Yes of course they have to be isometries. I meant: there is no way of decomposing an object in an infinite set of open sets and then recomposing them in a different way so that each peace has the same measure but the sum of the measures of all the pieces is different. If this were possible, the theory of integration would be inconsistent. — Mephist
I know your objection: if there is an infinite number of pieces the measure of each peace cannot be finite. OK, but you can build the limit of a sequence of decompositions, like you do with regular integration. — Mephist
I am not arguing that BT theorem is false, I am arguing that it works only because you perform the transformation on pieces that are not measurable. — Mephist
If the pieces were made using the decomposition in open sets, as with regular integration, it couldn't work. I know that you can even define a Lebesgue integral that is working on sets that are not open: this is not a necessary condition, but is a sufficient condition to preserve additivity. — Mephist
Yes, ok. — fishfry
Sure. Agreed. All open sets are measurable. — fishfry
My objection was only this one: BT doesn't make integration inconsistent. — Mephist
You can reason about infinitesimal parts and be confident of the fact that integration works, if you decompose the object in open sets. — Mephist
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