Almost everywhere (ae) Homogeneous function of degree $0$ equals to a constant for ae $x \in (0,\infty)$ provided $ f $ is measurable?

Is an almost everywhere (ae) Homogeneous function $f$of degree $0$ equals to a constant for almost every $x \in (0,\infty)$ given that $ f $ is measurable?

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 true?

I know it is true if $ f $ is locally integrable see here

I encountered this problem while studying bounded linear operators $ T:L^2 \to L^2$

Solution 1:

Lemma $1$ $\quad A\subset\mathbb{R}$ is measurable and $m(A)>0$. then $m\left(\mathbb{R}-\bigcup_{q\in\mathbb{Q}}q\cdot A\right)=0$.

${ Proof}$ It's sufficient to show $m\left([1/n,n]-\bigcup_{q\in\mathbb{Q}}qA\right)=0$, where $n\in\mathbb{Z}-\{0\}$.

$\forall \alpha<1$, $\exists$ an interval $I$ s.t. $m(I\cap A)>\alpha\cdot m(I)$. Obviously, $[1/n,n]\subset\bigcup_{q\in\mathbb{Q}}qI$. By arranging the intervals properly, we can find finitely many $\{q_k\}_{k=1}^N\subset\mathbb{Q}$ such that the corresponding $\{I_k\}_{k=1}^N:=\{q_kI\}_{k=1}^N$ satisfies

$$[1/n,n]\subset\bigcup_{k=1}^N I_k,\quad \sum_{k=1}^n m(I_k)\leq3n.$$

Then we have \begin{eqnarray*} [1/n,n]-\bigcup_{q\in\mathbb{Q}}q_kA&\subset&[1/n,n]-\bigcup_{k=1}^Nq_kA\\ &\subset&\left\{[1/n,n]-\bigcup_{k=1}^NI_k\right\}\bigcup\left\{\bigcup_{k=1}^N[(q_kA)^c\cap I_k]\right\}\\ &\subset&\left\{\bigcup_{k=1}^N[(q_kA)^c\cap I_k]\right\} \end{eqnarray*}

Thus, $$m\left\{[1/n,n]-\bigcup_{q\in\mathbb{Q}}\{qA\}\right\}\leq m\left\{\bigcup_{k=1}^N[(q_kA)^c\cap I_k]\right\}\leq (1-\alpha)3n.$$

Let $\alpha\to 1$, so we proved the lemma, and directly have lemma 2.

Lemma $2$ $\quad r\in\mathbb{R},$ $m\{f\leq(\geq)r\}>0\Rightarrow f\leq(\geq)r\ a.e.$

$ Proof:$ $f\overset{a.e.}{\leq} r$ on $q\cdot\{f\leq r\}$ because of homogeneity, where $q\in\mathbb{Q}$. Unit of all $q\cdot\{f\leq r\}$ with respect to $q$ covers $\mathbb{R}$ almost everywhere due to lemma 1, so we proved lemma 2.

Let $R_1:=\{r\in\mathbb{R}:m\{f\leq r\}>0\},\ R_2:=\{r\in\mathbb{R}:m\{f\geq r\}>0\}$. $f\overset{a.e.}{\equiv}\inf R_1=\sup R_2$.

Seems we don't need any extra condition here, just measurability.