Upper bound "square" of Lebesgue measure of set

Question

Let $\lambda$ be the Lebesgue measure and $A$ a set with $\lambda(A) < \varepsilon$. Consider the set $A^2 = \{a \cdot a \ | \ a \in A\}$. Can we produce an upper bound on $\lambda(A^2)$ in terms of $\varepsilon$?

Answer 1

No. If $A=(n,n+\frac 1 {\sqrt n})$ then $A^{2}=(n^{2},(n+\frac 1 {\sqrt n} )^{2})$. $\lambda (n,n+\frac 1 {\sqrt n}) \to 0$ but $\lambda (n^{2},(n+\frac 1 {\sqrt n})^{2}) \to \infty$.