Vitali Set Existence Theorem

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This proof is about Vitali Set Existence Theorem. For other uses, see Vitali Theorem.

Theorem

There exists a set of real numbers which is not Lebesgue measurable.


Proof

Lemma

For all real numbers in the closed unit interval $\mathbb I = \closedint 0 1$, define the relation $\sim$ such that:

$\forall x, y \in \mathbb I: x \sim y \iff x - y \in \Q$

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

That is, $x \sim y$ if and only if their difference is rational.

Then $\sim$ is an equivalence relation.

$\Box$


Let $\map \mu X$ denote the Lebesgue measure of a set $X$ of real numbers.

We have that:

$(1): \quad$ $\map \mu X$ is a countably additive function
$(2): \quad$ $\map \mu X$ is translation invariant
$(3): \quad$ From Measure of Interval is Length, $\map \mu {\closedint a b} = b - a$ for every closed interval $\closedint a b$.

For all real numbers in the closed unit interval $\mathbb I = \closedint 0 1$, define the relation $\sim$ such that:

$\forall x, y \in \mathbb I: x \sim y \iff x - y \in \Q$

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

That is, $x \sim y$ if and only if their difference is rational.

By the Lemma, $\sim$ is an equivalence relation.

For each $x \in \mathbb I$, let $\eqclass x \sim$ denote the equivalence class of $x$ under $\sim$.


Let us use the Axiom of Choice to create a set $M \subset \mathbb I$ containing exactly $1$ element from each equivalence class.

Hence:

for each $x \in \R$ there exists a unique $y \in M$ and $r \in \Q$ such that $x = y + r$.

Let:

$M_r = \set {y + r: y \in M}$

for each $r \in \Q$.

Thus $\R$ is partitioned into countably many disjoint sets:

$(1): \quad \ds \R = \bigcup \set {M_r: r \in \Q}$


Aiming for a contradiction, suppose $M$ were Lebesgue measurable.

First, $\map \mu M = 0$ is impossible, because from $(1)$ this would mean $\map \mu \R = 0$.

Suppose $\map \mu M > 0$.

Then:

\(\ds \closedint 0 2\) \(\supseteq\) \(\ds \bigcup \set {M_r: r \in \Q \land 0 \le r \le 1}\)
\(\ds \leadsto \ \ \) \(\ds \map \mu {\closedint 0 2}\) \(\ge\) \(\ds \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }\) Measure is Monotone
\(\ds \) \(=\) \(\ds \sum_{\substack {r \mathop \in \Q \\ 0 \mathop \le r \mathop \le 1} } \map \mu {M_r}\) Axiom $(2)$ of Measure
\(\ds \) \(=\) \(\ds \infty\) as each $M_r$ would have to have the same measure as $M$

So $\map \mu M > 0$ is also impossible.

Hence $M$ is not Lebesgue measurable.

$\blacksquare$


Axiom of Choice

This theorem 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.


Also presented as

The proof of the Vitali Set Existence Theorem is presented in Thomas J. Jech: The Axiom of Choice using the following argument:


Aiming for a contradiction, suppose $M$ were Lebesgue measurable.

First, $\map \mu M = 0$ is impossible, because from $(1)$ this would mean $\map \mu \R = 0$.

Suppose $\map \mu M > 0$.

Then:

\(\ds \map \mu {\closedint 0 1}\) \(\ge\) \(\ds \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }\)
\(\ds \) \(=\) \(\ds \sum_{\substack {r \mathop \in \Q \\ 0 \mathop \le r \mathop \le 1} } \map \mu {M_r}\)
\(\ds \) \(=\) \(\ds \infty\) as each $M_r$ would have to have the same measure as $M$

So $\map \mu M > 0$ is also impossible.

Hence $M$ is not Lebesgue measurable.


It needs to be pointed out that this:

$\ds \map \mu {\closedint 0 1} \ge \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$

is not immediately obvious, as it is not the case that:

$\ds \closedint 0 1 \supseteq \bigcup \set {M_r: r \in \Q \land 0 \le r \le 1}$

However, because $\map \mu X$ is translation invariant, $M_r$ has the same measure as $M_0$.

Hence the result.

$\blacksquare$


Also see

The set $M$ defined here is an example of a Vitali set: a subset of the real numbers which has no Lebesgue measure.


Source of Name

This entry was named for Giuseppe Vitali.


Linguistic Note

The term Vitali Set Existence Theorem was invented by $\mathsf{Pr} \infty \mathsf{fWiki}$ as it is difficult to establish a universally-accepted name for it.

As such, it is not generally expected to be seen in this context outside $\mathsf{Pr} \infty \mathsf{fWiki}$.

Some sources refer to this as the Vitali Theorem, but this can easily be confused with the result known on $\mathsf{Pr} \infty \mathsf{fWiki}$ as Vitali's Theorem, which is a completely different result.

However, as both results are in the field of measure theory, it is not possible to distinguish them by context.

Many sources quote and analyse the result without giving it a name.


Sources

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