1. A compact, contented set

Thanks to Laurent's answer to my first post, i was encouraged to resume reading my Edwards, which i had all but given up, due to the many difficult gap-fillers the reader was required to supply. Sure enough, it didn't take long before i've reached another impasse. Consider the attached proof of Proposition 6.1. The last sentence delivers nonchalantly the following opaque claim: "A+ is a compact contented set". Do you understand why this should be so? (Please note that the definition for an "absolutely integrable" function is given on the left leaf)

P.S.
The excerpt is taken form C.H. Edwards' "Advanced Calculus of Several Variables", Dover, 1994, which is an unabridged, corrected republication of the work first published by Academic Press, New York, 1973. The book is available for sale in such stores as Amazon and Barnes & Noble (to name a few).

2. Originally Posted by itai
Consider the attached proof of Proposition 6.1. The last sentence delivers nonchalantly the following opaque claim: "A+ is a compact contented set". Do you understand why this should be so? (Please note that the definition for an "absolutely integrable" function is given on the left leaf)
I may be mistaking, but isn't it because $\displaystyle A^+\subset B^\varepsilon\cup A$ and both $\displaystyle A$ and $\displaystyle B^\varepsilon$ are compact contented? However, I could'nt find a definition of "compact contented" on the internet so I can't be sure; is it what is usually called "compact"?

3. Some relevant definitions and theorems

Hi Laurent,

My reply is in the attached pdf file. Unfortunately, i couldn't get the mathematical symbols to display properly, when i copied the text from the pdf document directly to this message.

Thanks.

4. Originally Posted by itai
Hi Laurent,

My reply is in the attached pdf file. Unfortunately, i couldn't get the mathematical symbols to display properly, when i copied the text from the pdf document directly to this message.

Thanks.
You're right, I gave it too quick a look. And it doesn't seem either correct as such or even simple to correct.