7-8

Question: Prove the following converse of Theorem 7.31: If f is of bounded variation on [a,b], and if the function V(x)=V[a,x] is absolutely continuous on [a,b], then f is absolutely continuous on [a,b].
Solution: 
Since f is of bounded variation on [a,b], then we have
V(x)=V[a,x]\leq V[a,b]<\infty
for all a\leq x\leq b.
Thus, V(b_i)-V(a_i) is well-defined and so is \sum |V(b_i)-V(a_i)|. 
Since the function V(x)=V[a,x] is absolutely continuous on [a,b], by the definition of absolutely continuous, for any \epsilon>0, there exists \delta>0 such that for any collection \{[a_i,b_i]\} of nonoverlapping subintervals of [a,b], 
\sum |V(b_i)-V(a_i)|<\epsilon\quad \text{if}\quad\sum(b_i-a_i)<\delta.
From Chapter 2, we know that |f(b_i)-f(a_i)|\leq V[a_i,b_i]. So, if \sum(b_i-a_i)<\delta,
\sum|f(b_i)-f(a_i)|\leq \sum V[a_i,b_i].
Since V[a_i,b_i]=V[a,b_i]-V[a,a_i]=V(b_i)-V(a_i), we then have 
\begin{align}
\sum|f(b_i)-f(a_i)|&\leq \sum|V(b_i)-V(a_i)|\\
&<\epsilon
\end{align}
when \sum(b_i-a_i)<\delta.
Thus, we show that f is absolutely continuous on [a,b].