# Existence of certain set

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!

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.
• @blst Use the fact that $$\sum_{n = 0}^{\infty} \frac{\epsilon}{2^n} = \epsilon$$. – user61527 Nov 9 '13 at 1:47