Understanding extended Riemann integral in Munkres. I'm self-studying Analysis on Manifolds by Munkres.  I understood the theory of the Riemann integral over bounded rectangles and more general rectifiable sets in $\mathbb{R}^n$.  In the part on improper integrals, I am becoming confused.
Munkres defines the extended or improper integral of a continuous function $f$ over an open set $A \subset \mathbf{R}^n$.  He chooses any sequence $C_N$ of compact rectifiable sets such that $A = \bigcup_N C_N$   and $C_N \subset \text{Int }C_{N+1}$ for all $N$ and states that the extended integral exists if and only if the sequence $\int_{C_N} |f|$ is bounded and 
$$\int_Af = \lim_{N \to \infty} \int_{C_N} f.$$
He states "...if the ordinary integral exists, then so does the extended integral and the two integrals are equal", but then " ...the extended integral may exist when the ordinary integral does not."  
This makes sense if you think about integrals over intervals in $\mathbf
{R}$.  If a function is unbounded or the interval is unbounded then the Riemann integral does not exist but the improper integral can. But Munkres claims this even if $A$ is a bounded, open set and $f:A \to \mathbf{R}$ is a bounded, continuous function.
How is this possible?
 A: This is true because unlike integrals over bounded intervals in $\mathbb{R}$, the nature of the boundary of $A \subset \mathbb{R}^n$ is a complicating factor for Riemann integration.
Even if $A$ is a bounded, open set and $f$ is bounded and continuous on $A$, the ordinary Riemann integral may not exist, if $A$ is not rectifiable in the sense that the measure of the boundary $\partial A$ is not zero.
The integral is defined using an enclosing rectangle $R$ as 
$$\int_A f = \int_R f \chi_A,$$
and unless $f(x)$ approaches zero at essentially every boundary point from the interior, $f \chi_A$ will fail to be Riemann integrable if $M(\partial A) \neq 0$.  In general, if a function is bounded and continuous inside a bounded, open set, it may not be extendable as a continuous function to the closure, e.g., $f(x) =\sin x^{-1}$ on $(0,1).$ This is never an issue for $\mathbb{R}$  since the boundary of an open interval  is always a zero-measure set.
On the other hand, the extended integral exists if  $\int_C f^+$ and $\int_C f^-$ are bounded for every compact rectifiable set $C \subset A$. These certainly exist when $f$ is continuous, in which case  and the extended integral is
$$\int_A f = \sup_{C \subset A}\int f^+ - \sup_{C \subset A}\int f^- .$$
This will always hold for bounded continuous $f$ on a bounded open set $A$.
