Let $g:\mathbb{R}\to\mathbb{R}$ be a measurable function such that $g(x+y) =g(x)+g(y).$ Then $g(x) = g(1)x$ . Let $g:\mathbb{R}\to\mathbb{R}$ be a measurable function such that 
  $$g(x+y) =g(x)+g(y).$$
  How to prove that $g(x) = cx$ for some $c\in \mathbb{R}?$




The main thing to do here relies upon the fact that such function should be continuous and therefore by natural argument the answer will follow.

Using this
Additivity + Measurability $\implies$ Continuity 

Therefore I found out that there is nothing missing in this question.
For $x_1,....,x_n \in \mathbb R$ we have, by induction, $g(x_1+...+x_n)=g(x_1)+...+g(x_n)$

Let $c=g(1)$.

Then prove in the following order:


$g(n)=cn$   $\quad$for all $ n \in \mathbb N$,
$g(1/n)=c\frac{1}{n}$ $\quad$ for all $ n \in \mathbb N$,
$g(\frac{m}{n})=c\frac{m}{n}$ $\quad$ for all  $n,m \in \mathbb N$,
$g(r)=cr$  $\quad$ for all $r \in \mathbb Q$.


Now use that $f$  is measurable to show $g(x)=cx$ $\quad$ for all $x \in \mathbb Q$.
For $x_1,....,x_n \in \mathbb R$ we have, by induction, $g(x_1+...+x_n)=g(x_1)+...+g(x_n)$

Let $c=g(1)$.

Then prove in the following order:


$g(n)=cn$   $\quad$for all $ n \in \mathbb N$,
$g(1/n)=c\frac{1}{n}$ $\quad$ for all $ n \in \mathbb N$,
$g(\frac{m}{n})=c\frac{m}{n}$ $\quad$ for all  $n,m \in \mathbb N$,
$g(r)=cr$  $\quad$ for all $r \in \mathbb Q$.


Now use that $f$  is measurable to show $g(x)=cx$ $\quad$ for all $x \in \mathbb Q$.