Let $A\subset(1, \infty)$ satisfy the following property : $x \in A$ iff $x^2 \in A$. What are the possible values of $m(A)$ if $A$ is measurable?

Question

Here I'm assuming $A$ is (Lebesgue) measurable, and $m(A)$ is its usual Lebesgue measure. It didn't take me long until I felt that either $m(A)=0$ or $m(A)=\infty$. I'm not sure if that's indeed the case (and even so, if my proof is correct), so I decided to post it here to double check. My idea was as follows:

Firstly, notice that the set $A=\{2^{2^k} : k \in \mathbb{Z}\}$ satisfies the above property. Since $A$ is countable, we can have $m(A)=0$. However, I claim that, if $m(A)>0$, then in fact $m(A)=\infty$. To see why, notice that by countable additivity the positivity of $m(A)$ implies that $A \cap (n, n+1]$ has positive measure for some $n \in \mathbb{N}$. Fix some $n \in \mathbb{N}$ for which this is true, and let $I:=(n, n+1]$. Since $$n^{2^j} > \left(\frac{n+1}{n} \right)^{2^k}$$ for sufficiently large $j \in \mathbb{N}$, we know that, given any $k \in \mathbb{N}$, we can find $j \in \mathbb{N}$ for which the sets $(n^{2^k}, (n+1)^{2^k}]$ and $(n^{2^{k+j}}, (n+1)^{2^{k+j}}$ are disjoint. Because of the stated property of $A$, there are as many elements of $A$ in $(n^{2^k}, (n+1)^{2^k}]$ as there are in $(n^{2^{k+j}}, (n+1)^{2^{k+j}}$. So the measures of $A$ restricted to these intervals coincide. Following this reasoning, we can find infinitely many intervals of this form, say, $\{I_l\}_{l \in S}$, where $I_l:=(n^{2^l}, (n+1)^{2^l}]$ and $S$ is a countable subset of $\mathbb{N}$. Therefore, we have that $$m(A) \geq m(\bigcup (A \cap I_l))=\sum_{l \in S} m(A \cap I_l)=\sum_{l \in S} m(A \cap I)=\sum_{l=1}^{\infty} m(A \cap I)=\infty$$ since we are adding $m(A \cap I)>0$ infinitely many times. Therefore, any set $A$ with the above property either is null or has infinite measure. $\blacksquare$

Is my proof correct?

Answer 1

The proof by Ramiro works fine, but since I see from the comments that you haven't gotten to integration yet, you can also reason as follows.

The Lebesgue measure can be defined as the infimum of the sum of lengths of open intervals covering a set. Therefore if $0<m(A)<\infty$, then there are intervals $(a_i,b_i)$ with $$\bigcup_i (a_i,b_i)\supseteq A$$ and $$\sum_{i} b_i-a_i<2m(A)\text{.}$$ But then if $x\in A$, $x^2\in A$, which implies $x^2\in (a_i,b_i)$ for some $i$, and therefore $x\in (\sqrt{a_i},\sqrt{b_i})$. We therefore have $$\bigcup_i (\sqrt{a_i},\sqrt{b_i})\supseteq A$$ and so $$m(A)\leq \sum_i \sqrt{b_i}-\sqrt{a_i}\leq \sum_{i} \frac{b_i-a_i}{\sqrt{b_i}+\sqrt{a_i}} < \sum_i\frac{b_i-a_i}{2}<m(A),$$ a contradiction.

Answer 2

Let $A\subset(1, \infty)$ be a measurable set satisfying the following property : $x \in A$ iff $x^2 \in A$. Then $m(A)=0$ or $m(A)=\infty$.

Here is a simple proof.

Proof: Since $x \in A$ iff $x^2 \in A$, note that the functions $x\mapsto 1_A(x)$ and $x\mapsto 1_A(x^2)$ are identical. Then, by the Theorem of Change of Variable, we have $$ m(A)= \int_1^\infty 1_A(y)dy = \int_1^\infty 1_A(x^2)(2x)dx = \int_1^\infty 1_A(x)(2x)dx \geqslant \int_1^\infty 1_A(x)(2)dx = \\ = 2 m(A) $$ So $2 m(A) \leqslant m(A)$. So $m(A)=0$ or $m(A)=\infty$. $\square$