proof of pseudoparadox in measure theory

Since this paradoxMathworldPlanetmath depends crucially on the axiom of choiceMathworldPlanetmath, we will place the application of this controversial axiom at the head of the proof rather than bury it deep within the bowels of the argument.

One can define an equivalence relationMathworldPlanetmath on by the condition that xy if and only if x-y is rational. By the Archimedean property of the real line, for every x there will exist a number y[0,1) such that yx. Therefore, by the axiom of choice, there will exist a choice function f:[0,1) such that f(x)=f(y) if and only if xy.

We shall use our choice function f to exhibit a bijection between [0,1) and [0,2). Let w be the “wrap-around function” which is defined as w(x)=x when x0 and w(x)=x+2 when x<0. Define g:[0,1) by


From the definition, it is clear that, since xf(x) and w(x)x, g(x)x. Also, it is easy to see that g maps [0,1) into [0,2). If f(x)2x, then g(x)=2x-f(x)2x<2. On the other hand, if f(x)>2x, then g(x)=2x+2-f(x). Since 2x-f(x) is strictly negative, g(x)<2. Since f(x)<1, g(x)>0.

Next, we will show that g is injectivePlanetmathPlanetmath. Suppose that g(x)=g(y) and x<y. By what we already observed, xy, so y-x is a non-negative rational numberPlanetmathPlanetmath and f(x)=f(y). There are 3 possible cases: 1) f(x)2x2y In this case, g(x)=g(y) implies that 2x-f(x)=2y-f(x), which would imply that x=y. 2) 2x<f(x)<2y In this case, g(x)=g(y) implies 2+2x-f(x)=2y-f(x) which, in turn, implies that y=x+1, which is impossible if both x and y belong to [0,1). 3) 2x<2y<f(x) In this case, g(x)=g(y) implies that 2x+2-f(x)=2y+2-f(x), which would imply that x=y. The only remaining possibility is that x=y, so g(x)=g(y) implies that x=y.

Next, we show that g is surjectivePlanetmathPlanetmath. Pick a number y in [0,2). We need to find a number x[0,1) such that w(2x-f(y))=y. If f(y)+y<2, we can choose x=(f(y)+y)/2. If 2f(y)+y, we can choose x=(f(y)+y)/2-1.

Having shown that g is a bijection between [0,1) and [0,2), we shall now completePlanetmathPlanetmathPlanetmathPlanetmath the proof by examining the action of g. As we already noted, g(x)-x is a rational number. Since the rational numbers are countableMathworldPlanetmath, we can arrange them in a series r0,r1,r2 such that no number is counted twice. Define AiC1 as


It is obvious from this definition that the Ai are mutually disjoint. Furthermore, i=1Ai=[0,1) and i=1Bi=[0,2) where Bi is the translate of Ai by ri.

Title proof of pseudoparadox in measure theory
Canonical name ProofOfPseudoparadoxInMeasureTheory
Date of creation 2013-03-22 14:38:43
Last modified on 2013-03-22 14:38:43
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 10
Author rspuzio (6075)
Entry type Proof
Classification msc 28E99
Related topic ProofOfVitalisTheorem