Measurabilty question

Question

Consider following problem:
Let $f:S^1 \rightarrow S^1$ be a rotation. Show that a subset $M \subset S^1$ is measurable if and only $f(M)$ is measurable.

Answer 1

Let us recapitulate here

Consider the unit circle $S^1 \subset \mathbb C$ as a submanifold of $\mathbb R ^2.$ Choose an atlas for $S^1$. Choose a partition of unity $\{p_i\}_{i \in I}$ that is subordinate to the domains of the charts of that atlas (i.e. $\operatorname{supp}(p_i) \subset V_i$ for a domain $V_i$). Choose an embedding $\gamma: \Omega \rightarrow \mathbb R^2$ with an open $\Omega \subset \mathbb R$. We call a subset $M \subset S^1$ measurable if for all $i \in I$ the function $(\chi_M \cdot p_i )\circ \gamma \cdot \sqrt{g^\gamma}: \mathbb R \rightarrow \mathbb R$ is measurable with respect to the Lebesgue measure on $\mathbb R$. Here $\chi_M$ denotes the characteristic function of $M$ and $g^\gamma$ is the Gram determinant of $d \gamma$. (One can show that this definition is independent of the choices made.)

We want to prove that

Let $f:S^1 \rightarrow S^1$ be a rotation. Show that a subset $M \subset S^1$ is measurable if and only $f(M)$ is measurable.

Proof: We can write $(\chi_M \cdot p_i )\circ \gamma \cdot \sqrt{g^\gamma} = \chi_{\gamma^{-1}(M)}\cdot (p_i \circ \gamma)\cdot \sqrt{g^\gamma}$. The Gram determinant is non-zero because $\gamma$ is an embedding. Because of that and because the product of measurable functions is measurable, $(\chi_M \cdot p_i )\circ \gamma \cdot \sqrt{g^\gamma}$ is measurable if and only if $\chi_{\gamma^{-1}(M)}\cdot (p_i \circ \gamma)$ is measurable.

So we have:

1. $M$ is measurable if and only if, for all $i \in I$, $\chi_{\gamma^{-1}(M)}\cdot (p_i \circ \gamma)$ is measurable (which independs of the athas chosen, the $\{p_i\}_{i \in I}$ partition of unity that is subordinate to the domains of the charts of the atlas and the embedding $\gamma: \Omega \rightarrow \mathbb R^2$).

In the same way,

2. $f(M)$ is measurable if and only if, for all $i \in I$, $\chi_{\gamma^{-1}(f(M))}\cdot (p_i \circ \gamma)$ is measurable, (which independs of the athas chosen, the $\{p_i\}_{i \in I}$ partition of unity that is subordinate to the domains of the charts of the atlas and the embedding $\gamma: \Omega \rightarrow \mathbb R^2$).

It is easy to see that $f^{-1}\circ \gamma$ is an embedding. So by item 1, we have that $M$ is measurable if and only if, for all $i \in I$, $\chi_{\gamma^{-1}(f(M))}\cdot (p_i \circ f^{-1}\circ \gamma)$ is measurable. But $\{p_i \circ f^{-1}\}_{i \in I}$ is a partition of unity that is subordinate to the domains of the charts of the atlas rotated by $f$. Let, for all $i\in I$, $p_i'= p_i \circ f^{-1}$. Then, using item 2, we have that $M$ is measurable if and only if, for all $i \in I$, $\chi_{\gamma^{-1}(f(M))}\cdot (p_i'\circ \gamma)$ is measurable if and only if $f(M)$ is measurable. So, $M$ is measurable if and only if $f(M)$ is measurable.