppose $F$ is differentiable at every point of $[a,b]$ with $F'(x) = f(x)$ there. No integrability of $f$ is assumed, and boundedness is assumed only on one side at a time. Then, with $\underline{\int}$ and $\overline{\int}$ the lower and upper Riemann integrals of Rudin's Definition 6.2 (the case $\alpha(x) = x$):

1. if $f$ is bounded below on $[a,b]$, then

2. if $f$ is bounded above on $[a,b]$, then

3. consequently, if $f$ is bounded below and $f \in \mathcal{R}$ on $[a,b]$, then

The mechanism is the mean value theorem applied on each subinterval of a partition $a = x_0 \le x_1 \le \dots \le x_n = b$: there is $t_i \in [x_{i-1}, x_i]$ with $F(x_i) - F(x_{i-1}) = f(t_i)\,\Delta x_i$. If $f$ is bounded below, the infimum $m_i$ of $f$ on the subinterval is a genuine infimum and $m_i \le f(t_i)$, so summing and telescoping gives $L(P,f) \le F(b) - F(a)$ for every partition $P$; taking the supremum over $P$ gives (1). Symmetrically, boundedness above gives $F(b) - F(a) \le U(P,f)$ and hence (2).

Taken together, (1) and (2) contain Rudin's Theorem 6.21 for bounded $f$: if $f$ is bounded and integrable, the upper and lower integrals agree and are therefore both equal to $F(b) - F(a)$. Each one-sided hypothesis is needed for its own half: in this formalization suprema and infima of unbounded sets take a default value, and an increasing everywhere differentiable $F$ with unbounded derivative shows that (2) fails without boundedness above, its reflection $-F$ that (1) fails without boundedness below.