Let $C$ be a (Fat) Cantor set in $[0,1]$ with positive measure $a\in (0,1)$ and the set $A=\{(x,y) \in [0,1] \times [0,1]: x-y \in C\} \subseteq \Bbb{R}^2$. Prove that $m_2(A)>0$ where $m_2$ is the Lebesgue measure on the plane.
Here is my attempt: Lets denote $m$ the Lebesgue measure on the line.
$A$ is a compact subset of the plane thus Borel measurable.
Now By Fubini we have that: $m_2(A)=\int_{\Bbb{R}}m(A^y)dm(y)$
and we can observe that $$A^y:=\{x \in \Bbb{R} : (x,y) \in A\}=[0,1] \cap (C+y)$$ for every $y \in [0,1]$ for $x,y \notin [0,1]$ we have $A^y=\emptyset$
Also from the fact that $(E \cap F)-z=(E-z) \cap (F-z)$ we see that $$m(A^y)=m(A^y-y)=m([0,1-y] \cap C)$$
For $y \in (0,a/2)$ we have that $1-y>1-a/2$ thus $[0,1-a/2] \subseteq [0,1-y]$ and also $m(C \cap [0,1-a/2])>0$ because if not,then we would have that $$a=m(C)=m(C \cap [1-a/2,1]) \leq m([1-a/2,1])=a/2<a$$ which is a contradiction.
Now $$m_2(A) \geq \int_{[0,1]}m(A^y-y)dm(y)$$ $$\geq \int_{[0,a/2)}m(A^y-y)dm(y)$$ $$ =\int_{[0,a/2]}m([0,1-y] \cap C)dm(y) \geq \frac{a}{2}m(C \cap [0,1-a/2])>0$$
Does this solution have any mistakes? If not,is there another solution less complicated than this?
I have checked your solution and think that it is correct and that it also presents a good level of detail for each step.
A possible alternative approach (not sure if this is actually simpler in the end) would be to start with making a variable substitution $u:=x, v:=x-y$. (Since the question is only about $m_2(A)>0$ we can ignore the constants that arise from the determinant. Also this constant is $1$ if I am not mistaken.) Then the new domain could is maybe a little bit simpler. However, I do not see a need to replace your solution.