Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a càdlàg function with finite total variation, i.e. $$ V_f(t)\doteq\sup_{n}\sum_{j=1}^n\left|f(t_{j,n})-f(t_{j-1,n})\right|<\infty $$ where $0=t_{0,n}<t_{1,n}<\cdots<t_{n,n}=t$ is any partition of the interval $[0,t]$. I have found the following result: $$ V_f(t)= V_f(t-)+\left| f(t)-f(t-)\right|\quad (1) $$ where $f(t-)=\lim_{s\rightarrow t-}f(s)$ indicates the left-limit. The equation $(1)$ says that the jumps of the total variation coincide with those of the function $|f|$.
Any suggestion or reference on where I can find the proof?
Take any partition of $[0,t]$, $0=t_{0}<\cdots<t_{n}=t$. Let $t_{n-1}<s<t$. Then \begin{align*} \sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})| & \leq\sum_{i=1}^{n-1}|f(t_{i} )-f(t_{i-1})|+|f(s)-f(t_{n-1})|+|f(t)-f(s)|\\ & \leq V(s)+|f(t)-f(s)|\leq V(t-)+|f(t)-f(s)|. \end{align*} Letting $s\rightarrow t^{-}$, we get $$ \sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|\leq V(t-)+|f(t)-f(t-)|. $$ Taking the supremum over all partitions gives $$ V(t)\leq V(t-)+|f(t)-f(t-)|. $$ Conversely, given $\varepsilon>0$, find $\delta>0$ such that $$ V(t-)\leq V(s)+\varepsilon,\quad|f(t-)-f(s)|\leq\varepsilon $$ for all $s\in\lbrack t-\delta,t)$. Take any partition of $[0,s]$, $0=t_{0}<\cdots<t_{n}=s$. Then $$ \sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|+|f(t)-f(s)|\leq V(t). $$ Taking the supremum over all partitions gives $$ V(s)+|f(t)-f(s)|\leq V(t). $$ Now for $s\in\lbrack t-\delta,t)$, $$ V(t-)-2\varepsilon+|f(t-)-f(t)|\leq V(s)+|f(t)-f(s)|\leq V(t), $$ where we used the fact that \begin{align*} |f(t-)-f(t)| & \leq|f(t-)-f(s)|+|f(t)-f(s)|\\ & \leq|f(t)-f(s)|+\varepsilon. \end{align*} Letting $\varepsilon\rightarrow0$ gives $$ V(t-)+|f(t-)-f(t)|\leq V(t), $$