Does Absolute Integrability Imply Riemann Integrability?

In John B. Conway's book he defines Riemann integrability, and then defines improper integrals based off of that. He says $$f:(a,b)\to\mathbb{R}$$ is improperly integrable if it is integrable on all $$[c,d]\subseteq(a,b)$$ and $$\lim_{c\to a^+}\lim_{d\to b^-}\int_c^df$$ exists and is finite.

Without defining any other form of integrability he claims that if $$f:(a,b)\to\mathbb{R}$$ is absolutely integrable (that is $$|f|$$ is integrable), then it is integrable. Consider

$$f:=\begin{cases}1&x\in\mathbb{Q}\\-1&x\not\in\mathbb{Q}\end{cases}.$$

This modified Dirichlet function is absolutely integrable on any finite $$(a,b)$$ for $$|f(x)|\equiv 1$$. Conway's claim asserts that $$f$$ is then integrable. If $$[c,d]\subseteq(a,b)$$, then would not $$U(f|_{[c,d]})=(d-c)$$ and $$L(f_{[c,d]})=-(d-c)$$? In which case $$f$$ is not Riemann Integrable on $$[c,d]$$, hence not improperly integrable on $$(a,b)$$.

In fact, if $$|f|$$ is improperly integrable in the above sense, then $$f$$ isn't necessarily integrable on compact sets $$[c,d] \subset (a,b)$$. You have already mentioned the typical example. Probably, Conway had in mind that $$f$$ is continuous on $$(a,b)$$, or piecewise continuous. Then $$f$$ is improperly integrable, if $$|f|$$ is improperly integrable.
Note that you can replace $$\mathbb{Q}$$ by a non-measurable set $$E$$, say a Vitali-set, to get a function with is non-measurbale, but |f| is measurable.