Let $\beta\in\mathbb{R}$. Prove that for any set $A\subset\mathbb{R}^n$ we have $\lambda_n(A)=\lambda_n(A+\beta)$; in other words, if we translate a set the translated set will have the same Lebesgue measure as the initial set.
Attempt: By definition of Lebesgue measure, we have \begin{equation} \lambda_n(A):=\inf\left\{\sum_i(b_i-a_i):A\subset\bigcup_i[a_i,b_i)\right\}. \end{equation}
Note that $A\subset\bigcup_i[a_i,b_i)$ is equivalent to $A+\beta\subset\bigcup_i[a_i+\beta,b_i+\beta)$, and using our definition for Lebesgue measure above we have (using the fact that $(a_i+\beta)-(b_i+\beta)=(a_i-b_i)$)
\begin{equation} \lambda_n(A+\beta):=\inf\left\{\sum_i(b_i-a_i):A+\beta\subset\bigcup_i[a_i+\beta,b_i+\beta)\right\}. \end{equation}
But I cannot see how to demonstrate the equivalence to $\lambda_n(A)$ using just the definition.
Well you can prove that $\lambda_n (A) \leq \lambda_n(A+\beta)$, since on the right hand side you are taking an infimum over a potentially smaller set of sums of countable collections of covering sets. The same reasoning shows that $\lambda_n (A+\beta) \leq \lambda_n((A+\beta) - \beta)$.