Does Absolute Improper Integrability imply Lebesgue Integrability? These is the definition I use: 
Definition 1: An improper integral will be an integral of the form: 
$$\int_a^b f(x) \, dx $$
Where $f$ is Riemann Integrable, on every finite subinterval, $[s,t] \subseteq (a,b)$. ($a,b$ can be $ \pm \infty$). If for some $c \in (a,b)$, the limits exists, $$ \int_a^b f(x) \, dx := \lim_{s \rightarrow a } \int_s^c f (x) \, dx + \lim _{ t \rightarrow b} 
\int_c^t f(x) \, dx . $$ 
An improper integral is \textit{absolutely convergent} if $\int_a^b |f(x)| \,dx $ converges.

Question: I know if $f$ is Riemann Integrable on $[a,b]$ then (i) it is measurable, and (ii) its integral conincides with the Lesbgue
  Integral. But if $f$ is improperly Riemann Integrable, then is it
  even measurable in the first place?

 A: Since $f$ and, hence, $|f|$ is Riemann integrable on $[a + 1/n,b-1/n]$, for every positive integer $n$ the restriction of $|f|$ to any such interval is measurable. 
Hence, for every $\alpha \in \mathbb{R}$, the set $E_{\alpha,n}=\{x \in [a+1/n,b-1/n]:|f(x)| > \alpha \}$ is measurable. Consequently, for every $\alpha$ we have measurability of 
$$E_\alpha = \{x \in [a,b]:|f(x)| > \alpha \} = \bigcup_{n=1}^\infty E_{\alpha,n} \cup F,$$
since $F$ is either $\phi$, $\{a\}$, $\{b\}$, or $\{a,b\}$ which is a measure-zero measurable set and, hence, $E_\alpha$ is a countable union of measurable sets.
Therfore, $f$ is measurable on $[a,b]$. 
Also, the Lebesgue integral of $|f|$ coincides with the improper Riemann integral of $|f|$. For one side (and similarly for the other) we have by the monotone convergence theorem,
$$\int_c^b |f(x)| \, dx = \lim_{t \to b-} \int_c^t |f(x)| \, dx = \lim_{t \to b-} \int_{[c,t]} |f| = \lim_{t \to b-} \int_{[c,b]} |f| \chi_{[c,b]} = \int_{[c,b]}|f|$$
Here I use $\int_c^d g(x) \, dx$ to denote a Riemann integral and $\int_{[c,d]} g$ to denote a Lebesgue integral.
Now, the Lebesgue and improper Riemann integrals of $f$ can be shown equivalent using the dominated convergence theorem.
