Rotation send measure zero set to measure zero

Question

$m_2$ is the Lebesgue measure on $\mathbb R^2$. Suppose U is a zero set. $T:\mathbb R^2 \to \mathbb R^2$ is a rotation. Show $T(U)$ is a zero set. What I tried is to show the outer measure defined by open rectangle cover is equivalent to the outer measure defined by closed ball cover and closed ball is invariant under rotation, but it seems rather hard to show the equivalence.

Answer 1

Here is a long winded answer that doesn't use the usual $\det$ formula.

This shows that $m(TA) = m(A)$ for all measurable sets $A$.

Let $R_n = \{ [x, x+{1 \over 2^n}) \times [y, y+{1 \over 2^n}) | x,y \in {1 \over 2^n} \mathbb{Z} \}$. It should be clear that $R_n$ forms a partition of $\mathbb{R}^2$. It should also be clear that if $m>n$ and $S_n \in R_n, S_m \in R_m$, then either $S_m \subset S_n$ or $S_m \cap S_n = \emptyset$.

We see that $[0,1)^2$ can be written as the disjoint union of $(2^n)^2$ elements of $R_n$, each of which are translations of each other.

Define $\tau(A) = m(TA)$. It is easy to check that $\tau$ is a translation invariant measure since $m$ is.

Let $c=\tau([0,1)^2)$. From above, we see that if $S \in R_n$, then $\tau(S) = {1 \over (2^n)^2 } c = c m(S)$.

Since $T$ is invertible, we see that $U$ is open iff $TU$ is open.

Suppose $U$ is open. We want to show that $U$ can be written as a countable disjoint collection of elements of the $R_n$. Let ${\cal V}_1 = \{ S \in R_1 | S \subset U \}$, $W_1= \cup_{S \in {\cal V}_1} S$, ${\cal V}_{n+1} = \{ S \in R_{n+1} | S \subset U \setminus W_n\}$, $W_n = \cup_{S \in {\cal V}_1 \cup \cdots \cup {\cal V_n}} S$. Let ${\cal V}$ be the countable pairwise disjoint collection ${\cal V} = \cup_n {\cal V_n}$. We see that $U = \cup_{S \in {\cal V}} S$.

Then we have $m (TU) = \tau(U) = \sum_{S \in {\cal V}} \tau(S) = c \sum_{S \in {\cal V}} m(S) = c m(U)$. (Note that this implies $m(U) = c m(T^{-1} U)$.)

For any measurable set $A$, we have $m(A) = \inf \{ m(U) | U \text{ open}, A \subset U \}$, so there exists open $U_k$ such that $A \subset U_k$ and $m(U_k) \to m(A)$. Then $TA \subset T U_k$ and so $m(TA) \le c m(A)$. Similarly, suppose $TA \subset V_k$ and $m(V_k) \to m(TA)$, then $A \subset T^{-1} V_k$ and so $m(A) \le {1 \over c} m(TA)$.

Hence $m(TA) = c m(A)$ for all measurable $A$.

Finally, since $T$ is a rotation, we have $m(T B(0,1)) = m(B(0,1))$, so $c=1$.