Since my book talks about the Vitali set in a concise way, I decided to make those few words explicit. I want to propose it in the hope of not having committed inaccuracies.
Let $X=[0,1)$. For every $x,y\in [0,1)$ we define the following operation: $$x\dot{+}y:=\begin{cases} x+y & \text{if $x+y<1$} \\ x+y-1 & \text{if $x+y\ge1$} \end{cases}$$
Let $E\subseteq [0,1)$, define for each $y\in [0,1)$ $$E\dot{+} y:=\bigg\{e\dot{+}y\;\bigg|\;e\in E\bigg\}=\bigg\{e+y\;\bigg|\;e\in[0,1-y)\bigg\}\bigcup\bigg\{e+y-1\;\bigg|\;e\in[1-y,1)\bigg\}$$
We denote with $\mathcal{L(}\mathbb{R})$ the Lebesgue $\sigma-$algebra.
Theorem. There are subsets of $\mathbb{R}$ that are not Lebesgue measurable. That is $$\mathcal{L(}\mathbb{R})\subsetneq\mathcal{P}(\mathbb{R}).$$
Proof.
Step $0$. Consider the function $\{x\}\colon\mathbb{R}\to [0,1)$, defined as $x\mapsto x-[x].$ Observe that if $\tilde{q}\in\mathbb{Q}$, $\{\tilde{q}\}=\tilde{q}-[\tilde{q}]\in\mathbb{Q}\cap [0,1)$, in fact if were $\tilde{q}-[\tilde{q}]=r$, where $r\in\big(\mathbb{R}\setminus\mathbb{Q}\big)\cap [0,1)$, we would $\tilde{q}=r+[\tilde{q}]\in\mathbb{R}\setminus\mathbb{Q}$. Absurd!
Step $1$. Define the relation $\sim$ on $[0,1)$ in the following way: $$\text{For all}\; x,y\in [0,1)\quad x\sim y \iff x-y\in\mathbb{Q}.$$ Therefore if for each $\alpha\in [0,1)$ we place $$E_\alpha=\{y\in [0,1)\;|\; y-\alpha\in\mathbb{Q}\},$$ we have $$[0,1)=\bigcup_{\alpha\in [0,1)} E_\alpha.$$ For the Axiom of Choice, exists $V\subseteq [0,1)$ such that $$V\cap E_\alpha=\{x_\alpha\}\quad\text{for all}\;\alpha\in [0,1). $$
Let $\{q_n\}$ an enumeration of $\mathbb{Q}\cap [0,1)$ with $q_1=0.$ For each $n\in\mathbb{N}$ we define $$V_n:=V\dot{+} q_n:=\{x_\alpha\dot{+}q_n\;|\; x_\alpha\in V\}=\{x_\alpha+q_n\;|\; x_\alpha\in V\cap [0,1-q_n)\}\cup \{x_\alpha +q_n-1\;|\; x_\alpha\in V\cap [1-q_n,1))$$
Step $2$. Prove that $$[0,1)=\bigcup _{n=1}^\infty V_n.$$ Let $x\in [0,1)$, then $x\in E_\alpha$ for some $\alpha\in [0,1)$ $\Rightarrow$ $x-x_\alpha\in\mathbb{Q}\cap (-1,1).$ Then $$x=x_\alpha\dot{+} q_n\quad\text{for same}\; n\in\mathbb{N},$$ in fact we have:
If $x-x_\alpha\in\mathbb{Q}\cap [0,1)\Rightarrow x-x_\alpha= q_n\Rightarrow x=x_\alpha+q_n$, where $x_\alpha\in V\cap [0,1-q_n)$, in fact, if were $x_\alpha\in V\cap [1-q_n,1)\Rightarrow x=x_\alpha+ q_n\ge 1-q_n+q_n=1,$ absurd!
On the other hand, if $x-x_\alpha\in\mathbb{Q}\cap (-1,0)\Rightarrow x-x_\alpha=\tilde{q}$, where $\tilde{q}\in\mathbb{Q}\cap (-1,0).$ Now, $\{\tilde{q}\}=\tilde{q}-[\tilde{q}]=\tilde{q}+1:=q_n\in\mathbb{Q}\cap [0,1).$
Therefore $x-x_\alpha=q_n-1\Rightarrow x=x_\alpha+q_n-1$, where $x_\alpha\in V\cap [1-q_n,1)$ In fact, if were $x_\alpha\in V\cap [0.1-q_n)$ we would $x=x_\alpha+q_n-1<1-q_n+q_n-1=0$ absurd!
Then $x\in V_n$ for same $n\in\mathbb{N}.$
The other inclusion is obvious.
Step 3. We prove that $$V_n\cap V_m=\emptyset\quad\text{for}\;n\ne m.$$
We suppose for absurdity that for $n\ne m$, $y\in V_n\cap V_m$. Then $$y\in V_n\Rightarrow y=x_\alpha\dot{+} q_n;$$ and $$y\in V_m\Rightarrow y=x_\beta\dot{+} q_m,$$ where $x_\alpha, x_\beta\in V$ and $\alpha, \beta\in [0,1).$
Then $$x_\alpha\dot{+} q_n= x_\beta\dot{+} q_m,$$ from which we derive the following cases
Case 1. $[x_\alpha+q_n<1]$ Then $$x_\alpha+q_n=x_\beta+q_m\quad\text{or}\quad x_\alpha+q_n=x_\beta+q_m-1$$
Case 2. $[x_\alpha+q_n\ge 1]$ Then $$x_\alpha+q_n-1=x_\beta+q_m\quad\text{or}\quad x_\alpha+q_n-1=x_\beta+q_m-1.$$
In any case we have $x_\beta-x_\alpha\in\mathbb{Q}\Rightarrow x_\beta\sim x_\alpha\Rightarrow x_\beta \in E_\alpha\Rightarrow x_\beta=x_\alpha.$
Therefore $q_n=q_m$, in fact we have
$\bullet$ If $x_\alpha+q_n<1$ and $x_\alpha=x_\beta$ we have $$q_n-q_m=-1\;\text{(absurd)}\quad\text{or}\quad q_n-q_m=0$$
$\bullet\bullet$ If $x_\alpha+q_n\ge1$ and $x_\alpha=x_\beta$ we have $$q_n-q_m=-1\;\text{(absurd)}\quad\text{or}\quad q_n-q_m=0.$$
From the fact that $q_n=q_m$ we have that $n=m$.
Step $4$ Suppose that $V\in\mathcal{L}(\mathbb{R})$, then $V_n\in\mathcal{L}(\mathbb{R})$ for all $n\in\mathbb{N}$ and we have
\begin{split} 1=\lambda([0,1))=\sum_{n=1}^{\infty}\lambda(V_n)=\sum_{n=1}^{\infty}\lambda(V\dot{+}q_n)=\sum_{n=1}^{\infty}\lambda(V)=\begin{cases} 0 & \text{if $\lambda(V)=0$} \\ \infty & \text{if $\lambda(V)>0$} \end{cases} \end{split}
Absurd. Therefore $V\notin\mathcal{L}(\mathbb{R})$
Question Is there any inaccuracy? Could something be omitted? Unnecessary steps?
Thanks!