# Is continuity necessary to establish $f(x+r) =f(x) +f(r), r\in\Bbb Q \Rightarrow f(x+y) =f(x) +f(y)$

Let $$f:\mathbb{R} \to \mathbb{R}$$ be a function such that $$f(x+r) =f(x) +f(r)$$,$$\forall x\in \mathbb{R}$$ and $$\forall r \in \mathbb{Q}$$. I know that if $$f$$ were continuous, then we would have $$f(x+y) =f(x) +f(y)$$, $$\forall x, y \in \mathbb{R}$$ using the fact that the rational numbers are dense in the reals. Could the same thing be established without $$f$$'s continuity? I think it could not, but I can't find a counterexample.

• Let $f(x)=x$ if $x$ is irrational, $f(x)=0$ if $x$ is rational. – Wojowu Apr 21 at 10:06
• @Wojowu Exactly the same example I was thinking of...+1 – DonAntonio Apr 21 at 10:07
• This is true for measurable additive functions. – Dbchatto67 Apr 21 at 10:08
• @Wojowu thank you, beautiful counterexample! – Math Guy Apr 21 at 10:10
• @Wojowu Why are you answering in a comment? – Arthur Apr 21 at 10:10

No, continuity of $$f$$ is necessary here. Consider the following function: $$f(x)=\begin{cases}0 & \text{if x is rational,}\\ 1 & \text{if x is irrational.}\end{cases}$$ Clearly for any rational $$r$$, we have $$x$$ rational iff $$x+r$$ is rational, so $$f(x+r)=f(x)=f(x)+0=f(x+r)$$, but $$f(\sqrt{2}+\sqrt{2})=1\neq 2=f(\sqrt{2})+f(\sqrt{2})$$.

The value of $$f$$ on $$\mathbb Q$$ can be uniquely determined by the value of $$f(1)$$. That's because we can easily see that $$f(n)=nf(1);\\ f\left(\frac{1}{q}\right)=\frac{1}{q}f(1)$$ for $$n,q\in \mathbb Z^+$$.

However, the value of $$f(1)$$ cannot determine anything further, because the two identities above are as far as the condition $$f(x+r)=f(x)+f(r)$$ can go. For example, $$f(\sqrt 2)$$ cannot be determined from $$f(1)$$ because $$\sqrt 2$$ is irrational.

We can define $$f(\sqrt 2)$$. Then $$f(a+\sqrt{2})$$ where $$a,b\in \mathbb Q$$ is also defined.

So far $$f(\sqrt 3)$$ is still undefined. We can take $$f(\sqrt 3)$$ to be any value. Then $$f(a+\sqrt{3})$$ is defined, but not $$f(2\sqrt{3})=f(\sqrt{3}+\sqrt{3})$$, since one of the terms inside the bracket must be rational. You can continue to define as you like.

Generally, you can divide $$\mathbb R$$ into many equivalence classes. The equivalence relation is defined as follows: $$r_1 \sim r_2$$ iff $$r_2-r_1\in\mathbb Q$$.

Let's choose one representative from each equvilance class. Let those representatives be $$\{r_\lambda\}, \lambda\in\Lambda$$. For all possible set of real numbers $$\{b_\lambda\}$$, you can define $$f(r_\lambda)=b_\lambda$$. Define $$f(r)$$ for rational $$r$$ as above, and for any $$x\in \mathbb R$$, there exist $$\lambda\in\Lambda$$ such that $$x=r_\lambda+a, a\in \mathbb Q$$. We define $$f(x)=b_\lambda+f(a)$$.

Now we can easily show that $$f$$ is the function you require. Since you can simply take any values on each equivalence class, the function is very likely to be not additive. for example, if you let $$f(\sqrt2)=1,f(\sqrt 3)=1,f(\sqrt 2+\sqrt 3)=1$$, then $$f(\sqrt2)+f(\sqrt3)\neq f(\sqrt 2+\sqrt 3)$$