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For a subset $A$ of $\Bbb R$ and real numbers $a$ and $b$ define the set $$aA+b=\{ax+b:x\in A\}$$ Show that $m^{*}(aA+b)=|a|m^{*}(A)$ and if $A$ is Lebesgue measurable so is $aA+b$.

I don't know how to show the first except this

Since outer measure is translation invariant: $m^{*}(aA+b)=m^{*}(aA)$

And for the second one, for any set $E\subset \Bbb R$ we need to show $$m^{*}(E)=m^{*}[E\cap(aA+b)]+m^{*}[E\cap(aA+b)^c]$$ But I don't know how any help?

marya
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1 Answers1

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A translate of a measurable set is measurable i.e. For any $b\in \Bbb R$ if $A$ is measurable so is $A+b$.

Note that for any interval $(s,t)$; $m^{*}(c(s,t))=|c|m^{*}((s,t))=|c|(t-s)$ and

$A\subset \bigcup_{n=1}^{\infty}(a_n,b_n)$ holds if and only if $aA\subset \bigcup_{n=1}^{\infty}a(a_n,b_n)$ holds for each $a\in \Bbb R$ and since $m^{*}([a_n,b_n))=m^{*}((a_n,b_n))$, it follows that $m^{*}(aA)=|a|m^{*}(A)$ for each $a\in \Bbb R$. Consider the following useful identites $$E\cap aA=a((a^{-1}E)\cap A)\text{ and } E\cap (aA)^c=a((a^{-1}E)\cap A^c)\quad (a\neq0)$$ imply $$m^{*}(E\cap aA)+m^{*}(E\cap (aA)^c)=|a|[m^{*}((a^{-1}E)\cap A)+m^{*}((a^{-1}E)\cap A^c)]$$ which shows $A$ is measurable if and only if $aA$ is measurable.

albo
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