Vitali Set Existence Theorem/Also presented as

Theorem

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

Proof

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.

$\blacksquare$

Source of Name

This entry was named for Giuseppe Vitali.

Sources

• 1973: Thomas J. Jech: The Axiom of Choice ... (previous) ... (next): $1.$ Introduction: $1.2$ A nonmeasurable set of real numbers