A function $f : [a\,..b] \to \mathbb R$ is (uniformly) continuous if and only, if it admits a modulus of continuity. Absolutely continuous functions are uniformly continuous.
Is there a condition concerning the moduli of continuity of a function $f$ that precisely says that $f$ is absolutely continuous?
There is no such condition: Let $f : [a,b] \to \Bbb{R}$ be continuous and let $\omega : [0,\infty) \to [0,\infty), h \mapsto \sup_{x,y \in [a,b], |x-y|\leq h} |f(x) - f(y)|$ be the (minimal) modulus of continuity.
By the usual properties of (minimal) moduli of continuity for continuous functions defined on compact, convex sets, we have the following:
$\omega (0) = 0$ and $\omega$ is continuous.
$\omega$ is nondecreasing
$\omega$ is concave.
From continuity + concavity, it follows (see e.g. here: Prove that convex function on $[a,b]$ is absolutely continuous) that the derivative $\omega '$ exists almost everywhere and is nonincreasing and satisfies $\omega (y) - \omega(x) = \int_x^y \omega'(t) dt$ for all $x,y \in [0,\infty)$. Furthermore, since $\omega$ is nondecreasing, $\omega ' \geq 0$.
Next, it is not hard to see $\omega(h) = \omega(b-a)$ for all $h \geq b-a$, since there are no $x,y \in [a,b]$ with $|x-y| > b-a$. This implies $\omega ' (t) = 0$ for $t > b-a$, so that $\omega' \geq 0$ is in fact integrable (since $\int_0^\infty \omega' dt = \int_0^{b-a} \omega' dt = \omega(b-a) < \infty$). Hence, $\omega$ is absolutely continuous.
Finally, note for $x, h \in [0,\infty)$ that \begin{align*} 0 \leq \omega (x+h) - \omega(x) & = \int_x^{x+h} \omega'(t) dt \\ & = \int_0^h \omega' (t + x) dt \\ (\text{since } \omega' \text{nonincreasing and } x \geq 0)& \leq \int_0^h \omega ' (t) dt \\ &= \omega(h) - \omega(0) = \omega(h). \end{align*} This easily implies that the (minimal) modulus of continuity of $\omega$ is $\omega$.
Summary: What we have shown is that for an arbitrary (minimal) modulus of continuity $\omega$ of a continuous function $f$, there is an absolutely continuous function $g$ (in fact, we can take $g = \omega$) such that the modulus of continuity of $g$ is also $\omega$.
But since we can take $f$ to be not absolutely continuous, it is impossible to characterize absolute continuity just by looking at the associated (minimal) modulus of continuity (otherwise, $f$ would have to be absolutely continuous iff $g$ is).
EDIT: Finally, as we noted above, we have $\omega (x) = \omega(b-a)$ for all $x \geq b-a$. This implies that the modulus of continuity of $\omega |_{[0,b-a]}$ is also $\omega$.
Now, finally note that the map $[a,b] \to [0, b-a], x \mapsto x - a$ preserves distances. Hence, $g : [a,b] \to [0,\infty), x \mapsto \omega (x-a)$ is absolutely continuous and has (minimal) modulus of continuity $\omega$. Thus, it is not possible to characterize absolute continuity by the modulus of continuity, even if one restricts to the interval $[a,b]$. Of course, this only holds if $a < b$, since otherwise, one can not choose a continuous function $f : [a,b] \to \Bbb{R}$ which is not absolutely continuous.
