Prove that f(x)=kx

Observe that since f(x+x)=f(x)+f(x)=2f(x)

for any x, and any integer you have $f(nx) = nf(x)$

now pick a rational number p/q q non zero. we have that
$<br /> f(\frac{p}{q} x)=p f(\frac{1}{q} x)$ then multiply by q/q and use the first result

$=\frac{p}{q} f(\frac{q}{q} x) = \frac{p}{q} f(x)$

So we have proved that $f(ax)=af(x)$ for every rational number a.

In the next the continuity is crucial, let x be any real number, and $(a_n)_n$ sequence of rationals converging to x .

so $f(x) = \lim_n f(a_n) = \lim_n a_nf(1) =xf(1)$ since f is continuous,

hence $f(x)=Kx$ where K=f(1)

Edit: The above works only where f is continous so by Lusin's theorem f(x)=Kx on F where m(F) is small.
Im not sure how to prove that f is linear everywhere... let me think