Ex2-8

Assume f\in V(-\infty,\infty). Let v(x)=V(-\infty,x) for x\in(-\infty,\infty)
. It is clear that 
f(x)=v(x)-[v(x)-f(x)]Then we need to show that v(x) and v(x)-f(x) is increasing function. For any x_1<x_2, we have 
v(x_2)-v(x_1)=V[f;x_1,x_2]\geq 0
Since f(x) is of bounded variation, this implies v(x)is increasing bounded function. Moreover, f(x_2)-f(x_1)\leq |f(x_2)-f(x_1)|\leq V[f;x_1,x_2]=v(x_2)-v(x_1)So we find thatv(x_1)-f(x_1)\leq v(x_2)-f(x_2)
Since f(x) and v(x) both are of bounded variation, v(x)-f(x) is incresing bounded function. Hence f(x)
 is the difference of two increasing bounded functions.