let $f$ be a function of bounded variation on a closed and bounded interval $[a,b]$.then $f$ is measurable.
the proof is as follows: let $f$ be a function of bounded variation on a closed and bounded interval $[a,b],$ then by Jordan's theorem , $f$ is the difference of 2 increasing functions on $[a,b].$ EDIT: I found the proof here:
I think you need to be more careful. Do it from scratch: without loss of generality, $f$ is increasing. Fix $a\in \mathbb R$ and consider $f^{-1}((-\infty,a)).$ If this set is empty, then it is measurable. Otherwise, there is a $t\in f^{-1}((-\infty,a))$ and if $s\le t,$ then $f(s)\le f(t)<a$ so $s\in f^{-1}((-\infty,a))$ so $ f^{-1}((-\infty,a))$ is an interval. In fact, if $z:=\sup\{t:t\in f^{-1}((-\infty,a))\},$ then $f^{-1}((-\infty,a))$ is either $(\infty, z)$ or $(\infty, z]$ where $z$ may be $\infty$ (why?). It follows that $f^{-1}((-\infty,a))$ is measurable.
