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

Yes, ok.

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'm not objecting, I'm agreeing.

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

Yes of course! That's the point. Which reminds me of another reason the Vitali set is a warmup to B-T. The theorem only works because at least one of the pieces is nonmeasurable. If a set is measurable, an isometry must preserve its measure. If that's what you're saying, you're absolutely right.

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

Sure. Agreed. All open sets are measurable. Can I prove that? In the reals it's easy because every open set is a finite or countable union of open intervals, intervals are measurable, and countable unions of measurable sets are measurable. In the plane or 3-space there's an analogous theorem but I'm not sure exactly what the wording is. Cartesian products of intervals are measurable so n-rectangles are, and you can probably express any open set as a countable union of rectangles but that's the part I'm fuzzy on. So I'm pretty sure all open sets in n-space are measurable but I'm only certain of the proof in one dimension.

The axiom of choice lets Vitali cook up a nonmeasurable set, and that's what's happening here. The choice set on the orbits I believe isn't measurable, but don't quote me on that.

Yes, ok. — fishfry

:smile: :smile: :smile:

Sure. Agreed. All open sets are measurable. — fishfry

Yes!! I was starting to despair that there is a way to make me understand...

My objection was only this one: BT doesn't make integration inconsistent. You can reason about infinitesimal parts and be confident of the fact that integration works, if you decompose the object in open sets. That's all I wanted to say!!!

My objection was only this one: BT doesn't make integration inconsistent. — Mephist

Oh I see. You were concerned that you'd change the value of an integral by moving things around. But right, that can't happen because the pieces aren't measurable so they can't be integrated anyway. Good point.

You can reason about infinitesimal parts and be confident of the fact that integration works, if you decompose the object in open sets. — Mephist

Yes I see your concern now. Integration doesn't break and open sets are well-behaved. That's why B-T is a technical result that doesn't actually break anything important.