I'm trying understand this example, but I'm find very difficulties to understand it. Some notations and definitions used on the example are exposed below.
$\textbf{Notations:}$
$A \oplus B := \{ a + b \ ; \ a \in A, b \in B \}$ for $A, B \subset \mathbb{R}$.
$A \oplus b := A \oplus \{ b \}$ for $A \subset \mathbb{R}$ and $b \in \mathbb{R}$.
$\textbf{Definitions:}$
A subset $P \subset \mathbb{R}$ is said to be perfect if it is closed and ever point of $P$ is a limit point.
A subset $A \subset \mathbb{R}$ is said to be nowhere dense if its closure has an empty interior.
A subset $C \subset \mathbb{R}$ is said to be a Cantor-like set if $C$ is an uncountable, perfect, nowhere dense subset of $\mathbb{R}$.
$\textbf{Example 1.19:}$ There exists a set $G \in \mathcal{B} (\mathbb{R})$ such that $I \cap G$ and $I \cap G^c$ each have positive Lebesgue measure for any nonempty open interval $I$.
$\textbf{Proof:}$
For $\frac{1}{2} < \alpha < 1$, let $C$ be a Cantor-like subset of $[0,1]$ having Lebesgue measure $\alpha$ and let $\{ S_n \ ; \ n \in \mathbb{N} \}$ denote the open intervals removed during the construction of $C$. For each $n \in \mathbb{N}$, construct another Cantor-like subset in the interval $S_n$ having Lebesgue measure $\alpha \lambda(S_n)$ and let $\{ S_{n,m} \ ; \ m \in \mathbb{N} \}$ denote the open intervals removed during the process. For each $n \in \mathbb{N}$ and $m \in \mathbb{N}$, construct yet another Cantor-like subset in the interval $S_{n,m}$ having Lebesgue measure $\alpha \lambda(S_{n,m})$ and let $\{ S_{n,m,j} \ ; \ j \in \mathbb{N} \}$ denote the open intervals removed during the process. Continuing in this way, it follows that corresponding to any finite sequence $n_1, \cdots, n_j$ of positive integers, there exists an open subinterval $S_{n_1, \cdots, n_j}$ of $[0,1]$ on which is constructed a Cantor-like subset $C_{n_1, \cdots, n_j}$ having Lebesgue measure $\alpha \lambda(S_{n_1, \cdots, n_j})$ . Note that
$$\lambda \left( \bigcup_{n_1 = 1}^\limits{\infty} \bigcup_{n_2 = 1}^\limits{\infty} \cdots \bigcup_{n_j = 1}^\limits{\infty} S_{n_1, \cdots, n_j} \right) = (1 - \alpha)^j.$$
Let $A_1, A_2, \cdots$ be an enumeration of the sets in $\{ C, C_{n_1, \cdots, n_j} \ ; \ n_i, j \in \mathbb{N} \}$ and let $A = \cup_{n = 1}^{\infty} A_n$. Notice the $A_i$'s are disjoint and that $A \in \mathcal{B} (\mathbb{R})$. Further,
$$\lambda(A) = 1 - \sum_{j = 1}^\limits{\infty} (1 - \alpha)^j = \frac{2 \alpha - 1}{\alpha}.$$
Now, let $I$ be an open nonempty subinterval of $[0,1]$ and assume that $I \cap A_i$ is empty for each $i \in \mathbb{N}$. It thus follows that for any $j \in \mathbb{N}$ there exists positive integers $n_11, \cdots, n_j$ such that $I \subset S_{n_1, n_2, \cdots, n_j}$ and hence $I$ has zero Lebesgue measure. This contradiction implies that $I \cap A_i$ is nonempty for some $i \in \mathbb{N}$. Let $B_i = I \cap A_i$ and $D_i = I \cap \left( [0,1] - A_i \right)$, and note that $D_i$ nonempty. Let $x_1 \in B_i$ and note that since $A_i$ is perfect there exists some point $x_2 \neq x_1$ that is also element of $B_i$. Assume without loss of generality that $x_1 < x_2$. By construction there exists a nonempty interval $(a,b)$ disjoint from $A_i$ such that $x_1 < a$ and $b > x_2$. It follows that $D_i$ has Lebesgue measure positive. Further, $\lambda(D_i) < 1$, since $\lambda(D_i) \leq 1 - \lambda(A_i) \leq \alpha < 1$. Thus, since $\lambda(B_i) + \lambda(D_i) = 1$ , it follows that $B_i$ also has positive Lebesgue measure.
Finally, let $G := \cup_{n \in \mathbb{Z}} A \oplus n$ and note that $\lambda(G \cap I)$ and $\lambda(G^c \cap I)$ are each positive for any nonempty open subinterval $I$ of $\mathbb{R}$. $\square$
$\textbf{My doubts:}$
$(1)$ How can I Cantor-like subset in the interval $S_n$ having Lebesgue measure $\alpha \lambda(S_n)$?
$(2)$ Why is valid $\lambda \left( \bigcup_{n_1 = 1}^\limits{\infty} \bigcup_{n_2 = 1}^\limits{\infty} \cdots \bigcup_{n_j = 1}^\limits{\infty} S_{n_1, \cdots, n_j} \right) = (1 - \alpha)^j$?
$(3)$ Why is valid $1 - \sum_{j = 1}^\limits{\infty} (1 - \alpha)^j = \frac{2 \alpha - 1}{\alpha}$?
$(4)$ Why $D_i$ is nonempty?
$(5)$ Why exactly I can find an open interval $(a,b)$ disjoint from $A_i$ such that $x_1 < a$ and $b > x_2$?
$(6)$ Why $1 - \lambda(A_i) \leq \alpha$?
$\textbf{My attempts:}$
I will put my attempts just to my doubts that I thought something.
$(2)$ I think that I need to use an argument involving the continuity of $\lambda$ like this $\lambda \left( \bigcup_{n_1 = 1}^\limits{\infty} \bigcup_{n_2 = 1}^\limits{\infty} \cdots \bigcup_{n_j = 1}^\limits{\infty} S_{n_1, \cdots, n_j} \right) = \lim_\limits{k_1 \rightarrow \infty} \lim_\limits{k_2 \rightarrow \infty} \cdots \lim_\limits{k_j \rightarrow \infty} \lambda \left( \bigcup_{n_1 = 1}^\limits{n_1 = k_1} \bigcup_{n_2 = 1}^\limits{n_2 = k_2} \cdots \bigcup_{n_j = 1}^\limits{n_j = k_j} S_{n_1, \cdots, n_j} \right)$ and use the $\sigma$-additivity for the measure $\lambda$, but I'm stuck in how to proceed with this argument.
$(3)$ I just realized why it's true now, but I will put the explanation here if anyone has the same doubt that I had. It's just observe that $0 < 1 - \alpha < 1$ by the choice of $\alpha$ did on the beginning of the proof, then we can apply the formula for infinite geometric series.
$(4)$ Since $\text{int.} (\overline{B_i}) = \text{int.} (\overline{I \cap A_i}) \subset \text{int.} (\overline{I} \cap \overline{A_i}) \subset \text{int.} (\overline{I}) \cap \text{int.} (\overline{A_i}) = \emptyset$, $B_i$ is nowhere dense, which means that, for every $x \in B_i$, every open neighborhood $U_x$ of $x$ intersects $B_i$ and $[0,1] \backslash B_i$, in particular, $I \cap U_x$ is an open neighborhood of $x$ which intersects $B_i$ and $[0,1] \backslash B_i$, so there is a $y \in (I \cap U_x) \cap ([0,1] \backslash B_i) \subset D_i$, therefore $D_i \neq \emptyset$.
$(5)$ I think it's true because $A_i$ is a Cantor-like set and, by definition, it is perfect, then it is closed, which imply that $[0,1] \backslash A_i$ is open in $[0,1]$ and $D_i = I \cap ([0,1] \backslash A_i) \subset [0,1] \backslash A_i$ is open in $[0,1]$, because $I$ is an open nonempty subinterval of $[0,1]$ and $[0,1] \backslash A_i$ is open in $[0,1]$, but I'm stuck how to argue that $x_1 < a$ and $b > x_2$. I can see the reason of the existence of this open interval $(a,b) \subset [0,1]$, but I don't know how to argue formally the existence of it.