Bounded variation and total variation on an interval

Let $a\le b$ and let $f:[a,b]\to\mathbb R$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). If $a<b$ and $P=(n,t)$ is a partition of $[a,b]$ (Partition of $[a,b]$ as a finite strictly increasing list $a = t_0 < t_1 < \dots < t_n = b$, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions), the variation of $f$ over $P$ is

$V(f,P):=\sum_{i<n}|f(t_{i+1})-f(t_i)|.$

The sum is finite (Finite sums and finite products, by recursion, Laws of finite sums and finite products) and nonnegative (Absolute value in an ordered field). The set of all such sums is nonempty, since $[a,b]$ has the partition with point set $\{a,b\}$. The function $f$ has bounded variation on $[a,b]$ when this set of sums is bounded above (Lower bound, bounded below, bounded set). In that case its total variation is

$\operatorname{Var}_{[a,b]}(f):=\sup_P V(f,P).$

Completeness of $\mathbb R$ gives this supremum and Suprema and infima are unique makes it unique (Complete ordered field (least-upper-bound property)). On a singleton interval, by convention, $\operatorname{Var}_{[a,a]}(f):=0$; no partition from Partition of $[a,b]$ as a finite strictly increasing list $a = t_0 < t_1 < \dots < t_n = b$, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, whose standing hypothesis is $a<b$, is invoked.