debunking nonmeasurable set
Here i present an attack on the often used concept of a nonmeasurable set. In most real analysis textbooks , the standard example of a nonmeasurable set is a subset of the real line due to Vitali. All nonmeasurable sets are measurable-isomorphic to eachother and to Vitali , thus Vitali (nonmeasurable) sets are of fundamental importance. Vitali sets are used in set topology , set algebra and analysis , so the influence of Vitali sets / nonmeasurable sets on a big part of today's math is huge.
The famous Banach-Tarski paradox also uses nonmeasurable sets. A Vitali set is an example of a set of real numbers that is not measurable , found by Giuseppe Vitali in 1905. The Vitali theorem is the existence theorem that there are such sets. There are uncountably many Vitali sets , and their existence is proven on the assumption of the axiom of choice. So we already know we need the axiom of choice , and i already object to that , but even if i accept the axiom of choice i still have an argument against - called an attack - the concept of a nonmeasurable set. A Vitali set V is a subset of [0,1] which , for each real number r , contains exactly one number v such that v-r is rational( existance of that set follows from the axiom of choice ). The idea of Vitali's example is to express the unit interval I as a disjoint union of countable many mutually congruent sets A_k. The nonmeasurability of each A_k follows from the observation that I = U keZ A_k and that countable additivity of measure implies that 1 = m(I) = sum keZ m(A_k). Since each set A_k must have the same measure , the last equation shows that no nonnegative value can be assigned as the measure of each A_k ; in other words no real value can be assigned to m(A_k).
Basicly because sum kez is an infinite sum and m(A_k) = m(A_1) THUS => m(I) = 1 = oo * m(A_k).
The equation oo ( or countable infinity ) * x = 1
has no (real) solution for x , thus m(A_1) = x has no (real) solution.
If M(A_1) = 0 then m(I) = 0 instead of 1.
If M(A_1) = finite real then m(I) = oo instead of 1.
What about my counterargument (attack) to the concept of nonmeasurable set you might be wondering.
Simple : sum keZ m(A_k) = 1 AFTERALL.
Note that Z has the same cardinality as N and thus both Z and N are countable infinity.
We may thus replace sum keZ m(A_k) = 1 with
sum neN m(A_N) = 1
and even
sum neN m(A_1) = 1
We then arrive at the fundamental equation :
sum neN * x = 1.
And now comes the shock :
We can replace neN with lim n -> oo.
As we always do in calculus.
Thus we arrive at :
sum neN * x = 1 = lim n -> oo n * x = 1.
And the equation finally becomes :
lim n -> oo n * x = 1.
And that can be solved !!
x = 1/n lim n -> oo
Notice x IS NOT 0 , but an infinitesimal.
Thus the measure m of A_k is the infinitesimal m = 1/n lim n -> oo.
QED.
Copyright tommy1729
- update ! -
To support the above , there actually have been mathematicians working on infinitesimal measures despite its controversy.
As an example there's something called Loeb measure in non-standard analysis.
http://www.math.uiuc.edu/~loeb/
http://books.google.com/books?id=jcjBpRnMiecC
Also
www.math.upenn.edu/~pemantle/Hypreals%5B1%5D.rtf
states
<quote>
Bernstein and Wattenberg (1969). In an early paper, B & W construct a
measure that assigns a "probability" in *[0,1) to every subset of real
[0,1), even those that are not Lebesgue-measurable. (E.g., Vitali
sets get infinitesimal probability.)
</quote>
The reference is to
Bernstein, Allen R. and Wattenberg, Frank. (1969) "Nonstandard
measure theory." In W.A.J. Luxemburg, ed., Applications of Model
Theory to Algebra, Analysis, and Probability (NY: Holt, Rinehart and
Winston), pp. 171-186.
Now some people have claimed that this infinitesimal measure must be tested in the sense that one should be able to prove an integral with it that was earlier done with ordinary lesbesgue measure or even integrals defining the volume of a sphere or a cone.
But that is based on nothing apart from perhaps a misunderstanding.
Infinitesimal measure is not meant to " be an enemy " of lebesgue measure , not even an " alternative ".
I am not against Lebesgue-measure but rather extending it !
I am not selling a totally new theory of integration !
So there is no need to " defend " Lebesgue-measure or methods of integration ( such as the Riemann integral ).
Thank you for your attention.
tommy1729
- update(2) ! -
Some additional comments. My infinitesimal measure is written as dx in accordance with calculus.
In my opinion measure is relative (to something) rather than absolute.
According to AC , a nonmeasurable set cannot be explicitly constructed. However I can.
(since I do not accept AC)
Consider the interval [0,1] on the extended real line ( being [-oo,+oo] ).
Clearly the measure [0,1] with respect to the extended real line is not 0 and not 1 nor any real. It is "nonmeasurable" or in other words it has infinitesimal measure.
To show that last claim ( infinitesimal ! ) , remember that there always exists asymptoticly biholomorphic mappings of connected sets ( such as an interval on the (extended) reals )
that preserves measure (and cardinality) !! (see more formal explaination below)
Lets call that mapping f, where f maps the largest set 'A' asymptoticly biholomorphic to its considered connected subset 'B' while the measure of 'B' with respect to 'A' is identical to the measure of 'f(B)' with respect to 'f(A)'.
We know that there exists such a asymptoticly biholomorphic mapping f between [0,1] and the extended real line.
( notice open or closed has no influence on cardinality or measure since its just about 2 points ; the boundary values of the interval. Hence it does not matter if I speak of open or closed intervals and if f looses 1 or 2 points is hence also irrelevant ! )
Clearly the measure [0,1] with respect to the extended real line is "nonmeasurable" AND(!) f([-oo,+oo]) = [0,1] THUS the measure of [0,1] with respect to [-oo,+oo] is the same as the measure of f([0,1]) with respect to [0,1].
Hence this proofs that "nonmeasurable" means infinitesimal. QED
Thank you for your attention
tommy1729