Outer Measurable Sets

Recall from the Outer Measures on Measurable Spaces page that if we have the measurable space $(X, \mathcal P(X))$ then an outer measure on this space is a set function $\mu^* : \mathcal P(X) \to [0, \infty]$ with the following properties:

• 1) $\mu^*(\emptyset) = 0$.

• 2) If $A$ and $B$ are subsets of $X$ with $A \subseteq B$ then $\mu^* (A) \leq \mu^*(B)$.

• 3) If $(A_n)_{n=1}^{\infty}$ is ANY collection of subsets of $X$ then $\displaystyle{\mu^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} \mu^*(A_n)}$.

Recall the Lebesgue outer measure $m^*$. We defined a subset $E$ of $\mathbb{R}$ to be a Lebesgue measurable set if for all subsets $A$ of $\mathbb{R}$ we have that:

(1)

\begin{align} \quad m^*(A) = m^*(A \cap E) + m^*(A \cap E^c) \end{align}

We will now define when a set is outer measure with respect to a particular outer measure.

Definition: Let $(X, \mathcal P(X))$ be a measurable space with outer measure $\mu^*$. A subset $E$ of $X$ is said to be Outer Measurable or $\mu^*$-Measurable if for all subsets $A$ of $X$ we have that $\mu^* (A) = \mu^*(A \cap E) + \mu^*(A \cap E^c)$.

Note that if $A$ is any subset of $X$ then:

(2)

\begin{align} \quad A = \left ( A \cap E \right ) \cup \left ( A \cap E^c \right ) \end{align}

By definition, if $\mu^*$ is an outer measure on $(X, \mathcal P(X))$, then property (3) of outer measures gives us that:

(3)

\begin{align} \quad \mu^*(A) = \mu^*((A \cap E) \cup (A \cap E^c)) \leq \mu^*(A \cap E) + \mu^*(A \cap E^c) \end{align}

So to show that a subset $E$ of $X$ is outer measurable, we only need to prove that for all subsets $A$ of $X$ we have that:

(4)

\begin{align} \quad \mu^*(A) \geq \mu^*(A \cap E) + \mu^*(A \cap E^c) \end{align}