Suppose we have a set $A \subset \Bbb{R^2}$ that is Lebesgue measurable.Now define $B$ = {$(x,-y) : (x,y) \in A$). We want to find the Lebesgue Measure of B
It’s seems to me that $B$ has the same Lebesgue Measure than $A$, but I need help to make my “proof” formal enough :
$\lambda(A)^* $ = inf{$\sum_{n=1}^{\infty}v(I_n) : A \subseteq \cup_{n=1}^{\infty}I_n$}, with $I_n$ open. Take precisely the $I_n$’s such that the sum of the areas of the squares is the infimum of the set.
If $I_n = (a,b)$ with $a_1 \leq a \leq a_2$ ; $b_1 \leq b \leq b_2$, if we make $I_n’ = (a,-b)$ with $a_1 \leq a \leq a_2$ ; $-b_2 \leq -b \leq -b_1$ then $\cup_{n=1}^{\infty} I_n’$ is a cover of $B$ (how can I show this?)
Since any square $(c,d)$ in $\Bbb{R^2}$ has the same area that $(c,-d)$, $\lambda (A) = \lambda (B)$.
It’s my idea correct? How can I show that $\cup_{n=1}^{\infty} I_n’$ is a cover of $B$? Is there any other suggerencr to make this more rigorous?Thanks.