$f$ is measurable. Prove that $f$ has to be constant on $(0,∞)$. Let $f : (0,\infty) \to \mathbb{R}$ be measurable and $0 < \lambda < 1$. Suppose that
$$f(x + y) = \lambda f(x) + (1 − \lambda)f(y)$$
holds for any $x, y > 0$. Prove that $f$ has to be constant on $(0,\infty)$.

Lemma. For $n\in\mathbb N$ and $x\in\mathbb R$, we have  $f(nx)=f(x)$. In particular, $f(n)=f(1)$.
Proof by induction on $n$. For $n=1$ it is clear. If the statement is true for some $n\in\mathbb N$, then $$f((n+1)\cdot x)=\lambda \underbrace{f(nx)}_{=f(x)}+(1-\lambda)f(x)=f(x). \square$$
Hence, we have by the Lemma for all $x\in\mathbb R$ and $n\in\mathbb N$: 
\begin{gather}\tag 1\label 1 f(x+n)=\lambda f(x)+(1-\lambda)f(1)=f(x+2n)\\ f(x+2n)=\lambda f(x+n)+(1-\lambda) f(1)\tag 2 \label 2\end{gather}
and thus by dividing by $\lambda\neq 0$, $$f(x)=f(x+n).$$ From the Lemma and the condition on $f$ we have $$f(x+n)=\lambda f(x) + (1-\lambda) f(1).$$
Thus $$(1-\lambda) f(x)=(1-\lambda) f(1)$$ from which we get (since $\lambda\neq1$) that $f(x)=f(1)$ for all $x$ i.e. that $f$ is constant.
Note that the measurability of $f$ was not needed.
