I have the following question:

*If $f:[0,1]\to \mathbb{R}$ is a bounded continuous function of $\sigma$-finite variation in sense 1, then is it true that $f$ is approximately differentiable a.e. on $[0,1]$?*

The $\sigma$-finite variation in sense 1 is obtained in the following way. For any measurable set $E\subset [0,1]$, we define $$v_f^\delta(E)=\inf\{\sum_{i=1}diam f([a_i,b_i]):E\subset \cup_i[a_i,b_i], b_i-a_i<\delta\}$$ and $$v_f(E)=\lim_{\delta\to0}v_h^\delta(E).$$ We say that $f$ has finite variation on $E$ if $v_f(E)<\infty$. We say that $f:[0,1]\to \mathbb{R}$ has $\sigma$-finite variation in sense 1 if there exists a partition $\{E_i\}$ of $[0,1]$ such that $f$ has finite variation on each $E_i$.

Note that the answer to the question is positive if the following statement is true:

*Let $f:[0,1]\to \mathbb{R}$ be a bounded continuous function. Then $f$ has has $\sigma$-finite variation in sense 1 implies that $f$ has $\sigma$-finite variation in sense 2.*

For $E\subset [0,1]$, one defines another variation as $$v_f'(E)=\sup\sum|f(b_i)-f(a_i)|,$$ where the supremum is taken over all non-overlapping intervals $\{[a_k,b_k]\}$ whose end points belong to $E$. We say that $f$ has finite variation on $E$ in sense 2 if $v_f'(E)<\infty$. A bounded function $f:[0,1]\to \mathbb{R}$ is said to have $\sigma$-finite variation in sense 2 if $[0,1]=\bigcup_{i=1}^\infty E_i$ such that $f$ has bounded variation on each $E_i$ in sense 2.

Thanks for any suggestions and answers!