Naive Question About the set $\{x: 2^n < |f(x)|<2^{n+1}\}$

Question

Consider a function $f \in L^1$ and the set $$ A_n := \{x \in \mathbb{R}: 2^n < |f(x)| < 2^{n+1} \} $$ where $n \in \mathbb{Z}$. Is $A_n$ disjoint?

My thought: I know that the $f$ lies between $2^n$ and $2^{n+1}$ which does not overlap for all $n$, but the elements $x \in A_n$ may be overlapped (? this is the point I do not so sure). I believe I simply missing some understanding here, any comment is appreciated.

Thank you.

Answer 1

They are disjoint. Here's a quick proof by contradiction: Suppose $x \in A_n$ and $A_m$ for $m \neq n$. Then $|f(x)| \in (2^n, 2^{n+1})$ and $|f(x)| \in (2^m, 2^{m+1})$, which is impossible since those two intervals are disjoint.

Answer 2

Hint: Suppose $x \in A_n$ and $x \in A_m$. Then, what's the value of $f(x)$? Can you see the contradiction?