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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.

Guy Fsone
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  • This is called Cauchy's functional equation. For proofs, see https://en.wikipedia.org/wiki/Cauchy's_functional_equation and http://math.stackexchange.com/questions/359183/measurable-cauchy-function-is-continuous – Dominik Dec 21 '16 at 12:21
  • @Dominik "Cauchy's functional equation" is not a good name for the additivity equation, because it's not the only functional equation that Cauchy studied. If I remember right he also considered such functional equations as $f(xy)=f(x)f(y)$ and $f(xy)=f(x)+f(y)$ and $f(x+y)=f(x)f(y).$ – bof Dec 21 '16 at 13:14
  • @bof Sure Cauchy also studied other equations, but only the additive one is usually called Cauchy functional equation. – Dominik Dec 21 '16 at 13:17
  • people say this question miss something that is why they closed it . but the answer to this actually exists . by simply proving that such function are continuous. sad – Guy Fsone Oct 22 '17 at 16:45
  • The closure for missing context may or may not have been appropriate, but it's a duplicate of several linked questions, so why nominate for reopen? –  Oct 23 '17 at 18:49
  • Is that a new way to say duplicate?(duplicate of several linked question) :::)))I discover that link yesterday. – Guy Fsone Oct 23 '17 at 18:53
  • This answer is excellent, in my opinion. –  Oct 23 '17 at 19:00

1 Answers1

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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:

  1. $g(n)=cn$ $\quad$for all $ n \in \mathbb N$,

  2. $g(1/n)=c\frac{1}{n}$ $\quad$ for all $ n \in \mathbb N$,

  3. $g(\frac{m}{n})=c\frac{m}{n}$ $\quad$ for all $n,m \in \mathbb N$,

  4. $g(r)=cr$ $\quad$ for all $r \in \mathbb Q$.

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

Fred
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