I am reading the book Real Analysis by Folland, and I found that someone else asked the same question that I currently have in Question about positive variations of function
For a function $f:\mathbb{R} \to \mathbb{C}$ of bounded variation, let me first write $$ v(f) = \sup \left\{\sum_{1}^n [f(x_j)-f(x_{j-1})]^+: -\infty < x_0 <x_1 < \dots x_n = x , n \in \mathbb{N} \right \} + \frac{1}{2}f(-\infty)$$ and $$ u(f) = \sup \left\{\sum_{1}^n [f(x_j)-f(x_{j-1})]^+ + \frac{1}{2}f(x_0): -\infty < x_0 <x_1 < \dots x_n = x , n \in \mathbb{N} \right \} $$
In order to show $v(f) = u(f)$, I want to show that any upper bound of the set for $v$ is an upper bound of the set for $u$, and vice versa. But without knowing how $f$ behaves as $x \to -\infty$, how is this possible?
$\sum_{1}^n [f(x_j)-f(x_{j-1})]^+ + \frac{1}{2}f(x_0)\le u(f)$ for any partition $P$.
For $\epsilon>0$, there is an $x_0$ such that $f(-\infty)-\epsilon<f(x_0)<f(-\infty)+\epsilon$. Without loss of generality, we may assume that $P$ contains $x_0$. Then,
$\sum_{1}^n [f(x_j)-f(x_{j-1})]^+ + \frac{1}{2}f(-\infty)\le \sum_{1}^n [f(x_j)-f(x_{j-1})]^+ + \frac{1}{2}f(x_0)+\frac{1}{2}\epsilon\le u(f)+\frac{1}{2}\epsilon.$
This is true for all partitions $P$ so in fact, $v(f)\le u(f).$
Now fix a partition $P.$ Then, $\sum_{1}^n [f(x_j)-f(x_{j-1})]^+ + \frac{1}{2}f(-\infty)\le v(f)$
And for $\epsilon>0$, there is an $x_0$ such that $f(-\infty)-\epsilon<f(x_0)<f(-\infty)+\epsilon$. Without loss of generality, we may assume $P$ contains $x_0$. Then,
$\sum_{1}^n [f(x_j)-f(x_{j-1})]^++\frac{1}{2}f(x_0)\le \sum_{1}^n [f(x_j)-f(x_{j-1})]^+ + \frac{1}{2}f(-\infty)+\frac{1}{2}\epsilon\le v(f)+\frac{1}{2}\epsilon,$ which, on suping over all partitions $P$, implies that $u(f)\le v(f).$