Ex2-7

Chapter 2 Question 7 
Question: Suppose f is of bounded variation on [a,b]. If f is continuous at a point x_0 , show that V(x), P(x), and N(x) are also continuous at x_0. In particular, if f is continuous on [a,b], then so are V(x), P(x), and N(x). 
Proof: Given a partition \Gamma with two adjacent points x_0<x, we prove the inequality below:
V[x_0,x]\leq |f(x)-f(x_0)|+V[a,b]-S_\Gamma.
Indeed, by applying Theorem 2.2(ii) twice, we have V[a,b]-V[x_0,x]=V[a,x_0]+V[x,b]. Moreover, we can write S_\Gamma as
S_\Gamma=S_{\Gamma_1}+|f(x)-f(x_0)|+S_{\Gamma_2},
where \Gamma_1=\Gamma\cap [a,x_0] and \Gamma_2=\Gamma\cap [x,b]. Basically, we have split \Gamma into three parts: \Gamma_1, \{x_0,x\}, and \Gamma_2. Since \Gamma_1 is a partition of [a,x_0], we have V[a,x_0]\geq S_{\Gamma_1}. Similarly, V[x,b]\geq S_{\Gamma_2}. Finally, we have
S_\Gamma\leq V[a,x_0]+|f(x)-f(x_0)|+V[x,b]=|f(x)-f(x_0)|+V[a,b]-V[x_0,x].
Now we prove the statement. Given \varepsilon>0, x_0\in[a,b], we choose \delta>0 such that 
|f(x)-f(x_0)|<\frac\varepsilon2\quad\text{ whenever }|x-x_0|< \delta.
Next, we choose a partition \Gamma such that V[a,b]-S_\Gamma<\varepsilon/2. Without loss of generality, we might assume x_0\in \Gamma since a refinement of partition will increase the value of S_\Gamma. We can also assume two points adjacent to x_0 (if x_0 is not boundary point) or the one point adjacent to x_0 (if x_0 is boundary point) is contained in the open interval (x_0-\delta,x_0+\delta). If this is not true, we simply add more points in the partition. 
Let x_1 be the point in \Gamma after x_0, we have
\begin{align}|V(x_1)-V(x_0)|&=V[x_0,x_1]\\
&\leq |f(x_1)-f(x_0)|+V[a,b]-S_{\Gamma}\\
&< \dfrac\varepsilon2 +\frac\varepsilon2\\
&=\varepsilon.\end{align}
Similar result holds for x_2 being the point in \Gamma before x_0. 
By monotonicity of V(x), we conclude |V(x)-V(x_0)|<\varepsilon for all x\in[x_2,x_1] where [x_2,x_1] has x_0 as an interior point. Thus V is continuous at x_0.
By Theorem 2.6, we find that since V(x),f(x) are both continuous at x_0, so are 
P(x)=\frac12[V(x)+f(x)-f(a)]\quad\text{ and }\quad N(x)=\frac12[V(x)-f(x)+f(a)].