Vitali Set Existence Theorem

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

Theorem

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

Proof 1

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:

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

Hence $M$ is not Lebesgue measurable.

$\blacksquare$

Proof 2

We construct such a set.

For $x, y \in \hointr 0 1$, 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 $E \subset \hointr 0 1$ be a measurable set.

Let $E_1 = E \cap \hointr 0 {1 - x}$ and $E_2 = E \cap \hointr {1 - x} 1$.

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:

$\map m {E_1} + \map m {E_2} = \map m E$

We have:

$E_1 +_1 x = E_1 + x$

By Lebesgue Measure is Translation Invariant:

$\map m {E_1 +_1 x} = \map m {E_1}$

Also:

$E_2 +_1 x = E_2 + x - 1$

and so:

$\map m {E_2 +_1 x} = \map m {E_2}$

Then we have:

$\map m {E +_1 x} = \map m {E_1 +_1 x} + \map m {E_2 +_1 x} = \map m {E_1} + \map m {E_2} = \map m E$

So, for each $x \in \hointr 0 1$, the set $E +_1 x$ is measurable and:

$\map m {E + x} = \map m E$

Taking, as before, $x, y \in \hointr 0 1$, 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 $\hointr 0 1$ into equivalence classes.

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

Aiming for a contradiction, suppose $P$ is measurable.

Let $\set {r_i}_{i \mathop = 0}^\infty$ be an enumeration of the rational numbers in $\hointr 0 1$ 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 $x \in \hointr 0 1$ 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:

$\ds \bigcup P_i = \hointr 0 1$

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

But if this were the case, then:

$\ds m \hointr 0 1 = \sum_{i \mathop = 1}^\infty m \paren {P_i} = \sum_{i \mathop = 1}^\infty \map m P$

Therefore:

$\map m P = 0$ implies $m \hointr 0 1 = 0$

and:

$\map m P \ne 0$ implies $m \hointr 0 1 = \infty$

This contradicts Measure of Interval is Length.

So the set $P$ is not 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 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:

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 known as

The is also known as Vitali's Paradox.

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 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

• 1905: Giuseppe Vitali: Sul Problema della Misura dei Gruppi di Punti di Una Retta (Bologna: Tip. Gamberini e Parmeggiani )
• 1973: Thomas J. Jech: The Axiom of Choice ... (previous) ... (next): $1.$ Introduction: $1.2$ A nonmeasurable set of real numbers
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): measure
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): measure