Example Problems Regarding The Lebesgue Outer Measure

# Example Problems Regarding the Lebesgue Outer Measure

So far we have shown the following properties regarding the Lebesgue outer measure:

• 1. For any finite or countably infinite subset $E$ of $\mathbb{R}$:
(1)
\begin{align} \quad m^*(E) = 0 \end{align}
• $m^*$ has the monotonicity property. If $A \subseteq B$ then:
(2)
\begin{align} \quad m(A) \leq m(B) \end{align}
• 3. If $I$ is any interval then:
(3)
\begin{align} \quad m^*(I) = l(I) \end{align}
• 4. $m^*$ is translation invariant. For any subset $E$ of $\mathbb{R}$ and for any $a \in \mathbb{R}$ we have that:
(4)
\begin{align} \quad m^*(E + a) = m^*(E) \end{align}
• 5. $m^*$ is countably subadditive. For any sequence $(A_n)_{k=1}^{\infty}$ of subsets of $\mathbb{R}$ we have that:
(5)
\begin{align} \quad m^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} m^*(A_n) \end{align}

We now look at some example problems involving the Lebesgue outer measure.

## Example 1

Prove that $[0, 1]$ is an uncountable subset of $\mathbb{R}$.

If $[0, 1]$ were countable then $m^*([0, 1]) = 0$. But $[0, 1]$ is an interval and so $m^*([0, 1]) = l([0, 1]) = 1$. Therefore $[0, 1]$ must be an uncountable subset of $\mathbb{R}$.

## Example 2

Let $I$ be the set of all irrational numbers contained in the interval $[0, 1]$. Prove that $m^*(I) = 1$.

The set of all rational numbers contained in $[0, 1]$ is a countable set and so by countable subadditivity we have that:

(6)
\begin{align} \quad 1 = m^*([0, 1]) = m^*(I \cup ([0, 1] \setminus I)) \leq m^*(I) + m([0, 1] \setminus I) = m^*(I) \end{align}

But since $I \subseteq [0, 1]$ we have the monotonicity of the Lebesgue outer measure that:

(7)
\begin{align} \quad m^*(I) \leq m^*([0, 1]) = 1 \end{align}

Therefore we conclude that $m^*(I) = 1$.

## Example 3

Let $A, B \subseteq \mathbb{R}$. Prove that if $m^*(A) = 0$ then $m^*(A \cup B) = m^*(B)$.

Since $B \subseteq A \cup B$ we have by the monotonicity property of the Lebesgue outer measure that:

(8)
\begin{align} \quad m^*(B) \leq m^*(A \cup B) \end{align}

Furthermore by the countable subadditivity of the Lebesgue outer measure we have that:

(9)
\begin{align} \quad m^*(A \cup B) \leq m^*(A) + m^*(B) = 0 + m^*(B) = m^*(B) \end{align}

Therefore we conclude that $m^*(A \cup B) = m^*(B)$.