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.


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.

Thanks in advance!

  • 1
    $\begingroup$ Someone will probably give a nice self-contained answer, so I'm putting this in a comment. Try googling the three phrases (simultaneously) "Hamel basis" "additive function" "transfinite induction". I found this (see pp. 28-30) on the second page of results (10 results per page), and this wasn't the only result that looked promising. (Yes, I realize the hard part is knowing what to google!) $\endgroup$ Jan 2, 2013 at 22:56
  • $\begingroup$ Apparently I was on autopilot and wasn't thinking when I said to include "transfinite induction", as I see from Hagen von Eitzen's answer! $\endgroup$ Jan 2, 2013 at 23:01
  • $\begingroup$ Thanks Dave, really nice material! $\endgroup$
    – ougao
    Jan 2, 2013 at 23:48

1 Answer 1


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)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.