The Existence of a Non-Lebesgue Measurable Set

# The Existence of a Non-Lebesgue Measurable Set

So far we have noted that many subsets of $\mathbb{R}$ are Lebesgue measurable, so the next question we may ask is if every subset of $\mathbb{R}$ is Lebesgue measurable. The answer is no. We will now demonstrate the existence of a non-Lebesgue measurable set.

 Theorem 1: There exists a non-Lebesgue measurable set.
• Proof: Consider the closed interval $[0, 1]$. Define a relation $\sim$ on $[a, b]$ for all $x, y \in [0, 1]$ by:
(1)
• Note that $\sim$ is an equivalence relation on $[0, 1]$. To see this, we note that for all $x \in [0, 1]$ we have that $x - x = 0 \in \mathbb{Q}$ so $x \sim x$. This shows that $\sim$ is reflexive. Now let $x, y \in [0, 1]$ and suppose that $x \sim y$. Then $(x - y) \in \mathbb{Q}$. So clearly $(y - x) \in \mathbb{Q}$, and $y \sim x$. This shows that $\sim$ is symmetric. Now let $x, y, z \in [0, 1]$ and suppose that $x \sim y$ and $y \sim z$ then $(x - y) \in \mathbb{Q}$ and $(y - z) \in \mathbb{Q}$. So their sum $(x - z) \in \mathbb{Q}$ and $x \sim z$. This show that $\sim$ is transitive.
• Since $\sim$ is an equivalence relation on $[0, 1]$ we can partition $[0, 1]$ into equivalence classes of $\sim$ that are mutually disjoint. By the Axiom of Choice, we can choose a representative of each equivalence class. Let $N$ be a set of such representative from these equivalence classes.
• Now let $\{ r_1, r_2, ... \}$ be an enumeration of $\mathbb{Q} \cap [-1, 1]$. Consider the collection of sets:
(2)
\begin{align} \quad \{ N + r_i \}_{i=1}^{\infty} \end{align}
• Here $N + r_i = \{ y + r_i : y \in N \}$. Then the sets in $\{ N + r_i \}_{i=1}^{\infty}$ are mutually disjoint. To see this, let $i, j \in \mathbb{N}$ with $i \neq j$ and suppose that $x \in (N + r_i) \cap (N + r_j)$. Then there exists $a, b \in N$ such that $x = a + r_i$ and $x = b + r_j$. So $a + r_i = b + r_j$ which implies that $a - b = r_j - r_i \in \mathbb{Q}$, $r_j - r_i \neq 0$. So $a$ and $b$ must be in the same equivalence class. But this is a contradiction since $a \neq b$. So indeed the sets in $\{ N + r_i \}_{i=1}^{\infty}$ are mutually disjoint.
• Assume that $N$ is a Lebesgue measurable set. Then each of the translations $N + r_i$ is Lebesgue measurable. Furthermore:
(3)
\begin{align} \quad m \left ( \bigcup_{n=1}^{\infty} (N + r_i) \right ) = \sum_{i=1}^{\infty} m(N + r_i) = \sum_{i=1}^{\infty} m(N) = \infty \cdot m(N) \end{align}
• Now for each $i \in \mathbb{N}$ we have that $N + r_i \subset [-1, 2]$ (since $N \subset [0, 1]$ and $-1 \leq r_i \leq 1$). Therefore:
(4)
\begin{align} \quad \bigcup_{i=1}^{\infty} (N + r_i) \subset [-1, 2] \quad (*) \end{align}
• Furthermore, let $x \in [0, 1]$. Then $x$ belongs in an equivalence class $[y]$ where $y \in \mathbb{N}$. So there exists an $r_k \in \{ r_1, r_2, ... \}$ such that $x - r_k = y$. So $x = y + r_k$ which shows that $x \in (N + r_k)$. Therefore:
(5)
\begin{align} \quad [0, 1] \subset \bigcup_{i=1}^{\infty} (N + r_i) \quad (**) \end{align}
• By the monotonicity property of the Lebesgue measure and from $(*)$ and $(**)$ we have that:
(6)
\begin{align} \quad 1 = m([0, 1]) \leq \infty \cdot m(N) \leq m([-1, 2]) = 3 \end{align}
• But $m(N) \geq 0$. So the above inequality does not hold. So the assumption that $N$ was Lebesgue measurable is false. Hence $N$ is not Lebesgue measurable. $\blacksquare$