This is an exercise from a past exam I'm using to try to help me study. Let $f$ be measurable and bounded on $[0,1]$ satisfying $$f(x+y)=f(x)+f(y);\quad f(1)=1.$$ I'm trying to show that $f(x)=x$. We're given a hint to show it is continuous by using the hypothesis in a "mildly clever" way, show that it is the identity on the rational points, then extend by continuity.

I am able to show that the function is the identity on $[0,1]\cap\mathbb{Q}$ without any trouble, but I'm not sure how to show it's continuous. From there, showing the function is the identity would be easy since it follows from the fact that the rational numbers are dense in $\mathbb{R}$.

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    $\begingroup$ As far as I know, this problem is called the "Cauchy functional equation". The result of this exercise is a theorem due originally to Banach and Sierpinski. (Not 100% sure). $\endgroup$ – Giuseppe Negro Dec 13 '12 at 3:24

Appealing to Lusin's theorem and inner regularity of the Lebesgue measure, there is a closed set $E$ with $m(E) > 1 - \epsilon$ so that the restriction to $E$ of $f$ is uniformly continuous on $E$. Now, observe that for $\delta$ and $\epsilon$ sufficiently small, we have that $E \cap (E + \delta)$ is nonempty, so uniform continuity of $f$ on $E$ lets us conclude that for $x$ in the intersection $|f(x + \delta) - f(x)| = |f(\delta)|$ can be made arbitrarily small by choosing $\delta$ sufficiently close to zero, which implies continuity after appealing to linearity since for general $x$ we have $f(x + \delta) - f(x) = f(\delta)$.

As an aside this trick with Lusin's theorem pops up a lot with questions like this.

  • $\begingroup$ I understand it; thank you very much! $\endgroup$ – Clayton Dec 13 '12 at 4:28


It suffices to show continuity at $0$ because the function is a homomorphism.

Then because points with rational coordinates are dense in $\mathbb{C}$ (or $\mathbb{R}$ if you're working $f : \mathbb{R} \rightarrow \mathbb{R}$), for any disk $D$ around the origin you know that there is some $q$ so that $$f^{-1}(q + D)$$ has positive measure. Then by the Steinhaus theorem the difference set $$f^{-1}(q +D) - f^{-1}(q +D)$$ contains an open ball around the origin.

  • $\begingroup$ I like the argument very much; the simplicity is nice. I don't quite follow why we know the existence of $q$, though. Could you elaborate just a little bit? It seems perfectly true...just not sure I would be able to frame a rigorous argument. I followed everything after that, though. $\endgroup$ – Clayton Dec 13 '12 at 4:31
  • $\begingroup$ @Clayton Sure thing: there is such a $q$ because $q_n + D$ covers $\mathbb{C}$ by density of rationals. So this means $[0,1] \subseteq \bigcup f^{-1}(q_n + D)$ so at least one must have positive measure. $\endgroup$ – Deven Ware Dec 13 '12 at 4:42
  • $\begingroup$ Ah, I see it. Perfect; thanks. $\endgroup$ – Clayton Dec 13 '12 at 4:44

This is perhaps the simplest example of a phenomenon called automatic continuity: functions between nice enough topological spaces (say, Polish spaces) that preserve some algebraic structure (say, group homomorphisms) and have some minimal amount of regularity (say, Borel measurable) have to be continuous. You can read more about it in this MO question. The cited survey by Christian Rosendal (PDF) is quite nice.


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