Vitali Set Existence Theorem/Also presented as
Jump to navigation
Jump to search
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
- 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