11
$\begingroup$

A function $F:[0,1]\rightarrow\mathbb{R}$ satisfies Lusin's (N) property if for every measure zero set $A\subseteq [0,1]$, $F(A)$ has measure zero. (This includes the assertion that $F(A)$ is measurable.)

A function $F:[0,1]\rightarrow\mathbb{R}$ is absolutely continuous if for every $\varepsilon$ there is a $\delta$ such that whenever $(a_i,b_i)_{i<k}$ is a finite set of disjoint intervals with $\sum_i b_i-a_i < \delta$, then $\sum_i |F(b_i) - F(a_i)| < \varepsilon$.

It is well known that a function is absolutely continuous if and only if it is continuous, of bounded variation, and satisfies Lusin's (N) property.

Here's another property which as far as I know does not have a name, and which I'll here call the (N)-like property: for every $\varepsilon$ there is a $\delta$ such that whenever $(a_i,b_i)_{i<k}$ is a finite set of disjoint intervals with $\sum_i b_i - a_i <\delta$, if the intervals $(F(a_i),F(b_i))$ (indices reversed if necessary) are also disjoint, then $\sum_i |F(b_i) - F(a_i)| < \varepsilon$.

The (N)-like property implies Lusin's (N) property. It also implies continuity.

Question: Is the (N)-like property equivalent to Lusin's (N) property plus continuity?

I got interested in this property because of wanting to isolate bounded variation from the other aspects of absolute continuity using only properties with a small number of quantifiers. In the context of computable analysis, being absolutely continuous is a $\Pi_3$ property, having bounded variation is a $\Sigma_2$ property, and being continuous is a $\Pi_2$ property. To complete the analysis of the descriptive complexity of the different parts of absolute continuity, I was hoping to find a $\Pi_3$ equivalent to Lusin's (N) property which was independent of bounded variation, though entanglement with continuity was fine for me because all computable functions are continuous anyway. Aside from not knowing whether it is just Lusin's (N) property plus continuity, the (N)-like property works for this purpose, because:

  1. A function is absolutely continuous if and only if it is continuous, of bounded variation, and satisfies the (N)-like property; and

  2. The (N)-like property does not imply bounded variation. For example, $x^2 \sin (1/x^2)$ satisfies it.

$\endgroup$
  • 2
    $\begingroup$ I haven't had a chance to think about this, and now I see that Christian Remling has posted an answer. Since I've already tracked down a couple of places I was going to suggest looking, I may as well stick them here in case anyone is interested. For Luzin's (N) property, see my 24 November 2006 sci.math post and two others in the same thread. Also, my Bibliography for Singular Functions might have some use. $\endgroup$ – Dave L Renfro Apr 30 '14 at 20:16
9
$\begingroup$

This does not work. What you call (N) like, is a rephrasing of Banach's condition (S): For every $\epsilon >0$, there exists $\delta>0$ so that $|F(E)|<\epsilon$ whenever $|E|<\delta$. (This trivially implies (N) like; use regularity for the converse.)

These conditions are discussed in detail in Saks' book. The basic theorem is: (S) $\iff$ (N) and (T1) (Saks, IX.7.4)

(T1) denotes the condition that $F^{-1}(\{ x\})$ is finite for almost all $x$. This condition need not hold for continuous $F$'s satisfying (N); this is discussed in Saks, Section IX.6.

$\endgroup$
  • $\begingroup$ Could you please point out which book of Saks you referred to? $\endgroup$ – Changyu Guo Apr 30 '14 at 21:25
  • 2
    $\begingroup$ Saks, Theory of the integral, an old classic (available as a Dover edition, that's 10 bucks well spent). $\endgroup$ – Christian Remling May 1 '14 at 0:05
  • $\begingroup$ And what a marvelous book it is, both efficient and clear. I think I'll be buying a Dover copy myself. Thank you Christian! $\endgroup$ – Linda Brown Westrick May 1 '14 at 15:47

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.