Additivity of outer measure when one set is closed (Axler: Measure, Integration & Real Analysis)

Question

Suppose $A$ and $F$ are disjoint subsets of $\mathbb{R}$ and $F$ is a closed set. Then $|A \cup F| = |A| + |F|$, where $|X|$ denotes the outer measure of a set $X$. In his proof, Axler (2.63) shows that $ |A| + |F| \leq |G|$, where $G$ is an open cover of $|A \cup F|$. While this is quite clear, how can he conclude that $ |A| + |F| \leq |A \cup F|$?

I'd appreciate any help of yours.

Answer 1

Let $G = \bigcup_{k = 1}^\infty I_k$ where $I_1,I_2,\ldots$ is a sequence of open intervals containing $A \cup F$. Hence, $A \cup F \subseteq G$ and so $A \subseteq G - F$. By monotonicity of outer measure, $\vert A \vert \le \vert G - F \vert$. Since $F$ is a closed set and $G$ is an open set, $G - F$ is an open set. Thus, $G = F \cup (G - F)$ is a disjoint union and so, by 2.62, $\vert G \vert = \vert F \vert + \vert G - F \vert$. Technically speaking, we are disallowed from "subtracting $\vert F \vert$ from both sides" except in the case where $\vert F \vert$ is finite. Instead, we add $\vert F \vert$ to both sides of the inequality $\vert A \vert \le \vert G - F \vert$ to get $\vert A \vert + \vert F \vert \le \vert G - F \vert + \vert F \vert = \vert G \vert$.

So, we just showed that $\vert A \vert + \vert F \vert \le \vert G \vert$. At this point, Axler sweeps some technicalities under the rug and essentially "takes the infimum of both sides" to get $\vert A \vert + \vert F \vert \le \vert A \cup F \vert$. Let's look at the details. The outer measure of $A \cup F$ is defined to be \begin{align*} \vert A \cup F \vert &= \inf\bigg\{ \sum_{k = 1}^\infty \ell(I_k) : I_1,I_2,\ldots \text{ is a sequence of open intervals containing } A \cup F \bigg\}. \end{align*} We also note that $$\vert A \vert + \vert F \vert \le \vert G \vert = \bigg\vert \bigcup_{k = 1}^\infty I_k \bigg\vert \le \sum_{k = 1}^\infty \ell(I_k).$$

Hence, $\vert A \vert + \vert F \vert$ is less than or equal to every element of the set $$ \bigg\{ \sum_{k = 1}^\infty \ell(I_k) : I_1,I_2,\ldots \text{ is a sequence of open intervals containing } A \cup F \bigg\}. $$ Therefore, $\vert A \vert + \vert F \vert \le \vert A \cup F \vert$. $\blacksquare$

Answer 2

You already know that $ |A| + |F| \leq |G|$, where $G$ is an open cover of $|A \cup F|$. In other words: $ |A| + |F| \leq |G|$, where $G$ is an open set and $A \cup F \subseteq G $.

It follows that $ |A| + |F| \leq \inf \{|G| : G \text{ open set and } A \cup F \subseteq G \}$.

On the other hand, note that $|A \cup F| = \inf \{|G| : G \text{ open set and } A \cup F \subseteq G \}$.

So, we have $ |A| + |F| \leq |A \cup F|$.