Unclear steps in a Zorich proof (Measurable sets and smooth mappings)

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The discussion focuses on the proof steps in Zorich's Mathematical Analysis II, specifically regarding the compactness of the set ##\overline{E}_t## and the boundedness of the Jordan-measurable set ##E_x=\phi(E_t)##. Participants assert that the assumption of boundedness is crucial for understanding the properties of these sets. The conversation highlights the relationship between Jordan measure and compactness, emphasizing that a set with Jordan measure zero must indeed be bounded, thereby leading to compactness.

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  • Understanding of Jordan-measurable sets and their properties
  • Familiarity with compactness in the context of topology
  • Knowledge of measure theory, particularly Jordan measure
  • Basic concepts of mathematical analysis as presented in Zorich's Mathematical Analysis II
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  • Study the properties of Jordan-measurable sets in detail
  • Explore the implications of compactness in metric spaces
  • Review the definitions and properties of measure-zero boundaries
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Mathematics students, particularly those studying real analysis, researchers in measure theory, and anyone seeking to deepen their understanding of compactness and Jordan measure in mathematical proofs.

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From Zorich, Mathematical Analysis II, sec. 11.5.2:

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where as one can read from the statement, the sets could also be unbounded.

I do not report here the proof of the fact a), beacuse I have no doubt about it and one can, without the presence of dark steps in the reasoning, assume a) as prooved and pass to the analysis of b) and c).
The proposed proofs of these are the following:

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My questions are:

proof of b): why is ##\overline{E}_t## compact? In particular, I can't understand its boundedness.

proof of c): by hypothesis ##E_t## is Jordan-measurable (that is, it is bounded with measure-zero boundary). Then, why also should ##E_x=\phi(E_t)## be bounded, in general?
 
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I haven't thought about jordan measure ever before, but if ##\overline{E}_t## has jordan measure zero, does that imply it must be compact? It seems like a plausible consequence of the definition. I think if it's not bounded, then any finite set of rectangles you draw that cover the whole set must have infinite area, right?
 
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Right, thank you.
 

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