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Can one construct a set $E$, $m^*(rE)\neq rm^*(E)$.

Well , we know that if $E$ is measurable then $$m^*(rE)= rm^*(E)$$,so what is the case of a none-measurable set ?

Well,I think it holds for any set $E$, we assume that $$m^*(E)=\sum_n |I_n|$$ Where $I_n$ are some cubes which covers the $E$ , since for all $x\in E$, there will exist $x\in I_k$, thus $rx\in rE,rx\in rI_k$ , which implies $\{rI_n\}$ is an open cover of $rE$,thus $$m^*(rE)\leq rm^*(E)$$ On the other hand , assume $\{J_n\}$ is an open cover of $rE$ , one can use $1/r J_n$ to gain the reverse inequality.

I don't know whether there are any bugs in my proof , so , hope your nice answer, thank you in advance

Alexander Lau
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  • Theorem 1 on p. 57 (in Chapter III.2) of Kuczma's 1985 book says that if $f:{\mathbb R}^n \rightarrow {\mathbb R}^n$ is given by $f(x)=Lx+b,$ where $L$ is a non-singular $n \times n$ matrix and $b \in {\mathbb R}^{n},$ then for EVERY $E \subseteq {\mathbb R}^n$ we have the outer Lebesgue and inner Lebesgue measures of the image $f[E]$ equal, respectively, to $|\det L|$ times the outer Lebesgue and inner measures of $E.$ Unfortunately, I don't know a more accessible reference now, and standard texts seem to only consider this when $E$ is measurable. – Dave L. Renfro Apr 13 '19 at 15:00
  • I believe the result I just stated can also be found in Stanisław Łojasiewicz's 1988 (English translated book) An Introduction to the Theory of Real Functions, but I don't have a copy of this book to give a more precise reference. – Dave L. Renfro Apr 13 '19 at 15:07
  • thank both of you very much ！I'll go to find those books – Alexander Lau Apr 13 '19 at 15:27

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