I have the following sets: $$ \begin{align} A = \{(x,y)\in \mathbb{R}^{2}|x\geq0, y\geq 0, |x|^{3}+|y|^{3} < 1 \}, \\ B = \{(x,y)\in \mathbb{R}^{2}|x\leq0, y\geq 0, |x|^{3}+|y|^{3} < 1 \} \end{align} $$
and I have just shown that they are both Borel sets on $\mathbb{R}^{2}$ and I would now like to answer the following question:
"Show that $\lambda^{2}(A)$ = $\lambda^{2}(B)$ where $\lambda^{2}$ denotes the Lebesgue measure on $(\mathbb{R}^{2}),\mathscr{B}(\mathbb{R}^{2}))$. (Hint: You do not need to calculate $\lambda^{2}(A)$ or $\lambda^{2}(B)$)."
So how can I show that these measures are in fact the same without calculating them? The only difference between these two sets are that A can be sketched in the 1.quadrant and B in the 2.quadrant so there is a symmetry. Can I use this somehow and the fact that the Lebesgue measure is invariant under translation?
Any input?
The Lebesgue measure is not only invariant under translations, but also invariant under isometries, see here for example.
Note that this is also true for reflections and rotations (in fact, reflections and rotations are isometries).
There is no translation between $A$ and $B$, but there is an isometry. Can you find it?