Existence of Non-Measurable Subset of Real Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

There exists a subset of the real numbers which is not measurable.


Proof

We construct such a set.

For $x, y \in \left[{0 \,.\,.\, 1}\right)$, define the sum modulo 1:

$x +_1 y = \begin{cases} x + y & : x + y < 1 \\ x + y - 1 & : x + y \ge 1 \end{cases}$

Let $E \subset \left[{0 \,.\,.\, 1}\right)$ be a measurable set.

Let $E_1 = E \cap \left[{0 \,.\,.\, 1 - x}\right)$ and $E_2 = E \cap \left[{1 - x \,.\,.\, 1}\right)$.

By Measure of Interval is Length, these disjoint intervals are measurable.

By Measurable Sets form Algebra of Sets, so are these intersections $E_1$ and $E_2$.

So:

$m \left({E_1}\right) + m \left({E_2}\right) = m \left({E}\right)$


We have:

$E_1 +_1 x = E_1 + x$

By Lebesgue Measure is Translation-Invariant:

$m \left({E_1 +_1 x}\right) = m \left({E_1}\right)$

Also:

$E_2 +_1 x = E_2 + x - 1$, and so $m \left({E_2 +_1 x}\right) = m \left({E_2}\right)$

Then we have:

$m \left({E +_1 x}\right) = m \left({E_1 +_1 x}\right) + m \left({E_2 +_1 x}\right) = m \left({E_1}\right) + m \left({E_2}\right) = m \left({E}\right)$

So, for each $x \in \left[{0 \,.\,.\, 1}\right)$, the set $E +_1 x$ is measurable and:

$m \left({E + x}\right) = m \left({E}\right)$


Taking, as before, $x, y \in \left[{0 \,.\,.\, 1}\right)$, define the relation:

$x \sim y \iff x - y \in \Q$

where $\Q$ is the set of rational numbers.

By Difference is Rational is Equivalence Relation, $\sim$ is an equivalence relation.

As this is an equivalence relation we can invoke the fundamental theorem on equivalence relations.

Hence $\sim$ partitions $\left[{0 \,.\,.\, 1}\right)$ into equivalence classes.

By the axiom of choice, there is a set $P$ which contains exactly one element from each equivalence class.

Let $\left\{{r_i}\right\}_{i \mathop = 0}^\infty$ be an enumeration of the rational numbers in $\left[{0 \,.\,.\, 1}\right)$ with $r_0 = 0$.

Let $P_i := P +_1 r_i$.

Then $P_0 = P$.

Let $x \in P_i \cap P_j$.

Then:

$x = p_i + r_i = p_j + r_j$

where $p_i, p_j$ are elements of $P$.

But then $p_i - p_j$ is a rational number.

Since $P$ has only one element from each equivalence class:

$i = j$

The $P_i$ are pairwise disjoint.

Each real number $x \in \left[{0 \,.\,.\, 1}\right)$ is in some equivalence class and hence is equivalent to an element of $P$.

But if $x$ differs from an element in $P$ by the rational number $r_i$, then $x \in P_i$ and so:

$\displaystyle \bigcup P_i = \left[{0 \,.\,.\, 1}\right)$

Since each $P_i$ is a translation modulo $1$ of $P$, each $P_i$ will be measurable if $P$ is, with measure $m \left({P_i}\right) = m \left({P}\right)$.

But if this were the case, then:

$\displaystyle m \left[{0 \,.\,.\, 1}\right) = \sum_{i \mathop = 1}^\infty m \left({P_i}\right) = \sum_{i \mathop = 1}^\infty m \left({P}\right)$

Therefore:

$m \left({P}\right) = 0$ implies $m \left[{0 \,.\,.\, 1}\right) = 0$

and:

$m \left({P}\right) \ne 0$ implies $m \left[{0 \,.\,.\, 1}\right) = \infty$

This contradicts Measure of Interval is Length.

So the set $P$ is not measurable.

$\blacksquare$


Axiom of Choice

This proof depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.


Boolean Prime Ideal Theorem

This theorem depends on the Boolean Prime Ideal Theorem (BPI).

Although not as strong as the Axiom of Choice, the BPI is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.



Sources