TITLE: Almost everywhere (ae) Homogeneous function of degree $0$ equals to a constant for almost every $x \in (0,\infty)$?
QUESTION [5 upvotes]: Is an almost everywhere (ae) Homogeneous function of degree $0$ equals to a constant for almost every $x \in (0,\infty)$ ?

Let
   $f : \mathbb R \to \mathbb R$.
If $f(ax)=f(x)$ ae for any $a>0$
Then $f(x)=c$ for almost every $x \in (0,\infty)$, where $c$ is a constant.

Is the above even true by simply assuming $f$ is measurable?

REPLY [2 votes]: We can assume $f$ is bounded; if not, we can consider instead $\arctan f.$ The conclusion holds for $f$ iff it holds for $\arctan f.$
By the Lebesgue differentiation theorem there exists a measurable $E\subset (0,\infty)$ such that $(0,\infty)\setminus E$ has measure $0,$ and such that
$$\lim_{h\to 0}\frac{1}{h}\int_x^{x+h} f = f(x)\,\, \text {for all }x\in E.$$
Assume $x,y\in E.$ Let $a=y/x.$ Note that $ax = y\in E.$ Then
$$\frac{1}{h}\int_x^{x+h} f(t)\,dt = \frac{1}{h}\int_x^{x+h} f(at)\,dt = \frac{1}{ah}\int_{ax}^{ax+ah} f(s)\,ds.$$
As $h\to 0,$ left side converges to $f(x)$ and the right side converges to $f(ax) = f(y).$ Thus $f(x)=f(y).$ Therefore $f$ is constant on $E,$ which is the desired conclusion.