Proposition 2

Let $f$ be a real function which is finite and continuous on the closed bounded interval $I = [a, b]$ (where $a < b$*) and has a right derivative* (finite or not) at all the points of the relative complement in $[a, b)$ of a countable subset $A$ of this interval. If $f'_d(x) \geq 0$ at every point of $[a, b[$ not belonging to $A$, then one has $f(b) \geq f(a)$; if, further, $f'_d(x) > 0$ for at least one point of $[a, b[$, then $f(b) > f(a)$.

Let $\varepsilon > 0$ be arbitrary, and denote by $(a_n)_{n \geq 1}$ a sequence obtained by listing the countable set $A$. Let $J$ be the set of points $y \in I$ such that one has

(1)

for all $x$ with $a \leq x \leq y$, the sum in the second term of the right-hand side being taken over all indices $n$ for which $a_n < x$. We shall show that if $f'_d(x) \geq 0$ at every point of $[a, b[$ distinct from the $a_n$, then $J = I$.

It is clear that $J$ is not empty, since $a \in J$; moreover the definition of this set shows that if $y \in J$ one has $x \in J$ for $a \leq x \leq y$, so $J$ is an interval with left-hand endpoint $a$ (Gen. Top., IV, p. 336, prop. 1); let $c$ be its right-hand endpoint. One has $c \in J$; this is clear if $c = a$; if not, for every $x < c$ we have the inequality (1), and a fortiori

from which it follows, on letting $x$ tend to $c$ in this inequality (since $f$ is continuous), that $c$ satisfies (1).

This being so, we shall see that we must have $c = b$. Indeed, if one had $c < b$, then certainly one would have $c \notin A$; now $f'_d(c)$ exists, and since $f'_d(c) \geq 0$ by hypothesis, there exists a $y$ such that $c < y \leq b$ and such that for $c \leq x \leq y$ one has

from which, taking account of (1), where $x$ is replaced by $c$,

which signifies that $y \in J$, contradicting the definition of $c$. Thus we have $c = a_k$ for some index $k$; since $f$ is continuous at the point $a_k$ there is a $y$ such that $c < y \leq b$ and such that for $c < x \leq y$ one has

from which, taking account of (1), where $x$ is replaced by $c$, which again leads to a contradiction; we thus have $c = b$, and in consequence Since $\varepsilon > 0$ is arbitrary we deduce from (2) that $f(b) \geq f(a)$, which demonstrates the first part of the proposition.

We remark now that this result applied to an interval $[x, y]$ where $a \leq x < y \leq b$ proves that $f$ is increasing on I; if one had $f(b) = f(a)$ one could deduce that $f$ is constant on I, and then that $f'_d(x) = 0$ at every point of $[a, b[$; the second part follows from this.