7-5

Note that g is absolutely continuous. So g is of bounded variation on [a,b]. Then we know h=f-g is also of bounded variation on [a,b]. Then 
\begin{alignat}
{2}\int_a^b \phi df&=\lim_{P \to 0} \sum \phi(\epsilon)(f(x_i)-f(x_{i-1}))\\
&= \lim_{P \to 0} \sum \phi(\epsilon)(g(x_i)+h(x_i)-g(x_{i-1})-h(x_{i-1}))\\
&= \lim_{P \to 0} \sum \phi(\epsilon)(g(x_i)-g(x_{i-1}))+\lim_{P \to 0} \sum \phi(\epsilon)(h(x_i)-h(x_{i-1}))\\
&=\int_a^b \phi dg +\int_a^b \phi dh.
\end{alignat}
Then by Theorem 7.32 we get
\int_a^b \phi dg = \int_a^b \phi g' dx=\int_a^b\phi f'dx
since f'=g'+h'=g' almost everywhere on [a,b].
Combine these two results, we have 
\int_a^b \phi df = \int_a^b \phi f' dx + \int_a^b \phi dhThis complete the proof.