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This exercise appeared in my real analysis test last year, and is still puzzling me since then. Ironically, even the professor doubts if the b part is actually truth (still...)

Let $A \subset \mathbb{R}$

a) If $\displaystyle0, then for every $\alpha\in (0,1)$, exists an open interval $I$ such that $\displaystyle \alpha m^*(I) \leq m^*(A \cap I) $

b) If $ \displaystyle m^*(A \cap I) \leq \frac{m^*(I)}{2}$ for every open interval, then $m^*(A) = 0$

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    @PeteL.Clark No, it's fine. I'm just trying to be funny as well. I didn't even create the question, so I'm asking if "I" make the title better, will that affect your vote.2012-09-08

2 Answers 2

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Let $\varepsilon>0$ and $\{I_n\}_n$ a collection of open intervals which cover $A$ and such that $\sum_{j=1}^{+\infty}m^*(I_j)-m^*(A)\leq 2\varepsilon$. (the outer measure is finite). We have $m^*(A)\leq \sum_{j=1}^{+\infty}m^*(A\cap I_j)\leq \frac 12\sum_{j=1}^{+\infty}m^*(I_j)\leq \frac{m^*(A)}2+\varepsilon,$ hence $0\leq \frac{m^*(A)}2\leq \varepsilon$ and we are done.

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    We work with $A_n:=A\cap [-n,n]$ to see that $m^*(A\cap [-n,n]=0$ (it's what you did in your answer).2012-09-07
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Consider $A_n:=A\cap\subseteq [-n,n]$ for $n\in \mathbb N$. Let $\mu:=m^*(A_n)$. Then $\mu\le 2n<\infty$. If $\varepsilon>0$, then there exists a countable collection of Intervals $I_k$, $k\in \mathbb N$, such that $A\subseteq\bigcup_k I_k$ and $\sum_k m^*(I_k) < \mu+\varepsilon$. We obtain $\mu \le \sum_k m^*(I_k\cap A_n) \le \sum_k \frac 12 m^*(I_k)=\frac 12\sum_k m^*(I_k)<\frac{\mu+\varepsilon}2$, hence $\mu<\varepsilon$. Since $\epsilon>0$ was arbitrary, we conclude $\mu=0$, i.e. $m^*(A_n)=0$.

Finally, $A=\bigcup_n A_n$ implies $m^*(A)\le \sum_n m^*(A_n)=0$.