If A is Lebesgue measurable and x is an element of Rn, then the translation of A by x, defined by A + x = {a + x : a ∈ A}, is also Lebesgue measurable and has the same measure as A.

That's very strange; I just had precisely the same question as this in my Measure Theory assignment last week.

 

It's fairly easy. It's fairly easy to show that [imath]\mu^*(A) = \mu^*(A_{+a})[/imath]. Just consider the set of rectangles covering A and translate them by a, then prove that they provide a cover for A_+a.

 

Now if A is measurable, then for every epsilon > 0, there exists an elementary set B such that [imath]\mu(A \triangle B) < \epsilon[/imath] (triangle = symmetric difference of A and B). Consider B_+a and by using the definition you have there, show you get the same answer (i.e. [imath]\mu(A_{+a} \triangle B_{+a}) < \epsilon[/imath] - you need to use some set theoretic identities for this.

 

Hope this helps.