proof of pseudoparadox in measure theory

proof of pseudoparadox in measure theory

Since this

paradox

depends crucially on the

axiom of choice

, 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 relation

∼

on

ℝ

by the
condition that

x

∼

y

if and only if

x

-

y

is

rational

. By
the Archimedean property of the real line, for every

x

∈

ℝ

there will exist a number

y

∈

[

0

,

1

)

such that

y

∼

x

.
Therefore, by the axiom of choice, there will exist a

choice
function

f

:

ℝ

→

[

0

,

1

)

such that

f

⁢

(

x

)

=

f

⁢

(

y

)

if
and only if

x

∼

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

≥

0

and

w

⁢

(

x

)

=

x

+

2

when

x

<

0

. Define

g

:

[

0

,

1

)

→

ℝ

by

g

⁢

(

x

)

=

w

⁢

(

2

⁢

x

-

f

⁢

(

x

)

)

From the

definition

, it is clear that, since

x

∼

f

⁢

(

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

)

≤

2

⁢

x

, then

g

⁢

(

x

)

=

2

⁢

x

-

f

⁢

(

x

)

≤

2

⁢

x

<

2

. On the other hand, if

f

⁢

(

x

)

>

2

⁢

x

, then

g

⁢

(

x

)

=

2

⁢

x

+

2

-

f

⁢

(

x

)

. Since

2

⁢

x

-

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

<

y

. By what we already observed,

x

∼

y

, so

y

-

x

is a non-negative

rational number

and

f

⁢

(

x

)

=

f

⁢

(

y

)

. There are 3
possible cases: 1)

f

⁢

(

x

)

≤

2

⁢

x

≤

2

⁢

y

In this case,

g

⁢

(

x

)

=

g

⁢

(

y

)

implies that

2

⁢

x

-

f

⁢

(

x

)

=

2

⁢

y

-

f

⁢

(

x

)

, which would imply that

x

=

y

. 2)

2

⁢

x

<

f

⁢

(

x

)

<

2

⁢

y

In this case,

g

⁢

(

x

)

=

g

⁢

(

y

)

implies

2

+

2

⁢

x

-

f

⁢

(

x

)

=

2

⁢

y

-

f

⁢

(

x

)

which, in turn, implies that

y

=

x

+

1

, which
is impossible if both

x

and

y

belong to

[

0

,

1

)

. 3)

2

⁢

x

<

2

⁢

y

<

f

⁢

(

x

)

In this case,

g

⁢

(

x

)

=

g

⁢

(

y

)

implies that

2

⁢

x

+

2

-

f

⁢

(

x

)

=

2

⁢

y

+

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

∈

[

0

,

1

)

such that

w

⁢

(

2

⁢

x

-

f

⁢

(

y

)

)

=

y

.
If

f

⁢

(

y

)

+

y

<

2

, we can choose

x

=

(

f

⁢

(

y

)

+

y

)

/

2

. If

2

≤

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

⁢

…

such that no number is counted twice. Define

A

i

⊂

C

1

as

A

i

=

{

x

∈

[

0

,

1

)

∣

g

⁢

(

x

)

=

r

i

}

It is

obvious

from this definition that the

A

i

are

mutually
disjoint

. Furthermore,

⋃

i

=

1

∞

A

i

=

[

0

,

1

)

and

⋃

i

=

1

∞

B

i

=

[

0

,

2

)

where

B

i

is the translate of

A

i

by

r

i

.

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

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