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Lebesgue measure and the Banch-Tarski Paradox

  1. Aug 7, 2013 #1
    I have finally finished reading the proof of the Banach-Tarski paradox. I think the proof was the standard versoin I see around the internet, with the the group G of rotations phi and psi and so on. I was wondering how the proof fails as a result of lebesgue measure on the set?
  2. jcsd
  3. Aug 7, 2013 #2


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    I don't understand your question. The Banach-Tarski paradox is based on subdividing the sphere into a finite number of disjoint sets and assembling them into a set of different measure. This is possible only if the decomposed sets are non-measurable, since only rigid transformations (which are measure preserving) are allowed. Therefore if the decomposition was into measurable sets the recombination would have the same measure.
  4. Aug 7, 2013 #3
    Let me see if I have this right. The rigid motions are measure preserving in the case of the paradox because no measure is defined on the sets being moved but if we define a measure on the sets then the rigid motions, as constructed in the proof, no longer exists because they aren't measure preserving?
  5. Aug 7, 2013 #4
    I'm trying to understand which aspects of the proof were true before we had a measure which no longer are if we put a measure on the set.
  6. Aug 7, 2013 #5
    That's a weird question, so I don't know what you mean very well by this. Whether you put a measure on the set or not, the proof remains true.

    The proof splits up the unit ball in ##5## pieces, which can be reassembles to form ##2## balls. The problem is that these ##5## pieces are not Lebesgue measurable. The proof never shows that they are, and it can't show it since they're not.
  7. Aug 7, 2013 #6
    I can see that my questions are unclear but I'm not sure how to phrase them better, probably because my understanding of what I'm trying to say is pretty flakey but thanks for bearing with me. So if the 5 sets aren't measurable why does it make the proof false. What are the implications of a set not being lebesgue measurable? or in other words what does it mean about a set if it is not Lebesgue measurable. Seems to me it is more a reflection on the measure not being good enough rather than the proof.
  8. Aug 7, 2013 #7
    It doesn't make the proof false. The proof is true anyway you look at it.

    The Banach-Tarski paradox says the following: There are five sets ##K_i,~1\leq i\leq 5## such that ##\bigcup_i K_i## is the unit ball. And such that some other rearrangement gives two unit balls.

    This is a true theorem, and the proof is valid. The following would not be valid:

    There are five Lebesgue-measurable sets ##K_i,~1\leq i\leq 5## such that ##\bigcup_i K_i## is the unit ball. And such that some other rearrangement gives two unit balls.

    This would be false, and the proof never shows this anyway.

    You are right. The measure is not good enough. That is: not all sets are Lebesgue-measurable. The proof is perfectly fine though.
  9. Aug 7, 2013 #8
    If a set isn't measurable, it means that there is no sensible way to assign it a size. For example if you think of some common sets, the interval [2,5] has measure 3. That makes sense. And the union of the intervals [0,2] and [5,10] has measure 2 + 5 = 7.

    In measure theory they attempt to assign a sensible size to every possible set of real numbers. (Sensible has a technical definition: the measure has to satisfy some properties that we think a well-behaved measure should have).

    If you assume the Axiom of Choice, then you can prove that there is no measure on all the sets of real numbers. So we just accept that some sets are not measurable.

    One of the properties of measurable sets is that they are translation invariant. This means that if you have a shape in the plane, such as a circle, its area stays the same if you move the circle to a different place in the plane.

    In the BT paradox, we move around some shapes; yet their total size changes. It follows that at least one of the pieces must not be measurable. And since the proof specifically invokes the Axiom of Choice to prove the existence of a set whose members we could not possibly write down; we are pretty sure that set's the culprit.
  10. Aug 8, 2013 #9
    That makes total sense. It has clicked into place
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