Let $A$ and $B$ be subsets of the unit square $[0,1]^{2}$ such that:
$A\cap B = \emptyset$
$A\cup B = [0,1]^{2}$.
$A$ and $B$ are both dense in $[0,1]^{2}$.
Suppose there exists a continuous function $f:[0,1]^{2}\to[0,1]$ such that:
$f(A)\cap f(B) = \emptyset$
$f(A)\cup f(B) = [0,1]$.
$f(A)$ and $f(B)$ are both dense in $[0,1]$.
$f(A)$ and $f(B)$ are both Borel and so Lebesgue measurable.
$\lambda_{1}(f(A)) = 1$, $\lambda_{1}(f(B)) = 0$ (where $\lambda_{1}$ is the Lebesgue measure on $\mathbb{R}$.
Can we say anything about $\lambda_{2}(A)$ and $\lambda_{2}(B)$ (where $\lambda_{2}$ is the Lebesgue measure on $\mathbb{R}^{2}$)? This makes sense to talk about as $f$ is continuous and so the preimages $A$ and $B$ are Borel.
This arises from a homework question that was more specific about what $A$ and $B$ were. The answer was that $\lambda_{2}(A) = 1$ and $\lambda_{2}(B) = 0$. Is it true in this more general situation though?