Question 10 page 43 measure theory for Cohn Prove that if $A$ is Lebesgue measurable set then $\{(x,y):x\in A \}$ is Lebesgue measurable subset of $\mathbb{R}^2$. I tried to prove by contradiction but I did not get the result.

Thanks.
if $m(A)=0$ then $m(A \times R)=0$ also

if $m(A)$ is finite then $m(A \times R) := \infty $

if $m(A) : = \infty$ so $m(A \times R)$ because $m(R) :=\infty$.
if $m(A)=0$ then $m(A \times R)=0$ also

if $m(A)$ is finite then $m(A \times R) := \infty $

if $m(A) : = \infty$ so $m(A \times R)$ because $m(R) :=\infty$.