Nonmeasurable Sets

Let $\mu$ and $\mu^*$ denote the Lebesgue measure and Lebesgue outer measure for sets of $\mathbb{R}$ respectively. We prove that there exists a set $E \subset [0, 1]$ that fails to be measurable.

Preliminary

Suppose that $E \subset \mathbb{R}$ is lebesgue measurable and that $\mu(E) < \infty$.

Lemma 1. $\forall \epsilon > 0, \exists$ an open set $G \supset E$, s.t., $\mu(G \setminus E) < \epsilon$.

Proof. Via the definition of the Lebesgue measure $\mu(\cdot)$, there exists a sequence of open intervals $\{ I_n \}$, such that $ \sum_{n \ge 1} \mu (I_n) \le \mu(E) + \epsilon. $

Define $G \doteq \cup_{n \ge 1} I_n$. Then $G$ is an open set and $G \supset E$. Via monotonicity and countably sub-additivity of Lebesgue measure,

As $E$ is measurable and $\mu(E) < \infty$, $ \mu( G \setminus E) = \mu(G) - \mu(E) \le \epsilon. $

$\square$

Let $G_\delta$ denote the collection of Borel sets that are countable intersection of open sets.

Lemma 2. $E = A \setminus Z$, where $A$ is a $G_\delta$ set and $\mu(Z) = 0$.

Proof. For each $k \in \mathbb{N}_+$, there exists an open set $G_k$, s.t.,

1. $G_k \supset E$ and,
2. $\mu(G_k \setminus E) < 1 / k$.

It follows that

1. $A \doteq \cap_{k \ge 1} G_k \supset E$ and,
2. $Z \doteq A \setminus E$ has measure zero.

$\square$

We are ready to prove that Lebesgue measure is translation invariant.

Lemma 3. $\forall x \in \mathbb{R}$, the set $E + x \doteq \{ y + x : y \in E \}$ is measurable and $\mu(E) = \mu (E + x)$.

Proof. As Lebesgue outer measure is translation invariant, it holds that

It is left to prove that $E + x$ is Lebesgue measurable.

We can rewrite $E = A \setminus Z$, where $A = \cap_{k \ge 1} G_k$ is a countable intersection of open sets and $\mu(Z) = 0$. Since $\{ G_k \}$ are open, $\{ G_k + x \}$ are also open sets. Further, $Z + x$ also has outer measure $0$ and is therefore measurable.

It follows that $E + x$ is measurable as

$\square$

Nonmeasurable Set

For each $x, y \in [0, 1]$, denote $x \sim y$ if $|x - y| \in \mathbb{Q}$. Observe that $\sim$ is an equivalent relation, i.e., it is reflexive, symmetric, and transitive. For each $x \in [0, 1]$, define the equivalent class of $x$ as

It is easy to see that the collection of all possible equivalent classes forms a partition of $[0, 1]$. Via Axiom of Choice, we can select a representative number from each equivalent class. Denote the resulting set as $V$.

Theorem. $V$ is not measurable.

Proof. We list all rational numbers in $[-1, 1]$ as $r_1, r_2, \ldots$.

For each $y \in [0, 1]$, $\exists$ $x \in V$, s.t. $y - x$ is a rational number between $[-1, 1]$. Therefore,

Further, as $V$ contains exactly one representative number from each equivalent class, the sets $\{ V + r_n : n \in \mathbb{N}_+ \}$ are disjoint.

Assuming that $V$ is measurable, the lemmas in the previous section imply that $V + r_n$ is measurable and $\mu(V + r_n) = \mu(V)$, for each $n \in \mathbb{N}_+$.

Then $\cup_{n \ge 1 } \big( V + r_n \big)$ is also measurable. Via $\sigma$-additivity of $\mu$, it holds that

which is impossible since $\sum_{n \ge 1} \mu(V)$ is either infinity or zero.

$\square$

Reference

[1] 实变函数论，周民强.