Analysis question on outer measure

Question

Suppose $A \subseteq \mathbb{R}$ and $t\in \mathbb{R}$, define $tA=\{ta: a\in A\}$. Show that $|tA|=|t||A|$ (where $|t|$ is absolute value and $|tA|$ and $|A|$ are outer measure).

My thoughts: Since $tA$ and $A$ are both subsets of $\mathbb{R}$, by the subadditivity of outer measure we know that $|tA \cup A| \leq |tA| + |A|$. Because $A \subset tA\cup A$, we have $|A|\leq |tA \cup A| \leq |tA|+|A|$. So from this we can say $|tA\cup A| =|tA|=|A|$. I don't know why $|tA|$ is supposed to equal $|t||A|$ and not just $|A|$.

Answer 1

The key idea is that $l(tI) = |t|l(I)$, everything bubbles up from this fact.

Let $C_A=\{ \{ I_k \} | I_k \text{ is an open interval cover of } A\}$. Note that $t C_A = C_{tA}$ (with multiplication defined appropriately).

Define $\phi(\{ I_k \} ) = \sum_k l(I_k)$ and note that $\phi(t\{ I_k \} ) = \sum_k l(tI_k) = \sum_k |t|l(I_k) = |t| \phi(\{ I_k \} )$.

Hence $\{ \phi(\{ J_k \} ) | \{ J_k \} \in C_{tA} \} = \{ \phi(t\{ I_k \} ) | \{ I_k \} \in C_{A} \} =\{ |t|\phi( \{ I_k \} ) | \{ I_k \} \in C_{A} \} =|t|\{ \phi( \{ I_k \} ) | \{ I_k \} \in C_{A} \} $

In particular, taking the $\inf$ of both sides gives $m^* (tA) = |t| m^*(A)$.

Answer 2

Let $\{I_k\}_{k=1}^{\infty}$ be an open interval cover of $A$. Then the outer measure of $A$ be given by the following: $$m^*(A) = \inf\{\sum_{k=1}^{\infty}\ell(I_k): A \subseteq \bigcup_{k=1}^{\infty} I_k\}.$$ Notice, we can write the same for the new set $tA$, that is, we can write: $$m^*(tA) = \inf\{\sum_{k=1}^{\infty}\ell(tI_k): A \subseteq \bigcup_{k=1}^{\infty} tI_k\}.$$ However, notice that $\ell(tI_k) = |t|\ell(I_k)$ (for example think of $I_k = (-1,1)$ and $t = -2$.) so we can write: $$m^*(tA) = \inf\{\sum_{k=1}^{\infty}|t|\ell(I_k): A \subseteq \bigcup_{k=1}^{\infty} tI_k\}.$$ $$m^*(tA) = \inf\{|t|\sum_{k=1}^{\infty}\ell(I_k): A \subseteq \bigcup_{k=1}^{\infty} tI_k\}.$$ $$m^*(tA) = |t|\inf\{\sum_{k=1}^{\infty}\ell(I_k): A \subseteq \bigcup_{k=1}^{\infty} tI_k\}.$$ It should be clear that $$\inf\{\sum_{k=1}^{\infty}\ell(I_k): A \subseteq \bigcup_{k=1}^{\infty} I_k\} = \inf\{\sum_{k=1}^{\infty}\ell(I_k): A \subseteq \bigcup_{k=1}^{\infty} tI_k\}.$$

Hence, $$m^*(tA) = |t|m^*(A)$$ as desired.