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The set of rational numbers $\mathbb{Q}$ is an unbounded subset of $\mathbb{R}$ with Lebesgue outer measure zero. In addition, $\mathbb{R}$ is an unbounded subset of itself with Lebesgue outer measure $+\infty$. Therefore the following question came to my mind: is there an unbounded subset of $\mathbb{R}$ with positive Lebesgue outer measure?

If there is, can you give me an example?

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$[0,1]\cup\mathbb Z$ –  Mariano Suárez-Alvarez Jan 17 '11 at 21:53
    
Thank you, Mariano! I see, the idea is to "add" an unbounded countable set to a set with positive Lebesgue outer measure. –  ragrigg Jan 17 '11 at 21:57

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up vote 6 down vote accepted

I guess you mean with positive and finite outer measure. An easy example would be something like $[0,1]\cup\mathbb{Q}$. But perhaps you also want to have nonzero measure outside of each bounded interval? In that case, consider $[0,1/2]\cup[1,1+1/4]\cup[2,2+1/8]\cup[3,3+1/16]\cup\cdots$. If you want the set to have positive measure in each subinterval of $\mathbb{R}$, you could let $x_1,x_2,x_3,\ldots$ be a dense sequence (like the rationals) and take a union of open intervals $I_n$ such that $I_n$ contains $x_n$ and has length $1/2^n$.

On the other hand, it is often useful to keep in mind that every set of finite measure is "nearly bounded". That is, if $m(E)<\infty$ and $\epsilon>0$, then there is an $M\gt0$ such that $m(E\setminus[-M,M])<\epsilon$. One way to see this is by proving that the monotone sequence $(m(E\cap[-n,n]))_{n=1}^\infty$ converges to $m(E)$.

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An interesting property of the first answer is that it is also dense. –  Vitaly Lorman Jan 17 '11 at 22:10
    
The third example is dense and open. –  Jonas Meyer Jan 17 '11 at 22:19
    
Nice examples! –  ragrigg Jan 17 '11 at 22:20

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