Is this a case of the extended Riemann integral? Let $A \subset \mathbb{R}^n$ be an open subset, and let $f: A \to \mathbb{R}$ be a function that is locally bounded and whose discontinuity set has measure zero.
Consider $A = C_1 \cup C_2 \cup ...,$ where $C_i$ are compact and rectifiable (measurable), and each $C_i \subset int C_{i +1}$.
Prove that $f$ is integrable on $A$ if and only if the sequence $\int_{C_i}|f|$ is bounded, and show that:
$\int_Af = \lim_{i\to \infty}\int_{C_i}f.$
I am trying to prove the above but it's proving to be a bit tricky. Is this a case of the extended Riemann integral?
Please help.
 A: That result does pertain to the extended integral of $f$ over an open set $A \subset \mathbb{R}^n$. Given that $f$ is Riemann integrable over any compact rectifiable subset $D  \subset A$, the extended integral, if it exists, is defined as $\sup_{D \subset A} \int_D f$ if $f$ is nonnegative. In general, if $f$ is not nonegative, then the same definition applies to the positive and negative parts $f^+ = \max(f,0)$ and $f^-= \max (-f,0)$ and the extended integral is then
$$\int_A f = \int_A f^+ - \int_A f^- = \sup_{D \subset A}\int_D f^+ - \sup_{D \subset A}\int_A f^- .$$
To prove this, start by assuming that $f$ is nonnegative with $f = |f|$. Note also  that the sequence $\int_{C_i} f$ is increasing since $E \subset F \implies \int_E f \leqslant \int_Ff$ for Riemann integrals.
Forward Implication
Suppose $f$ is integrable in the extended sense over $A$.  
We also have 
$$\tag{*}\int_{C_i}f \leqslant \sup_{D \subset A} \int_Df = \int_Af.$$
Therefore, the sequence$\int_{C_i}f$ is bounded.
Reverse Implication
On the other hand suppose the sequence$\int_{C_i}f$ is bounded.
Any compact, rectifiable set $D \subset A$ is covered by the interiors of the sets $C_i$ since $A = \bigcup_i C_i$. Since the $C_i$ are increasing and $D$ is compact, we have $D \subset C_n$ for some $n$, and
$$\int_D f \leqslant \int_{C_n} f \leqslant \lim_{i \to \infty} \int_{C_i} f.$$
The limit on the RHS exists (finite) because the sequence is increasing and bounded by hypothesis.  Therefore, 
$$\tag{**} \int_Af = \sup_{D\subset A} \int_D f \leqslant \lim_{i \to \infty} \int_{C_i} f.$$
and the extended integral exists.
We have shown that $\int_A f$ exists if and only if the sequence $\int_{C_i} f$ is bounded.  Either both or none of these conditions are satisfied. When both are satisfied, we have by (*)
$$\lim_{i \to \infty} \int_{C_i} f \leqslant \int_A f.$$
This in conjunction with (**) implies 
$$\lim_{i \to \infty} \int_{C_i} f = \int_A f.$$
To generalize for functions that may not be nonnegative, use this result and the fact that
$$0 \leqslant f^+, \, f^- \leqslant |f| = f^+ + f^-.$$
See if you can finish from here.  
