Proof of a result about Riemann Stieltjes integral and functions of bounded variation I stumbled across the following result which I couldn't prove. I would gladly accept some hints.

Let f be a bounded function defined on $[a,b]$ and let $α$ be a function of bounded variation on [a,b]. Furthermore let $V(x)=Vα(a,x)$ be the total variation function on α. Then if $f$ is Riemann-Stieltjes integrable with respect to α on $[a,b]$ then $f$ is Riemann-Stieltjes integrable with respect to $V$ and $V−α$ on $[a,b]$.

I know that whenever $f(x)$ is continuous, $α$ of bounded variation on [a,b] then $f(x)$ is Riemann-Stieltjes integrable with respecto to α.
Thanks!
 A: We have $\vert f\vert \le M<\infty$ on $[a,b].$ 
Let $\epsilon >0$ and choose a partition $P = \left \{ a = x_0 < x_1 < \cdots < x_n = b \right \}$ such that
$\tag 1 V(b) <\sum_{k=1}^{n}\vert \Delta \alpha_k\vert +\frac{\epsilon }{4M}$ 
and 
$\tag 2 \sum_{k=1}^{n}\vert f(t_k')-f(t_k)\vert  \vert \Delta \alpha_k\vert  <\frac{\epsilon }{4}$.
Now choose the $t'_k\le t_k$ so that 
$\sum_{k=1}^{n}\vert M_k(f)-m_k(f)\vert \vert \Delta \alpha_k \vert<\sum_{k=1}^{n}\vert f(t_k')-f(t_k)\vert \vert \Delta \alpha_k \vert+\frac{\epsilon }{4}$.
Then, 
$U(P, f, V ) − L(P, f, V )=\sum_{k=1}^{n}(M_k(f)-m_k(f))\Delta V=\sum_{k=1}^{n}(M_k(f)-m_k(f))(\Delta V-\vert \alpha_k\vert)+\sum_{k=1}^{n}(M_k(f)-m_k(f))\vert \alpha_k\vert<$
$2M\sum_{k=1}^{n}(\Delta V_k-\vert\Delta \alpha_k\vert )+\frac{\epsilon }{2}=2M(V(b)-\sum_{k=1}^{n}\vert \Delta \alpha_k \vert)+\frac{\epsilon }{2}<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon,$
which shows that $f$ is R-S integrable with respect to $V$. The second part is now immediate.
edit: to prove 2)., define $S=\left \{ k\in P:\Delta \alpha_k\ge 0 \right \}$ and $T=\left \{ k\in P:\Delta \alpha_k< 0 \right \}$. Then, choose $t_k,t'_k$ so that $f(t_k)-f(t'_k)>M_k(f)-m_k(f)-\epsilon /8V(b)$ whenever $k\in S$, and $f(t'_k)-f(t_k)>M_k(f)-m_k(f)-\epsilon/8V(b)$ whenever $k\in T.\ $
Then, 
$\sum_{k=1}^{n}\vert M_k(f)-m_k(f)\vert \vert \Delta \alpha_k \vert<\sum_{k\in S}(f(t_k)-f(t'_k))\vert \Delta \alpha_k\vert+\sum_{k\in T}(f(t'_k)-f(t_k))\vert \Delta \alpha_k\vert+$
$\epsilon /8V(b)\sum_{k=1}^{n}\vert \Delta \alpha_k \vert =\sum_{k=1}^{n}(f(t_k)-f(t'_k))\Delta \alpha_k +$
$\epsilon /8V(b)\sum_{k=1}^{n}\vert \Delta \alpha_k \vert <\epsilon /8+\epsilon /8=\epsilon/4.$
Thus $\tag 3 \sum_{k=1}^{n}\vert f(t_k')-f(t_k) \vert \Delta\alpha_k\vert  <\sum_{k=1}^{n}\vert M_k(f)-m_k(f)\vert \vert \Delta \alpha_k \vert<\epsilon /4.$ 
