Prove that for any $k \in \Bbb{R}$ we have$ \quad \mu^{*}(kA) = |k|\mu^{*}A, $

Question

$A \in \Bbb{R}$. Prove that for any $k \in \Bbb{R}$ we have$ \quad \mu^{*}(kA) = |k|\mu^{*}A, $ where $kA:=\{ka: a \in A \}$

$\mu^{*}(kA)=inf |I_n|_{kA \subset \cup I_n}$

To bo honest I am not sure what to do now. Should I consider union of $kA$ ?

Answer 1

The case for $k<0$: For $\epsilon>0$, find a sequence $(I_{n})$ of intervals such that $\mu^{\ast}(A)+\epsilon>\displaystyle\sum_{n=1}|I_{n}|$ and that $A\subseteq\displaystyle\bigcup_{n=1}I_{n}$.

Expressing $I_{n}=(a_{n},b_{n})$, then $kI_{n}=(kb_{n},ka_{n})$. For all $x\in kA$, then $x/k\in A$ and hence there is some $n$ such that $x/k\in I_{n}$, so $x\in kI_{n}$, this shows that $kA\subseteq\displaystyle\bigcup_{n}kI_{n}$. Hence $\mu^{\ast}(kA)\leq\displaystyle\sum_{n}|kI_{n}|=|k|\sum_{n}|I_{n}|<|k|(\mu^{\ast}(A)+\epsilon)$. Since $\epsilon>0$ is arbitrary, then $\mu^{\ast}(kA)\leq|k|\mu^{\ast}(A)$.

Now $|k|\mu^{\ast}(A)=|k|\mu^{\ast}(k^{-1}kA)\leq|k|\cdot|k^{-1}|\mu^{\ast}(kA)=\mu^{\ast}(kA)$.