Meital Posted October 31, 2005 Share Posted October 31, 2005 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. Link to comment Share on other sites More sharing options...
Dave Posted October 31, 2005 Share Posted October 31, 2005 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. Link to comment Share on other sites More sharing options...
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