# proof of pseudoparadox in measure theory

One can define an equivalence relation  $\sim$ on $\mathbb{R}$ by the condition that $x\sim y$ if and only if $x-y$ is rational. By the Archimedean property of the real line, for every $x\in\mathbb{R}$ there will exist a number $y\in[0,1)$ such that $y\sim x$. Therefore, by the axiom of choice, there will exist a choice function $f\colon\mathbb{R}\to[0,1)$ such that $f(x)=f(y)$ if and 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” which is defined as $w(x)=x$ when $x\geq 0$ and $w(x)=x+2$ when $x<0$. Define $g\colon[0,1)\to\mathbb{R}$ 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 $f(x)\leq 2x$, then $g(x)=2x-f(x)\leq 2x<2$. 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 injective  . Suppose that $g(x)=g(y)$ and $x. By what we already observed, $x\sim y$, so $y-x$ is a non-negative rational number  and $f(x)=f(y)$. There are 3 possible cases: 1) $f(x)\leq 2x\leq 2y$ In this case, $g(x)=g(y)$ implies that $2x-f(x)=2y-f(x)$, which would imply that $x=y$. 2) $2x 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 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 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 $f(y)+y<2$, we can choose $x=(f(y)+y)/2$. If $2\leq f(y)+y$, we can choose $x=(f(y)+y)/2-1$.

Having shown that $g$ is a bijection between $[0,1)$ and $[0,2)$, we shall now complete    the proof by examining the action of $g$. As we already noted, $g(x)-x$ is a rational number. Since the rational numbers are countable  , we can arrange them in a series $r_{0},r_{1},r_{2}\ldots$ 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 mutually disjoint. Furthermore, $\bigcup_{i=1}^{\infty}A_{i}=[0,1)$ and $\bigcup_{i=1}^{\infty}B_{i}=[0,2)$ where $B_{i}$ is the translate of $A_{i}$ by $r_{i}$.

Title proof of pseudoparadox in measure theory ProofOfPseudoparadoxInMeasureTheory 2013-03-22 14:38:43 2013-03-22 14:38:43 rspuzio (6075) rspuzio (6075) 10 rspuzio (6075) Proof msc 28E99 ProofOfVitalisTheorem