Let $g : [a,b] \rightarrow \mathbb{C}$ be of bounded variation and let $1_A$ denote the indicator function of $A$ for measurable subsets of $[a,b]$.
We put
$$m_g(A) := \int 1_A dg,$$
i. e. $m_g(A)$ is the Lebesgue-Stieltjes integral of $1_A$ with respect to $g$. Then $m_g$ is a Borel measure on $[a,b]$.
I know that if $g$ is differentiable, then $m_g$ is absolutely continuous (=dominated by Lebesgue measure).
Now I wonder if this is always true? If not, what are conditions on $g$ that ensure that $m_g$ is absolutelly continuous?
Note: This is not a homework but I'm just interested and I am a little bit rusty on measure theory. Any reference or hint is welcome.
Lat $a<c<b$. If $g(x)=0$ for $x <c$ and $1$ for $x \geq c$ then $g$ is of bounded variation and $m_g$ is the delta measure at $c$. [This means $m_g(A)=1$ if $c \in A$ and $0$ otherwise]. This measure is singular w.r.t. Lebesgue measure. A sufficient condition for absolute continuity is $g$ continuously differentiable, as you have observed.
A necessary and sufficient condition is the so-called absolute continuity of the function $g$: For every $\epsilon >0$ there exists $\delta >0$ such that for any finite disjoint collection $(a_i,b_i): 1\leq i \leq n$ of intervals with $\sum (b_i-a_i) <\delta $ we have $\sum |g(b_i)-g(a_i)| <\epsilon$.