Real Analysis: Riemann Measurable

[datenshinoai]

Homework Statement

Assume S contained in R2 is bounded. Prove that if S is Riemann measurable, then so are its interior and closure

2. The attempt at a solution

Proof:

If S is Riemann measurable, its boundary is a zero set. Since the boundary of each open U in the int(S) is part of the boundary of S, this means that the boundary of the int(S) is also a zero set. Since S is bounded, so is its interior. Thus int(S) is Riemann measurable.

Since the boundary of S is the closure minus the interior and the boundary of S is a zero set, the closure must also be a zero set. So it is also Riemann measurable.

QED.

At first I thought this was fine, but then I ran into trouble when I try to prove that if the interior and the closure is Riemann integrable then so is S

Any help is appreciated!

 

[gammamcc]

This is the same as Jordan measure right? Note that every rectangular cover of S also covers closure of S. Every union of rectangles in intS is also in S, every such cover of cl(S) is well-approximated by a cover of S, etc. (apply epsilon notation). That is, try to show that from any admissible sets where outer and inter measures converge the produce the measure of S, you may construct admissible sets to produce the measure of int(S) or of cl(S).

 

[camac014]

There are actual formulas for the boundary of a set S, namely that bd(S)=cl(S)-int(S).
Try to use these to prove that bd(int(S)) and bd(cl(S)) are also zero sets given that bd(S) is a zero set.