Absolute continuity on an open interval of the real line?

Solution 1:

(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.