I'm having trouble understanding this proof:
Lemma: Let $\lambda$ be the Lebesgue measure restricted to the Borel sets of $[0,1]$. Then there exists a decreasing sequence $(E_n)_{n\ge0}$ of subsets of $[0,1]$ such that each of them is $\lambda$-thick and $\bigcap E_n = \emptyset$.
Proof: Let $E_n = [0,1]\cap\{f\ge n\}$, where $f$ is a noncontinuous solution $f:\mathbb{R} \to \mathbb{R}$ of $f(x+y) = f(x) + f(y)$.
A continuous solution of that functional equation is a line through the origin.
Let $\sim$ be the equivalence relation on $\mathbb{R} \setminus \{0\}$ such that $x \sim y \leftrightarrow \exists q\in\mathbb{Q}$ s.t. $x = qy$.
Let $A$ be a set of representatives from each equivalence class. Let $g:A\longrightarrow\mathbb{R}$ be a non-constant "assignment of slopes". Let $f:\mathbb{R} \setminus \{0\} \to \mathbb{R}$ s.t. $x \mapsto x\cdot g(y)$ where $x \sim y\in A$, let $f(0) = 0$.
To put it roughly, a noncontinuous solution is a stack of lines through the origin.
a) Is this description of a noncontinuous solution correct?
No it's not! As TCL helped me realize, this construction does satisfy the desired functional equation only for $x$ and $y$ in the same equivalence class. I need to require something more from g.
b) Why does the proof work? I fail to see why are the $E_n$ measurable, or why are they $\lambda$-thick.
They don't need to be measurable! In fact, the concept of thickness is introduced to get around measurability issues.
Let's construct $f$ this way: let $H$ be an Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$, and $a\in H$. Let $g(x) = x$ for $x\in H \setminus \{a\}$ and $g(a) = 0$. Let's extend this function to $f:\mathbb{R} \to \mathbb{R}$ by linearity using the Hamel basis.
Using this $f$ we have that $E_0 = [0,1]$, $E_1 = \{1\}$, which clearly isn't $\lambda$-thick. Where's the flaw?
The flaw is that $E_1 \ne \{1\}$ because... Hamel bases don't work that way :) See the comments for details.
I give up. I will ask my professor about this.