On second thought, the original solution I posted is more than you need if you are willing to cite Rudin.
Assume that $f : [a,b] \to \mathbb R$ is nondecreasing. Extend $f$ to all of $\mathbb R$ by $f(x) = f(b)$ if $x > b$ and $f(x) = f(a)$ if $x < a$.
Define $f_n(x) = n[f(x+1/n) - f(x)]$. Then each $f_n \ge 0$ on $\mathbb R$ and $f_n \to f'$ almost everywhere. Fatou's lemma implies that
$$ \int f' \, dx \le \liminf_n \int f_n \, dx.$$
The proof finished rather quickly since if $\frac 1n < b-a$ you get
$$\int f_n \, dx = f(b) - f(a)$$
so that $f' \in L^1$.
It is well-known that $f$ is absolutely continuous on $[a,b]$ if and only if it is continuous, has bounded variation, and carries sets of measure zero to sets of measure zero.
Assuming that $f$ is differentiable everywhere in $[a,b]$ you get continuity for free, and since $f$ is monotone its variation is simply $|f(b) - f(a)|$.
It's not too hard to show using a Vitali covering argument that if $E \subset [a,b]$, $f$ is differentiable at every point of $E$, and $|f'(x)| \le k$ for all $x \in E$, then $m^*(f(E)) \le k m^*(E)$ where $m^*$ is the Lebesgue outer measure. Consequently if $f$ is differentiable at every point of $[a,b]$ and $N \subset [a,b]$, then $m^*(N) = 0$ implies $m^*(f(N)) = 0$.
This gives you AC as needed.