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In classical real analysis I've only seen absolute continuity defined for functions on compact interval $[a,b]$, where the two equivalent definitions are: $f:[a,b]\rightarrow\mathbb{R}$ is AC if

(1) Given $\epsilon > 0$ there is a $\delta > 0$ such that $\sum_{i=1}^n |f(y_i)-f(x_i)|< \epsilon$ for every finite collection of nonoverlapping intervals $( (x_i,y_i) )_{i=1}^n $ each contained in $[a,b]$ with $\sum_{i=1}^n |y_i-x_i|< \delta$. Or,

(2) $f'$ exists a.e. on $[a,b]$, $f'$ is integrable on $[a,b]$, and

$f(x) = \int_a^x f'(y) dy + f(a)$ for all $x \in [a,b]$.

Is there an accepted definition for absolute continuity of a function on an open, possibly unbounded, interval $(a,b)$ where $-\infty \leq a < b \leq \infty$?

It seems that definition (1) extends easily to this case if we replace $[a,b]$ by $(a,b)$. If we call this condition (1'), then it's easy to show (1') is equivalent to (see answer by Jonas below). A natural extension of (2) to this case would be

(2') $f$ is AC on an open set $U$ if, for all compact intervals $[c,d] \subset (a,b)$, $f$ is AC in the sense of (2) on $[c,d]$.

Which of these is the best extension of the definition? I don't know enough about the notion of absolute continuity of measures to know if my extended definitions are consistent with that generalization as well.

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Greg: en.wikipedia.org/wiki/Absolute_continuity says that the its enough for $I$ to be an interval. In that case i can take $I=(-\infty,\infty)$. –  anonymous Sep 11 '10 at 18:57
    
Actually I am not able to answer your question but maybe these two posts Absolute Continuity I unapologetic.wordpress.com/2010/07/01/absolute-continuity-i and Absolute Continuity II unapologetic.wordpress.com/2010/07/02/absolute-continuity-ii from John Armstrong's blog (The Unapologetic Mathematician) help you. –  Américo Tavares Sep 11 '10 at 19:35
    
@Americo Tavares: I think he doesn't want to make use of Measure theory, as stated in the last line of the problem. –  anonymous Sep 11 '10 at 19:52
    
@Chandru1: You are right! –  Américo Tavares Sep 11 '10 at 19:57
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1 Answer

up vote 1 down vote accepted

(1') is not equivalent to (2'). For example, $f(x)=x^2$ satisfies (2') on $(-\infty,\infty)$ but not (1'). It is not even uniformly continuous.

Condition (2'), being AC on all compact subintervals, is a condition I have at least seen used, and it is the right one if you want to extend the equivalence to being an indefinite integral. Namely, it is equivalent to:

(3) If $a\lt c \lt b$, then $f(x)=f(c)+ \int_c^x f'$ for all $x\in(a,b)$.

This is only a slight modification of (2), which must be made because $f$ may be unbounded or otherwise undefined at $a$ and $f'$ may not be integrable on $(a,x)$, even if $f$ satisfies the stronger condition (1') (e.g., $f(x)=x$ on $(-\infty,\infty)$).

But you asked if this is the "best" definition or if there is an "accepted" definition, and of that I am not sure. I have not seen anyone write "$f$ is AC on $(a,b)$" when they mean (2') holds. For the case $(-\infty,\infty)$, I have seen simply "f is AC on bounded intervals". On the Wikipedia page, they use the phrase "locally absolutely continuous" in the section on the relationship to measures.

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Thanks Jonas. Still helping me even after you've graduated! –  Greg O. Sep 13 '10 at 15:46
    
Ah, I was wondering whether you were that Greg O.! You're welcome. It would still be interesting to see whether (1') has any uses, being strictly between Lipschitz continuous and indefinite integral, but I don't expect anything. –  Jonas Meyer Sep 13 '10 at 16:27
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