My Question
Define subsets $G$, $G_0$, and $G_1$ of $\mathbb{R}$ by\begin{align*} G &= \{x:x=r+n\sqrt{2}\ \text{for some $r$ in $\mathbb{Q}$ and $n$ in $\mathbb{Z}$}\},\\ G_0 &= \{x:x=r+2n\sqrt{2}\ \text{for some $r$ in $\mathbb{Q}$ and $n$ in $\mathbb{Z}$}\},\ \text{and}\\ G_1 &= \{x:x=r+(2n+1)\sqrt{2}\ \text{for some $r$ in $\mathbb{Q}$ and $n$ in $\mathbb{Z}$}\}.\end{align*}Define a relation $\sim$ on $\mathbb{R}$ by letting $x \sim y$ hold when $x-y\in G$; the relation $\sim$ is then an equivalence relation on $\mathbb{R}$. Use the axiom of choice to form a subset $E$ of $\mathbb{R}$ that contains exactly one representative of each equivalence class of $\sim$. Let $A = E + G_0$ (that is, let $A$ consist of the points that have the form $e+g_0$ for some $e$ in $E$ and some $g_0$ in $G_0$).
I got confused by the book's remark that "the set $A$ defined above is not Lebesgue measurable: if it were, then both $A$ and $A^c$ would include (in fact, would be) Lebesgue measurable sets of positive Lebesgue measure". Could someone please explain why this is true?
Background Information
The above remark is made after the follow proposition:
Proposition 1.4.11$\quad$There is a subset $A$ of $\mathbb{R}$ such that each Lebesgue measurable set that is included in $A$ or in $A^c$ has Lebesgue measure zero.
Proof$\quad$ Define subsets $G$, $G_0$, and $G_1$ of $\mathbb{R}$ by\begin{align*} G &= \{x:x=r+n\sqrt{2}\ \text{for some $r$ in $\mathbb{Q}$ and $n$ in $\mathbb{Z}$}\},\\ G_0 &= \{x:x=r+2n\sqrt{2}\ \text{for some $r$ in $\mathbb{Q}$ and $n$ in $\mathbb{Z}$}\},\ \text{and}\\ G_1 &= \{x:x=r+(2n+1)\sqrt{2}\ \text{for some $r$ in $\mathbb{Q}$ and $n$ in $\mathbb{Z}$}\}.\end{align*}One can prove that $G$ and $G_0$ are subgroups of $\mathbb{R}$ (under addition), and $G_0$ and $G_1$ are disjoint, that $G_1 = G_0 + \sqrt{2}$, and that $G = G_0 \bigcup G_1$. Define a relation $\sim$ on $\mathbb{R}$ by letting $x \sim y$ hold when $x-y\in G$; the relation $\sim$ is then an equivalence relation on $\mathbb{R}$. Use the axiom of choice to form a subset $E$ of $\mathbb{R}$ that contains exactly one representative of each equivalence class of $\sim$. Let $A = E + G_0$ (that is, let $A$ consist of the points that have the form $e+g_0$ for some $e$ in $E$ and some $g_0$ in $G_0$).
We now show that there does not exist a Lebesgue measurable subset $B$ of $A$ such that $\lambda(B)>0$. For this let us assume that such a set exists; we will derive a contradiction. Proposition 1.4.10 implies that there is an interval $(-\epsilon,\epsilon)$ that is included in $\text{diff}(B)$ and hence in $\text{diff}(A)$. Since $G_1$ is dense in $\mathbb{R}$, it meets the interval $(-\epsilon,\epsilon)$ and hence meets $\text{diff}(A)$. This, however, is impossible, since each element of $\text{diff}(A)$ is of the form $e_1-e_2+g_0$ (where $e_1$ and $e_2$ belong to $E$ and $g_0$ belongs to $G_0$) and so cannot belong to $G_1$ (the relation $e_1-e_2+g_0=g_1$ would imply that $e_1=e_2$ and $g_0=g_1$, contradicting the disjointness of $G_0$ and $G_1$). This completes our proof that every Lebesgue measurable subset of $A$ must have Lebesgue measure zero.
One can check that $A^c = E + G_1$ and hence that $A^c = A + \sqrt{2}$. It follows that each Lebesgue measurable subset of $A^c$ is of the form $B+\sqrt{2}$ for some Lebesgue measurable subset $B$ of $A$. Since $A$ has no Lebesgue measurable subsets of positive measure, it follows that $A^c$ also has no such subsets, and with this the proof is complete.
The definition of $\text{diff}$ is the following:
Definition$\quad$ Let $A$ be a subset of $\mathbb{R}$. Then $\text{diff}(A)$ is the subset of $\mathbb{R}$ defined by\begin{align*}\text{diff}(A) = \{x-y:x \in A\ \text{and}\ y \in A\}.\end{align*}
Proposition 1.4.10 is the following:
Proposition 1.4.10$\quad$Let $A$ be a Lebesgue measurable subset of $\mathbb{R}$ such that $\lambda(A) > 0$. Then $\text{diff}(A)$ includes an open interval that contains 0.
Any help will be really appreciated!