I need to find, for given $0<\alpha<1$, closed subset $C \subseteq [0,1]$ that satisfies $\lambda(C)=\alpha$ ($\lambda$ stands for Lebesgue measure) and includes no non-empty open set.
It's easy to find closed set without open subsets (countable union of points) but it's measure is zero, so I suposse I need uncountable set of points, maybe in interval $[0,\alpha]$, but I have no idea for such construction.
Best Answer
The following is from an old homework assignment. I've left the question and answer format intact, since it does not address only OP's question.
Construct a closed set $\hat{\mathcal{C}}$ so that at the $k$th stage of the construction one removes $2^{k-1}$ centrally situated open intervals each of length $l_k$ with
$$l_1 + 2l_2 + \cdots + 2^{k-1}l_k < 1$$
a) If $l_j$ are chosen small enough, then $$ \sum_{k=1}^\infty 2^{k-1}l_k < 1 $$ In this case, show that $m(\hat{\mathcal{C}}) > 0$, and in fact, $$m(\hat{\mathcal{C}}) = 1 - \displaystyle\sum_{k=1}^\infty 2^{k-1}l_k$$
Proof: First, we claim that $\hat{\mathcal{C}}$ is measurable. Denote by $O_k$ the union of the open sets removed from $[0,1]$ at step $k$ of the construction. Since the union of an arbitrary number of open sets is open, each of the $O_k$ is open. Furthermore, $O = \displaystyle\bigcup_{k=1}^\infty O_k$ is open. Now, we have that $\hat{\mathcal{C}} = [0,1] \setminus O$ is closed and therefore measurable (as all closed sets are measurable).
Now, to determine $m(\hat{\mathcal{C}})$, observe that both $O$ and $\hat{\mathcal{C}}$ are measurable ($O$ is measurable because it is open) and disjoint and that $O \cup \hat{\mathcal{C}} = [0,1]$. Then \begin{align*} m([0,1]) &= m(O) + m(\hat{\mathcal{C}})\\ m(\hat{\mathcal{C}}) &= m([0,1]) - m(O)\\ \end{align*} Furthermore observe that all of the $O_k$ are open (and so measurable) and disjoint with $O = \displaystyle\bigcup_{k=1}^\infty O_k$. If we further break the $O_k$ into their constituent open subsets, these properties still hold. Hence \begin{align*} m(\hat{\mathcal{C}}) =& m([0,1]) - m(O)\\ =& m([0,1]) - \displaystyle\sum_{k=1}^\infty m(O_k)\\ =& 1 - \displaystyle\sum_{k=1}^\infty 2^{k-1}l_k \end{align*}
b) Show that if $x \in \hat{\mathcal{C}}$, then there exists a sequence of points $\{x_n\}_{n=1}^\infty$ such that $x_n \notin \hat{\mathcal{C}}$, yet $x_n \rightarrow x$ and $x_n \in I_n$, where $I_n$ is a sub-interval in the complement of $\hat{\mathcal{C}}$ with $|I_n| \rightarrow 0$.
Proof: Observe first that since $\displaystyle\sum_{k=1}^\infty 2^{k-1}l_k < 1$, the tail of the series must go to zero. That is, for any $\epsilon > 0$, there exists $N$ such that $l_n < \epsilon$ for all $n \geq N$. Now, let $x \in \hat{\mathcal{C}}$. Let $\hat{\mathcal{C}}_k$ denote the $k$ stage of the construction. For each $k$, $x$ belongs to some closed subset $S_k$ of $C_k$. Let $I_k$ be the open interval removed from $S_k$ to proceed to the next step of the construction. We take any $x_k \in I_k$ to form our sequence $\{x_n\}_{n=1}^\infty$. Clearly, each $x_k$ belongs to an sub-interval in the complement of $\hat{\mathcal{C}}$. Furthermore, $|I_k| = l_k \rightarrow 0$. It remains to show that $x_n \rightarrow x$.
From the construction of $\hat{\mathcal{C}}_k$ and our selection of $x_n$, it is clear that $$|x - x_n| < |I_n| + |S_n|.$$ By our previous observation, we know that $|I_n| = l_n \rightarrow 0$. Now \begin{align*} |S_n| &= \frac{1 - \displaystyle\sum_{k=1}^n 2^{k-1}l_k}{2^n}\\ &\leq \frac{1}{2^n}\\ &\rightarrow 0 \text{ as } n \rightarrow \infty \end{align*} Hence, $|x - x_n| \rightarrow 0$. That is, $\{x_n\}_{n=1}^\infty$ converges to $x$.
c) Prove as a consequence that $\hat{\mathcal{C}}$ is perfect and contains no open interval.
Proof: To see that $\hat{\mathcal{C}}$ is perfect, let $\epsilon > 0$ be given and consider $B(x,\epsilon)$ for any $x \in \hat{\mathcal{C}}$. We can find $N \in \mathbb{N}$ such that $S_N \subset B(x,\epsilon)$. Now, this interval must have two endpoints $a_N$ and $b_N$ (one of which could possibly be equal to $x$). By the construction of $\hat{\mathcal{C}}$, we know that the endpoints of any interval are never removed, and so $a_N, b_N \in \hat{\mathcal{C}}$. Furthermore, we have that $a_N, b_N \in S_N \subset B(x,\epsilon)$. Therefore, $x$ is not isolated.
Suppose, to the contrary, that there exists an open interval $O \in \hat{\mathcal{C}}$. Then, for any $x \in O$, there exists $\epsilon_0$ such that $B(x,\epsilon_0) \subseteq O$. Let $\epsilon < \epsilon_0$. Then, there can be no sequence $\{x_n\}_{n=1}^\infty$ of the type described in part (b) whose limit is $x$, since $B(x,\epsilon_0) \subseteq \hat{\mathcal{C}}$ implies that $|x - x_n| > \epsilon_0 > \epsilon$ for all $n$. This contradicts the conclusion of part (b), and so it must be that $\hat{\mathcal{C}}$ contains no open interval.
d) Show also that $\hat{\mathcal{C}}$ is uncountable. (Just for fun)
Proof: We claim that $\hat{\mathcal{C}}$ is in one-to-one correspondence with infinite ternary strings containing only 0s and 2s, and so is uncountable.
($\Rightarrow$) Let $x \in \hat{\mathcal{C}}$. We build a ternary string for $x$ of the desired form as follows. Consider $\hat{\mathcal{C}}_1$. When we remove the centrally situated open interval, it must be that $x$ belongs to either the left closed subinterval (in which case let first digit of the ternary string for $x$ be 0) or the right closed subinterval (in which case let first digit of the ternary string for $x$ be 2). Next, consider $\hat{\mathcal{C}}_2$. The interval of $\hat{\mathcal{C}}_1$ to which $x$ currently belongs will be divided into three subintervals, and so we append a 0 to the ternary string for $x$ if it belongs to the leftmost subinterval or a 2 if it belongs to the rightmost subinterval. Continuing in this way, we see that $x$ has an associated ternary string containing only the digits 0 and 2.
($\Leftarrow$) Let $s$ be an infinite ternary string containing only 0s and 2s. We associate can with $s$ an $x \in \hat{\mathcal{C}}$ as follows. If the first digit of $s$ is 0, we choose the left subinterval of $\hat{\mathcal{C}}_1$. If the first digit of $s$ is 2, we choose the rightmost subinterval of $\hat{\mathcal{C}}_1$. When we form $\hat{\mathcal{C}}_2$, the interval we have just chosen will be subdivided into three subintervals. If the second digit of $s$ is 0, we select the leftmost subinterval. If the second digit of $s$ is 2, we select the rightmost subinterval. Continue in this way. Since each $x \in \hat{\mathcal{C}}$ belongs to a singleton set, we see that $s$ will specify some $x \in \hat{\mathcal{C}}$.