Sierpinski's construction of a non-measurable set

Here's Sierpinski's argument: Let $h:[\mathbb{R}]^{\omega} \to \mathbb{R}$ be any injection. Define $f:\mathbb{R} \to \mathbb{R}$ by $f(x) = h(E_x)$ where $E_x$ is the set of all reals which are at a rational distance from $x$. Note that $x - y$ is rational iff $f(x) = f(y)$. Towards a contradiction, suppose $f$ is Lebesgue measurable. Then $g(x) = f(x) - f(-x)$ is also measurable and is nonzero precisely at irrationals. Let $N = \{x: g(x) > 0\}$. Note that for any rational $r$, $g(r - x) = -g(x)$ so $x \in N \iff r - x \notin N$. But this contradicts Lebesgue density theorem.

I think the pages 249-250 are the most relevant source in ariane's pdf. Sierpinski outlines how to go from the cardinality hypothesis to the existence of a non-measurable set, as per Ashutosh's précis, although without specific reference to the Lebesgue density theorem; his proof invokes an argument about symmetry. He notes that the proof does not require Zermelo's axiom (well-ordering principle). Sierpinski proves his result for any function $f(x)$ satisfying the properties

$f(x) = f(y), x - y \in \mathbb{Q}, f(x) \neq f(y), x - y \notin \mathbb{Q}$.

Here is a fuller rendering of Sierpinski's argument from the French:

Suppose $f : [\mathbb{R}]^{\leq \omega} \rightarrow \mathbb{R}$. Let $E_x = \lbrace x+r : r \in \mathbb{Q} \rbrace$. So $E_x = E_y \Leftrightarrow x - y \in \mathbb{Q}$. For $x \in \mathbb{R}$, let $\varphi(x)= f(E_x)$. Note $\varphi(x) = \varphi(y) \Leftrightarrow x - y \in \mathbb{Q}$.

Claim: $\varphi(x)$ is a non-measurable function. In fact, any such function $\phi(x)$ satisfying $\phi(x) = \phi(y) \Leftrightarrow x - y \in \mathbb{Q}$ is non-measurable.

Proof: Suppose $\varphi(x)$ is measurable. Then $\psi(x) = \varphi(x) - \varphi(-x)$ is also measurable, and the set $N = \lbrace x : \psi(x) > 0 \rbrace$ is a measurable set. Let $Q = \lbrace y \in Irr : y \notin N \rbrace$. Note that for all $r \in \mathbb{Q}$ and $x \in Irr$, of the the two numbers $x$ and $2r - x$, one belongs to $N$ and the other to $Q$ (since $\psi(2r - x) = -\psi(x)$ for irrational $x$, while for rational $x, \psi(x) = 0$). So $N$ and $Q$ are symmetric images of each other, when one takes any point with rational abscissa as centre of symmetry.

Now let $(a,b)$ be any interval, and suppose $N \cap (a, b)$ is measurable. Let $(a_1, b_1)$ be an interval with rational endpoints such that $(a_1, b_1) \subseteq (a,b)$. Let $N_1 = N \cap (a_1, b_1), Q_1 = Q \cap (a_1, b_1)$. $N_1$ and $Q_1$ are measurable, being symmetric images and so have the same measure, which is half the length of the interval $(a_1, b_1)$, since the points in $(a_1, b_1)$ belonging neither to $N$ nor $Q$ are countably many.

So one can decompose $(a,b)$ into two sets which have the same measure on each rational interval. It follows easily (without using the axiom of Mr Zermelo) that we have reached a contradiction. So $\varphi$ cannot be measurable.

A reprint of the mentioned paper can be found in: W. Sierpinski, Oeuvres choisies Tome II, PWN--Editions Sci. Pologne, Warszawa, 1975, pp. 208-255