This was written in Freitag's (p196)
A continuous function $f:[a,b) \rightarrow \mathbb{C}$ is called improperly integrable iff the limit $$ \lim_{t\rightarrow b} \int_a^t f(x) \, dx $$ exists. It is (improperly) absolutely integralbe, if the above limit exists with $f$ replaced by $|f|$. Absolute integrability implies integrability (?)
What I don't understand is why the last holds bolded line holds.
My thoughts: I don't even see how this holds if $f$ were real. If we define $G(t):= \int_a^t f(x) \, dx$, then $G(t)$ is a continuous (by FTC), and bounded (by absolutely integrability) function on $[a,b)$. We need to show $\lim_{t \rightarrow b} G(t)$ exists, but it can be the case that $G(t)$ is some function like $\sin (1/x)$.
Outline: imitate the proof for absolute convergence $\implies$ convergence for sums. Write down the Cauchy criterion for $g(x) = \int_0^x \lvert f \rvert$, and use it and the triangle inequality to show that $h(x) = \int_0^x f$ is Cauchy.