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When thinking about the proof that every absolutely continuous function has Lusin property (N) (i.e., it maps null sets to null sets) I had the feeling that this property might be useful:

Let $f\colon[a,b]\to\mathbb R$. I am interested in functions with the property that for each $\varepsilon>0$ there exists $\delta>0$ such that if $a\le a_1\le b_1 \le \dots \le a_n \le b_n \le b$ and $$\sum_{i=1}^n (b_i-a_i)<\delta$$ then also $$\sum_{i=1}^n |f(y_i)-f(x_i)|<\varepsilon$$ for any choice of $x_i,y_i\in [a_i,b_i]$.

This property implies absolute continuity, which is what we get for $x_i=a_i$ and $y_i=b_i$. Is this property equivalent to absolute continuity?

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  • $\begingroup$ To clarify your observation here: let $\omega f([c,d])$ denote the oscillation of $f$ on the interval $[c,d]$, i.e., $$\omega f([c,d]) = \sup \{|f(y)-f(x)|: c\leq x < y \leq d \}.$$ Then you have shown that the apparently stronger condition that $\sum_{i=1}^n \omega f([a_i,b_i]) < \epsilon$ does not in fact produce a stronger version of absolute continuity. Try the same thing with bounded variation. $\endgroup$ – B. S. Thomson Dec 9 '15 at 16:44
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It is trivial that this is the same thing as absolute continuity. (I should have noticed this immediately.)

I can simply use that $$\sum_{i=1}^n |y_i-x_i| \le \sum_{i=1}^n (b_i-a_i) < \delta$$ and then use the definition of absolute continuity for the points $x_i$, $y_i$ to get $$\sum_{i=1}^n |f(y_i)-f(x_i)|<\varepsilon.$$

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