Double Integrals over General Regions: "Well-Behaved" Curve

[BoundByAxioms]

"Well-Behaved" curve

I am reviewing/browsing through Stewart's Early Transcendentals and had a question about the section on Double Integrals over General Regions. It says that in order to integrate a function f over regions D of a general shape, we suppose D is a bounded region that can be enclosed in a rectangular region R. Then a new function F with domain R is defined :

F(x,y) = f(x,y) if (x,y) is in D, and F(x,y) = 0 if (x,y) is in R but not in D.

Then, a theorem is stated:

$$\int$$$$\int_{D}$$f(x,y)dA=$$\int$$$$\int_{R}$$F(x,y)dA.And then (finally, getting to my question) it says that if f is continuous on D and the boundary curve of D is "well behaved" (in a sense outside the scope of this book), then it can be shown that $$\int$$$$\int_{R}$$F(x,y)dA exists and therefore $$\int$$$$\int_{D}$$f(x,y)dA exists. What does it mean by well-behaved, and can someone offer an elementary explanation as to why the boundary curve of D must be well behaved in order for $$\int$$$$\int_{R}$$F(x,y)dA to exist, and why that implies that $$\int$$$$\int_{D}$$f(x,y)dA exists?

 

[Cvan]

As for a technical explanation, forgive me, for I have only completed a more basic course in multivariable calculus, and have not been formally introduced to the more rigorous proofs involving calculus.

As for the basic idea though, if the curve is well behaved (most likely implying that it has derivatives across your region R that are well defined), than the double integral (antiderivative over the region enclosed, D), follows by the fundamental theorem of calculus.

Essentially a fundamental theorem of calculus expanded to two variables.

 

[quasar987]

In my opinion, problems might arise in the case where the boundary of D has measure >0. Because recall the characterization of a Riemann integrable function... F:R-->R (bounded) is integrable iff its set of discontinuities is of measure 0. If the boundary of D has measure >0, unless f(x)=0 almost everywhere on the boundary of D, then F will not be integrable on R by the above characterization.