Let $F\in BV$ be real-valued, and denote by $T_F(x)$ the total variation function of $F$, $$T_F(x)=\sup\left\{\sum_1^n |F(x_j)-F(x_{j-1})|:n\in\mathbb N,-\infty<x_0<\cdots<x_n=x\right\}.$$ Then we can write $F=\frac12 (T_F+F)-\frac12(T_F-F)$, where $\frac12 (T_F\pm F)$ are increasing. Now, Folland claims in real analysis text (page 103) that $$\frac12 (T_F\pm F)=\sup\left\{\sum_1^n [F(x_j)-F(x_{j-1})]^\pm :x_0<\cdots<x_n=x\right\}\mp\frac12 F(-\infty),$$where $x^+=\max(x,0)=\frac12 (|x|+x)$ and $x^-=\max(-x,0)=\frac12(|x|-x)$. Could somebody help me verify this equality? (Note that in the book it says $\pm\frac12 F(-\infty)$, but the errata says it should be $\mp\frac12 F(-\infty)$.)
Attempt: Let's just assume $\frac12 (T_F+F)$. Then $$[F(x_j)-F(x_{j-1})]^+=\frac12 |F(x_j)-F(x_{j-1})|+\frac12 (F(x_j)-F(x_{j-1})).$$Now, $\sum_1^n (F(x_j)-F(x_{j-1}))=F(x)-F(x_0)$. But this is where I'm stuck. How do I simplify $$\sup\left\{\frac12\sum_1^n |F(x_j)-F(x_{j-1})|+\frac12 (F(x)-F(x_0)):x_0<\cdots<x_n=x\right\}?$$
I know of $\sup (A+B)=\sup A+\sup B$ and $\sup cA=c\sup A, c>0$, but I'm a bit uncertain about these identities and if they apply here, since the above is a set of functions. We also have $\sup_A (f+g)\leq \sup_A f+\sup_A g$, but we don't want inequality.