Prove that f(x)=kx

f is a Lebesgue measurable function on E,m(E) is finite.f(x+y)=f(x)+f(y). Prove that f(x)=kx.
By Lusin's theorem,I can find a closed set F,f is continuous on F and m(E-F)is very small.I don't know what to do next...

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


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


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


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


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


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


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

hence$\displaystyle f(x)=Kx$ 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