Littlewood's First Principle

One very important theorem regarding Lebesgue measurable sets is as follows. If $E$ is a Lebesgue measurable set with finite Lebesgue measure $m(E) < \infty$ then $E$ is almost equal to the union of a finite collection of open intervals. This theorem is called Littlewood's First Principle and is outlined below.

Theorem 1 (Littlewood's First Principle): Let $E$ be a Lebesgue measurable set with $m(E) < \infty$. Then for all $\epsilon > 0$ there exists a finite collection of open intervals $\{ I_1, I_2, ..., I_N \}$ such that $\displaystyle{m \left ( E \Delta \bigcup_{n=1}^{N} I_n \right ) < \epsilon}$.

If $A$ and $B$ are sets then $A \Delta B$ denotes the Symmetric Difference of $A$ and $B$. It is defined as the collection of elements $x$ such that $x \in A$ or $x \in B$ but $x \not \in A \cap B$. Hence $A \Delta B = (A \setminus B) \cup (B \setminus A)$ or equivalently, $A \Delta B = (A \cup B) \setminus (A \cap B)$.

• Proof: Let $E$ be a Lebesgue measurable set with $m(E) < \infty$. Let $\epsilon > 0$ be given. By the definition of the Lebesgue measure being an infimum, we have that there exists an open interval cover $\{ I_1, I_2, ... \}$ of $E$ with $\displaystyle{E \subseteq \bigcup_{n=1}^{\infty} I_n}$ and such that:

(1)

\begin{align} \quad \sum_{n=1}^{\infty} l(I_n) < m(E) + \frac{\epsilon}{2} \quad \Leftrightarrow \quad \sum_{n=1}^{\infty} l(I_n) - m(E) < \frac{\epsilon}{2} \quad (*) \end{align}

• Since $m(E) < \infty$, the inequality above implies that $\displaystyle{\sum_{n=1}^{\infty} m(I_n) = \sum_{n=1}^{\infty} l(I_n) < \infty}$. Furthermore, by Countable Subadditivity of the Lebesgue Outer Measure we have that $\displaystyle{m \left (\bigcup_{n=1}^{\infty} I_n \right ) \leq \sum_{n=1}^{\infty} m(I_n) = \sum_{n=1}^{\infty}l(I_n) < \infty}$. So by The Excision Property of Lebesgue Measurable Sets and from $(*)$ above we have that:

(2)

\begin{align} \quad m \left ( \left ( \bigcup_{n=1}^{\infty} I_n \right ) \setminus E \right ) = m \left ( \bigcup_{n=1}^{\infty} I_n \right ) - m(E) \leq \sum_{n=1}^{\infty} l(I_n) - m(E) \overset{(*)} < \frac{\epsilon}{2} \end{align}

• Now, for any $N \in \mathbb{N}$ since $\displaystyle{\bigcup_{n=1}^{N} I_n \subseteq \bigcup_{n=1}^{\infty} I_n}$ we have that $\displaystyle{\left ( \bigcup_{n=1}^{N} I_n \right ) \setminus E \subseteq \left ( \bigcup_{n=1}^{\infty} I_n \right ) \setminus E}$. So by the monotonicity property of the Lebesgue measure we have that:

(3)

\begin{align} \quad m \left ( \left ( \bigcup_{n=1}^{N} I_n \right ) \setminus E \right ) \leq m \left ( \left ( \bigcup_{n=1}^{\infty} I_n \right ) \setminus E \right ) < \frac{\epsilon}{2} \quad (\dagger) \end{align}

• We have already remarked that $\displaystyle{\sum_{n=1}^{\infty} m(I_n) < \infty}$, so, this series converges and there exists an $N \in \mathbb{N}$ such $\displaystyle{\sum_{n=N+1}^{\infty} m(I_n) < \frac{\epsilon}{2}}$.

• Since $\displaystyle{E \subseteq \bigcup_{n=1}^{\infty} I_n}$ we have that:

(4)

\begin{align} \quad E \setminus \left ( \bigcup_{n=1}^{N} \right ) \subset \left ( \bigcup_{n=1}^{\infty} I_n \right ) \setminus \left ( \bigcup_{n=1}^{N} I_n \right ) \subseteq \bigcup_{n=N+1}^{\infty} I_n \quad (**) \end{align}

• So once again by the monotonicity property of the Lebesgue measure we have that:

(5)

\begin{align} \quad m \left ( E \setminus \left ( \bigcup_{n=1}^{N} I_n \right ) \right ) \leq m \left (\bigcup_{n=N+1}^{\infty} I_n\right ) \leq \sum_{n=N+1}^{\infty} m(I_n) < \frac{\epsilon}{2} \quad (\dagger \dagger ) \end{align}

• Finally, by using $(\dagger)$ and $(\dagger \dagger)$ we have that:

(6)

\begin{align} \quad m \left ( E \Delta \bigcup_{n=1}^{N} I_n \right ) = \underbrace{m \left ( E \setminus \left ( \bigcup_{n=1}^{N} I_n \right ) \right )}_{< \frac{\epsilon}{2} \: \mathrm{by} \: (\dagger \dagger)} + \underbrace{m \left ( \left ( \bigcup_{n=1}^{N} I_n \right ) \setminus E \right )}_{< \frac{\epsilon}{2} \: \mathrm{by} \: (\dagger)} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}