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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.

Thanks in advance!

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    $\begingroup$ 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!) $\endgroup$
    – Dave L. Renfro
    Commented 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! $\endgroup$
    – Dave L. Renfro
    Commented Jan 2, 2013 at 23:01
  • $\begingroup$ Thanks Dave, really nice material! $\endgroup$
    – ougao
    Commented Jan 2, 2013 at 23:48

1 Answer 1

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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)

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