If $f$ is measurable then it is constant.

Question

Let $f:(0,\infty)\to \Bbb R$ be such that $2f(x+y)=f(x)+f(y) \forall x,y\in(0,\infty)$. Prove that if $f$ is measurable then it is constant.

Answer 1

Suppose $f : (0,+\infty) \to \mathbb R$ satisfies $$f(x) + f(y) = 2f(x+y)\tag{*}$$ for all $x,y > 0$.

We claim $f(2a) = f(a)$ for all positive $a$. Indeed, \begin{align} f(a)+f(a) &= 2 f(a+a),\quad \text{by (*)} \\ 2f(a) &= 2 f(2a) \\ f(a) &= f(2a). \end{align}

Now we have $$ f(x)+f(y) = 2f(x+y) = 2f\left(\frac{x+y}{2}\right) \\ \frac{f(x)+f(y)}{2} =f\left(\frac{x+y}{2}\right) $$ It is a result of Sierpinski that any Lebesgue measurable midpoint convex function is convex. Applying this to $f$ and $-f$ we conclude $f$ is affine on $(0,+\infty)$. But $f(1) = f(2)$, so $f$ is constant.