How to prove that a function is of bounded outside variation?

Question

Let $E$ be the Vitali set (or any other nonmeasurable subset of $[0,1]$ with respect of Lebesgue measure), so $E\subset [0,1]$ is nonmeasurable. Consider the following function $f\colon [0,1]\to l_{\infty}[0,1]$ $$f(t)=\begin{cases} \theta, &t\notin E \\ (1,1,\ldots,1,\dots) & t\in E \\ \end{cases}.$$ It turns out that $\|f(t)\|=\chi_{E}(t)$ which implies that $\|f(\cdot)\|$ is nonmeasurable and hence neither Riemann nor Lebesgue integrable. I want to prove that $f$ is Riemann integrable. To do so, I need to show that it is of bounded outer variation, i.e., $$\sup\Big\|\sum_{i}\big(f(d_i)-f(c_i)\big)\Big\|<\infty$$ where the supremum is taken over all finite collections $\{[c_i,d_i]\}$ of nonoverlapping intervals of $[0,1]$. How to prove that $f$ is of bounded outer variation? Nothing comes to my mind unfortunately.

Answer 1

The worst scenario is when $d_i\in E$ all the time and $c_i\notin E$ for all $i$. But this is impsoosible, as if $d_i\in E$ for all $i$, then $c_{i+1}=d_i\in E$ - contradiction. This implies that if we divide the partition more and more, there is a finite number of intervals $[c_i,d_i]$ where $c_{i}\notin E$ and $d_{i}\in E$.