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Problem: Let $0<a<1$. Let $\lambda$ be the Lebesgue measure on $\mathbb{R}$. Show that:

(i) There exists a closed set $A\subseteq[0,1]$, which does not contain any non-empty open sets, such that $\lambda(A)=a$.

(ii) There exists an open, dense set $B\subseteq[0,1]$ such that $\lambda(B)=a$.

Ideas: (i): This means the interior of A has to be the empty set. I'm stuck here.

(ii): If we have a set A like in (i) where $\lambda(A)=1-a$, then $B=[0,1]\setminus A$ is obviously open, and it is also dense, since the closure of $B$ would be $[0,1]$. Furthermore $\lambda(B)=\lambda([0,1]\setminus A)=\lambda([0,1])-\lambda(A)=1-(1-a)=a$.

As always, I don't want a full solution, just hints to guide me. Thanks in advance!

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Hint for (ii): Enumerate the rationals as $\{r_n\}$ and put little intervals of really quickly decreasing length around each rational. This gives density and openness.

For (i), try modifying the construction of the Cantor set to remove less at each step.

Search term for (i):

Fat Cantor set

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  • $\begingroup$ Can you elaborate on your hint for (ii)? The density is obvious. I assume I have to set the "rate of decreasing length" of the intervals to control the measure. $\endgroup$ – blst Nov 9 '13 at 1:45
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    $\begingroup$ @blst Use the fact that $$\sum_{n = 0}^{\infty} \frac{\epsilon}{2^n} = \epsilon$$. $\endgroup$ – user61527 Nov 9 '13 at 1:47
  • $\begingroup$ Of course. Why I did not see that is beyond me since I used it for (i). Thank you! $\endgroup$ – blst Nov 9 '13 at 1:50
  • $\begingroup$ @blst You're very welcome. $\endgroup$ – user61527 Nov 9 '13 at 1:50

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