Measurable functions with arbitrarily small periods are constant

Question

Let $f:\mathbb{R}\to \mathbb{R}$ be a measurable function that has arbitrarily small periods, that is, for any $\varepsilon>0$ there is $0<T<\varepsilon$ such that $f(x+T)=f(x)$ for all $x\in\mathbb{R}$. I need to show that $f$ is constant, at least almost everywhere.

Can someone find an elementary proof of this ?

One proof: I found a proof in which I used distribution theory. First replace $f$ by $f_A:= f\mathbf{1}_{\vert f\vert\leq A}$ for $A>0$ to get a bounded function. Therefore, $f$ belongs to $L_{loc}^1(\mathbb{R})$ and can be seen as an element of $\mathcal{D}'(\mathbb{R})$.

Take $T$ a period of $f$ and denote by $\tau_Tf:= f(\cdot-T)$ the translation operator by $T$. Take $\varphi$ as a test function, and compute \begin{aligned} \left<f, \varphi \right> &=\left<\tau_T f, \varphi \right> \\ &= \left< f,\tau_{-T} \varphi \right> \end{aligned} Thus, for $T$ arbitrarily small $$\left<f,\tau_T \varphi - \varphi \right> = 0.$$ Then, one uses the fact that if $\underset{n\to\infty}{\lim}T_n =0$, then $$\underset{T_n\to0}{\lim} \frac{\tau_{T_n}\varphi - \varphi}{T_n} = \varphi'\quad \text{in}\ \mathcal{D(\mathbb{R})}.$$
Passing to the limit in the above inequality one gets that $$\left<f', \varphi \right> = 0 $$ for any test function $\varphi$. Thus $$f' = 0\quad\text{in}\ \mathcal{D'}(\mathbb{R}).$$ By a classical result, this implies that $f$ is constant almost everywhere.

Motivation: In a paper called "On the approximation of Lebesgue integrals by Riemann sums", Jessen proves the following theorem. Let $f$ be a $L^1(\mathbf{T})$ function. Denote its Riemann sum $$f_n:= \frac{1}{n}\sum_{k=1}^n f(x+\frac{k}{n}).$$ Then $$\underset{n\to\infty}{\lim} f_{2^n}(x) = \int_0^1 f\quad a.e.x $$ Jessen considers $$\phi(x):=\overline{\lim} f_{2^n}(x)$$ which is a $2^{-n}$ periodic function for all $n$, hence an almost everywhere constant function. The proof reduces to show that this constant is $\int_0^1 f$.

Answer 1

If $f$ is periodic with period $p$, then $\int_x^{x+p} f(t) \, dt = \int_y^{y+p} f(t) \, dt$, for all $x$ and $y$.

By the Fundamental Theorem of Calculus for Lebesgue integration (see Royden 4th edition, Chapter 6, Theorem 14), $\lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t) \, dt = f(x)$ almost everywhere. Let $A$ be a set of full measure witnessing this.

Let $p_n$ be a sequence of periods converging to 0. For $x,y \in A$, we have

$$f(x) = \lim_{n \to \infty} \frac{1}{p_n}\int_x^{x+p_n} f(t) \, dt = \lim_{n \to \infty} \frac{1}{p_n}\int_y^{y+p_n} f(t) \, dt = f(y).$$

Answer 2

I think your question is equivalent to :

Let $f : \mathbb R \to \mathbb R$.$ f $ is measurable.

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.

You get a similar result for $ x \in (-\infty, 0) $, when $ a <0$

If I'm right then your answer is here and you don't need distribution theory.