proof of pseudoparadox in measure theory

Since this paradox depends crucially on the axiom of choice, we willplace the application of this controversial axiom at the head of theproof rather than bury it deep within the bowels of the argument.

One can define an equivalence relation $\sim$ on $ℝ$ by thecondition that $x\sim y$ if and only if $x-y$ is rational. Bythe Archimedean property of the real line, for every $x\in ℝ$ there will exist a number $y\in [0,1)$ such that $y\sim x$.Therefore, by the axiom of choice, there will exist a choicefunction $f:ℝ\to [0,1)$ such that $f(x)=f(y)$ ifand only if $x\sim y$.

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

$$g(x)=w(2x-f(x))$$

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

Next, we will show that $g$ is injective. Suppose that $g(x)=g(y)$ and . By what we already observed, $x\sim y$, so $y-x$ is a non-negative rational number and $f(x)=f(y)$. There are 3possible cases: 1) $f(x)\le 2x\le 2y$ In this case, $g(x)=g(y)$implies that $2x-f(x)=2y-f(x)$, which would imply that $x=y$. 2) In this case, $g(x)=g(y)$ implies $2+2x-f(x)=2y-f(x)$ which, in turn, implies that $y=x+1$, whichis impossible if both $x$ and $y$ belong to $[0,1)$. 3) 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 remainingpossibility is that $x=y$, so $g(x)=g(y)$ implies that $x=y$.

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

Having shown that $g$ is a bijection between $[0,1)$ and $[0,2)$, weshall now complete the proof by examining the action of $g$. As wealready noted, $g(x)-x$ is a rational number. Since the rationalnumbers are countable, we can arrange them in a series $r_0,r_1,r_2\mathrm{\dots }$ such that no number is counted twice. Define $A_i\subset C_1$ as

$$A_i=\{x\in [0,1)\mid g(x)=r_i\}$$

It is obvious from this definition that the $A_i$ are mutuallydisjoint. Furthermore, ${\bigcup }_{i=1}^{\mathrm{\infty }}{A}_i=[0,1)$ and${\bigcup }_{i=1}^{\mathrm{\infty }}{B}_i=[0,2)$ where $B_i$ is the translate of$A_i$ by $r_i$.