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.

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    $\begingroup$ A fat Cantor set would work for this purpose. $\endgroup$ – user61527 Dec 18 '13 at 18:32
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    $\begingroup$ @AustinMohr you should post this as answer $\endgroup$ – Norbert Dec 18 '13 at 19:44
  • $\begingroup$ @AustinMohr: If you wish add it as answer so I can accept it, if not I will write answer in couple days =) $\endgroup$ – Meow Dec 18 '13 at 20:37

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 $x \in \hat{\mathcal{C}}$.

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    $\begingroup$ Great =) I would just like to add that if we need $m(C)=\alpha$ we choose $l_k=\frac{(1-\alpha)}{2^(2k-1)}$. $\endgroup$ – Meow Dec 19 '13 at 8:50

Let cover rational numbers of $[0,1]$ by any open set $G \subset [0,1]$ such that $m(G)=1-\alpha$. Then $C=[0,1] \setminus G$ is a desired set.


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