Vitali Theorem

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Theorem

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


Proof

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 Non-Measurable Set of Real Numbers: Lemma 1, $\sim$ is an equivalence relation.

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


By the Axiom of Choice, it is possible to choose $1$ element from each equivalence class.

Thus is created a set $M \subset \mathbb I$ such that:

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:

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


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:

\(\displaystyle \map \mu {\closedint 0 1}\) \(\ge\) \(\displaystyle \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{r \mathop \in \Q \mathop \land 0 \mathop \le r \mathop \le 1} \map \mu {M_r}\)
\(\displaystyle \) \(=\) \(\displaystyle \infty\) 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.

$\blacksquare$


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


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.


Source of Name

This entry was named for Giuseppe Vitali.


Sources