$f$ 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 $(0,\infty)$.
 A: 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.
A: Measurability is actually not needed here for the conclusion to hold. Note that the hypothesis implies $f(2x)=f(x)$ and then by an easy induction $f(kx)=f(x)$ for $k>0$ integer.
Let $g(x)=f(x)-f(1)$, then:
$g(x+y)-(\lambda g(x)+(1-\lambda)g(y))=f(x+y)-f(1)-(\lambda f(x)+(1-\lambda)f(y))+f(1)=0$, so $g$ satisfies same relation but $g(1)=0$. Then $g(k)=0$ for $k>0$ integer and since $g(kx)=g(x)$, we get $g(\frac{p}{q})=g(q\frac{p}{q})=0$ for any $r=\frac{p}{q}>0$ rational, so $g(x+r)=\lambda g(x)$ for any $x>0, r>0$, $r$ rational. 
Since then $g(x+2r)=\lambda g(x)$ as $2r$ is rational too, but also $g(x+2r)=\lambda g(x+r)=\lambda^2 g(x)$, we get that $\lambda g(x)=\lambda^2 g(x)$, hence $g(x)=0$ for arbitrary positive $x$, hence $f(x)=f(1)$. Done!
