# non-continuous function satisfies $f(x+y)=f(x)+f(y)$

As mentioned in this link, it shows that For any $f$ on the real line $\mathbb{R}^1$, $f(x+y)=f(x)+f(y)$ implies $f$ continuous $\Leftrightarrow$ $f$ measurable.

But

how to show there exists such an non-measurable function satisfying $f(x+y)=f(x)+f(y)$?

I guess we may use the uniform bounded principal and the fact that $f$ is continuous iff it is continuous at zero under the above assumption.

Considering $\mathbb R$ as infinite-dimensional $\mathbb Q$ vector space, any linear map will do. For example, one can extend the function $$f(x)=42a+666b\quad \text{ if } x=a+b\sqrt 2\text{ with }a,b\in \mathbb Q$$ defined on $\mathbb Q[\sqrt 2]$ to all of $\mathbb R$, if one extends the $\mathbb Q$-linearly independent set $\{1,\sqrt 2\}$ to a basis of $\mathbb R$. (This requires the Axiom of Choice, of course)