TITLE: Absolute continuity on an open interval of the real line?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: (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.