Define $M:=\{(x,y) \in \mathbb R^{2}: x \in \mathbb Q\cap [0,1], y \in [0,1]\}$
Find $\lambda^{2}(M)$
Ideas:
Well, first I should show that $M \in \mathcal{B}^2$
Note that for $x \in \mathbb Q \cap [0,1]$
$\{x\}\times[0,1]\in\mathcal{B}^2$
And so it follows: $(\mathbb Q \cap [0,1])\times[0,1]=\bigcup_{x \in \mathbb Q \cap [0,1]}\{x\}\times[0,1]$ which is $\in \mathcal{B}^2$
Now onto $\lambda^{2}(M)$:
As proven above:
$\lambda^{2}(M)=\lambda^{2}(\bigcup_{x \in \mathbb Q \cap [0,1]}\{x\}\times[0,1])=\sum_{x \in \mathbb Q \cap [0,1]}\lambda^{2}(\{x\}\times[0,1])$
I assume that $\{x\}\times[0,1]$ is a hyperplane but I lack any idea to show that it is a hyperplane. Is $\dim (\{x\}\times[0,1])=1$ or $2$?
Secondly, out of interest's sake, if I have $A \in \mathcal{B}^{2}$, with $\lambda^{2}(A)=c,$ where $c \in \mathbb R_{>0}$
what can I say about $\lambda(A)$ as well as $\lambda^{3}(A)$?
Any guidance is greatly appreciated
I assume $\lambda^{3}(A)=0$ as $A=A\times \{0\}\in \mathcal{B}^3$ is a hyperplane in $\mathbb R^{3}$, is it not?
For the first question, $\{x\}\times[0,1]$ is not a hyperplane, but rather a subset of the hyperplane $\{x\}\times\Bbb R$ in $\Bbb R^2$. The dimension of $\{x\}\times\Bbb R$ is $1$.
Regarding the second part (I'm assuming that $\lambda^n$ denotes the Lebesgue measure on $\Bbb R^n$), if $\lambda^n(A) >0$ then we have $\lambda^m(A)=0$ for all $m>n$ because $A$ would be contained in a hyperplane of $\Bbb R^m$ .
On the other hand, $\lambda^r(A)$ doesn't really make sense if $r<n$ because $A$ is an object in $\Bbb R^n$ which is not contained in $\Bbb R^r$.