how to get this sequence of partitions of a interval $[a,b] $

Question

Let $f:[a, b] \rightarrow \mathbb{R}$ of bounded variation, let $P = \{ a= x_0<\cdots< x_{n} = b\}$ a partition of $[a,b]$ and define $r_{P}^{-} =\displaystyle \sum_{i = 0}^{n} (f(x_{i+ 1}) - f(x_i))^{-}$. Thus $r^{-}_P$ is the sum of negative terms of the series $\displaystyle\sum_{i = 0}^{n}[f(x_{i+ 1}) - f(x_i)].$

Define $V^{-}\{f,[a,b]\} = \sup\{r^{-}_P; P~\text{is a partition of}~[a,b]\}$. How to find a sequence of partitions $(P_k)$ such that $P_{k + 1}$ is a refinement of $P_k$ for all $k\in\mathbb{N}$ and $ \displaystyle\lim_{k\to \infty} r^{-}_{P_k} = V^{-}\{f,[a,b]\}$?

Answer 1

Construction by induction.

Star with $P_0 = \{a,b\}$. For $k=1$, by the definition of $V^{-}\{f,[a,b]\}$ there exists a partition $P_1$ of $[a,b]$ such that $r^{-}_{P_1}>V^{-}\{f,[a,b]\} - 1$. Then $P_1$ is a refinement of $P_0$ and $$V^{-}\{f,[a,b]\}\geq r^{-}_{P_1}>V^{-}\{f,[a,b]\} - 1.$$ For $k=2$ there exists a partition $P_2^*$ of $[a,b]$ such that $r^{-}_{P_2^*}>V^{-}\{f,[a,b]\} - \frac{1}{2}$. Take $P_2 = P_1\cup P_2^*$. Then $P_2$ is a refinement of $P_1$ and $$V^{-}\{f,[a,b]\}\geq r^{-}_{P_2}>V^{-}\{f,[a,b]\} - \frac{1}{2}.$$

Hope you got the idea.