Countable Subadditivity of the Lebesgue Outer Measure

# Countable Subadditivity of the Lebesgue Outer Measure

Recall from The Lebesgue Outer Measure page the Lebesgue outer measure function $m^* : \mathcal P (\mathbb{R}) \to [0, \infty) \cup \{ \infty \}$ is defined for all sets $E \in \mathcal P(\mathbb{R})$ by:

(1)
\begin{align} \quad m^*(E) = \inf \left \{ \sum_{n=1}^{\infty} l(I_n) : E \subseteq \bigcup_{n=1}^{\infty} I_n \: \mathrm{and} \: \{ I_n = (a_n, b_n) \}_{n=1}^{\infty} \right \} \end{align}

The Lebesgue outer measure has a very nice property known as countable subadditivity. If we have a countable collection of subsets of $\mathbb{R}$, say $(A_n)_{n=1}^{\infty}$, then the Lebesgue outer measure of the union $\displaystyle{\bigcup_{n=1}^{\infty} A_n}$ will be less than or equal to the sum $\displaystyle{\sum_{n=1}^{\infty} m^*(A_n)}$. We prove this below.

 Theorem: Let $(A_n)_{n=1}^{\infty}$ be a sequence of subsets of $\mathbb{R}$. Then $\displaystyle{m^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} m^*(A_n)}$.
• Proof: If $\displaystyle{\sum_{n=1}^{\infty} m^*(A_n) = \infty}$ then clearly $\displaystyle{m^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} m^*(A_n) = \infty}$.
• So assume that $\displaystyle{\sum_{n=1}^{\infty} m^*(A_n)}$ is finite. Let $\epsilon > 0$ be given.
• For every $n \in \mathbb{N}$, let $(I_{n,m})_{m=1}^{\infty}$ be an open interval cover of $A_n$, that is, $\displaystyle{A_n \subseteq \bigcup_{m=1}^{\infty} I_{n,m}}$. By definition, since $\displaystyle{m^*(A_n) = \inf \left \{ \sum_{s=1}^{\infty} l(I_s) : A \subseteq \bigcup_{s=1}^{\infty} I_n \: \mathrm{and} \: \{ I_s = (a_s, b_s) \}_{s=1}^{\infty} \right \}}$ and $(I_{n,m})_{m=1}^{\infty}$ is an open interval cover of $A_n$, for $\displaystyle{\epsilon_0 = \frac{\epsilon}{2^n} > 0}$ we have that:
(2)
\begin{align} \quad \sum_{m=1}^{\infty} l(I_{n,m}) &< m^*(A_n) + \epsilon_0 = m^*(A_n) + \frac{\epsilon}{2^n} \end{align}
• Now since $(I_{n,m})_{m=1}^{\infty}$ is an open interval cover of $A_n$ for each $n \in \mathbb{N}$ we have that $(I_{n,m})_{m,n=1}^{\infty}$ is an open interval cover of $\displaystyle{\bigcup_{n=1}^{\infty} A_n}$. Thus:
(3)
\begin{align} \quad m^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} \left ( \sum_{m=1}^{\infty} l(I_{n, m}) \right ) \leq \sum_{n=1}^{\infty} \left ( m^*(A_n) + \frac{\epsilon}{2^n} \right ) = \sum_{n=1}^{\infty} m^*(A_n) + \epsilon \sum_{n=1}^{\infty} \frac{1}{2^n} \end{align}
• Notice that the series $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{2^n}}$ converges to $1$. Therefore $\displaystyle{m^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} m^*(A_n) + \epsilon}$. Since $\epsilon > 0$ was arbitrary, we have that:
(4)
\begin{align} \quad m^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} m^* (A_n) \quad \blacksquare \end{align}
 Corollary: Let $E \in \mathcal P(\mathbb{R})$ be a countable set. Then $m^*(E) = 0$.
• Proof: Since $E$ is countable, $E = \{ E_1, E_2, ... \}$. Let $A_n = \{ E_n \}$ for each $n \in \mathbb{N}$. Then by countable subadditivity of the Lebesgue outer measure:
(5)
\begin{align} \quad m^*(E) = m^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} m^*(A_n) = \sum_{n=1}^{\infty} m^*(\{ E_n \}) = \sum_{n=1}^{\infty} 0 = 0 \end{align}
• Therefore $m^*(E) = 0$. $\blacksquare$