# Total variation of a cadlag function

Let $f: [0, +\infty) \rightarrow \mathbb{R}$ be a cadlag function with $\|f \|_{t, \text{var}} < +\infty$, where $\|f \|_{t, \text{var}} < +\infty$ denotes the total variation of $f$ over the interval $[0,t]$.

There is an exercise asking us to prove that the function $g: [0, \infty) \rightarrow \mathbb{R}; \, t \mapsto \|f \|_{t, \text{var}}$ is also cadlag. (However, I have only seen the proof concerning right-continuity.)

Also, how can we compute the jump of the total variation function? (i.e. express $g(t) - g(t-)$ in terms of $f$)

Hint: Fix $t_0\in [0,\infty)$. The same argument which is used to establish the right continuity of $g=\|f\|_{t,var}$ at $t_0$ can be used to establish the left continuity of $\|f_0\|_{t,var}$ at $t_0$, where $f_0$ is defined to be $f$ except at $t_0$ where it is defined as $\lim_{t\to t_0^-}f(t)$. Then you can show that $g(t)-g(t^-)=|f(t)-f(t^-)|$.
• Can you explain why $g(t)- g(t-) = | f(t)- f(t-)|$?
• By the left continuity of $\|f_0\|_{t,var}$ at $t_0$, choose $t_1<t_0$ such that $\|f_0\|_{t_0,var}-\|f_0\|_{t_1,var}<\epsilon$. Thus $f_0$ has total variation at most $\epsilon$ on the interval $[t_1,t_0]$. Now you can use the definition of total variation to show that $f$ has total variation at most $\epsilon+|f(t_0)-f_0(t_0)|$ on the interval $[t_1,t_0]$. Feb 28, 2015 at 16:16