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Let $C$ be the Cantor set as a compact Abelian topological group, isomorphic to countable product of $\mathbb{Z}/2\mathbb{Z}$.

Its normalized Haar measure is denoted by $\mu$.

Is there a positive continuous map $f:C \to \mathbb{Q}$ which is not a locally constant map but satisfy $\int_{C} f^{2}d\mu=1 $?

Is there a positive continuous map $f:C \to \mathbb{Q}$ which is not a locally constant map but satisfy $\int_{C} f^{3}d\mu=1 $?

In the other word does $\int_{C} f^{2}d\mu=1 $ or $\int_{C} f^{3}d\mu=1 $ imply that $f$ is locally constant?

This question can be considered as a generalization of the equation $\sum_{i=1}^{n} x_{i}^{2}=1$ or $\sum_{i=1}^{n} x_{i}^{3}=1$ on $\mathbb{Q}^{n}$.

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No. For example, let $S_n=\{x\in C\colon x_1=x_2=\ldots=x_{n-1}=0; x_n=1\}$. Now inductively choose a sequence of rationals $q_n$ such that $\sum_{n=1}^N q_n^2\mu(S_n)\in (1-2\cdot 3^{-N},1-3^{-N})$. You can check you have $q_n=\Theta(2^n/3^n)$, so that 1) $f$ takes countably many values; 2) $f(x)\to 0$ as $x\to 0$, whence $f$ is continuous.

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    $\begingroup$ The natural question follows: can we do a function which is locally constant around no point? $\endgroup$
    – Gro-Tsen
    Commented Feb 4, 2016 at 20:44
  • $\begingroup$ Clearly the exact same argument will produce a function that is locally constant around no point if instead of defining $f$ to be constant on the cylinder sets, one defines $f$ to be a locally constant multiple of $1+\sum_{n=1}^\infty x_n 2^{-n}$. $\endgroup$
    – Anthony Quas
    Commented Feb 4, 2016 at 21:45
  • $\begingroup$ @AnthonyQuas thank you for your answer. what is f precisely? $\endgroup$
    – Ali Taghavi
    Commented Feb 6, 2016 at 11:28
  • $\begingroup$ So here is a concrete example. Take $f$ to be $2^{-n}$ on $S_n$ for $n>0$ and $f=4/3$ on $S_0$. With a little more imagination, you could build $f$ that is $2^{-n}$ on each $S_n$ with $n>1$ simultaneously satisfying your 2 conditions. My argument shows that there are lots of these functions - the construction is very flexible. $\endgroup$
    – Anthony Quas
    Commented Feb 6, 2016 at 16:07

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