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! 


1 Answer 1


There exists a continuous bounded function $f \colon [0,1] \to \mathbb R$ with $\sigma$-finite variation in the sense 1 and with positive set of points where the function is not approximately differentiable.

An illustration of one such function is below. (I will only briefly comment on it. Please (come and) ask more details if you need.) Construct an oscillating function so that the set where it oscillates has positive measure, i.e. the length of the intervals where we define the function to be affine sum to something less than $1$.

    alt text (source)

This function has $\sigma$-finite variation in the sense 1: On each of the countably many intervals where the function is affine, the variation is clearly finite. On the remaining set where we "oscillate", the variation is also easily seen to be finite since on any given scale we have removed the set where we jump. Notice that this remaining set does not have finite variation in the sense 2 because for it removing the affine part does not remove the variation. The function is clearly not approximately differentiable in the oscillating part.


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