Let $f$ be a left continuous function of bounded variation on $[a,b]$ s.t. $f(a)=0$.
$f$ can be written as the difference $f=f_1-f_2$ of two non-decreasing functions on $[a,b]$.
$f_1=\frac{1}{2}(v(x)+f(x))$, $f_2=\frac{1}{2}(v(x)-f(x))$ , when $v(x)$ is the total variation of $f$ on $[a,x]$, and I've shown that $v(x)$ is also left continuous so $f_1$,$f_2$ are non decreasing, $f_1(a)=f_2(a)=0$ and left continuous. Let $m_{1}$, $m_{2}$ be Stieltjes-Lebesgue measures defined on $[a,b]$ by $m_{1}([a,x))=f_{1}(x)$, $m_{2}([a,x))=f_{2}(x)$. so,$m=m_{1}-m_{2}$ is a signed measure and $f(x)=m_{1}([a,x))-m_{2}([a,x))$. I don't understand why $m_{1}$ is singular to $m_{2}$? I believe that this is true because then $v(x)=m_{1}([a,x))+m_{2}([a,x))$ and the total variation of $f$ correspond to the total variation of $ m$.
Thank you.