Share
Explore BrainMass

Lebesgue Measurable Set

Let A be a set in R^n, we denote by A + x_o a parallel shift of A by x_o to A + x_o,
A + x_o = { x : x = y + x_o, y in A}.

Now, if A is a lebesgue measurable then show that

1). x_o + A is also lebesgue measurable

2). m(A) = m(x_o + A)

Can someone check my answer and tell me if it is correct or not?

My work:
since A is a subset of R^n then A^c ( A compliment) = R^n - A and therefore A^c union A = R^n
and (A + x_o ) union (A^c + x_o) = R^n
Since A+x_o in R^n and (A+x_o)^c = A^c + x_o
( to show (A + x_o)^c = A^c + x_o :
let z be element of (A+x_o)^c then z doesn't belong to A + x_o, since x_o is a point then z doesn't belong to x_o, and z doesn't belong to A so z belongs to A^c and therefore z belongs to A^c + x_o
To do the other way, it is almost the same just reverse the procedure)

Nowwe get the following
E = ( E intersection A) union ( E intersection A^c) , where E is a subset of R^n
and E = (E intersection ( A + x_o) ) union( E intersection ( A^c + x_o))
so
(E intersection A) union ( E intersection A^c) = ( E intersection ( A + x_o)) union
(E intersection ( A^c + x_o))

so A+x_o is lebesgue measurable

now for part 2
m(E intercetion A) union ( E intersection A^c)) =< m((E intersection ( A + x_o)) union( E intersection (A^c + x_o))

and

m( E intersection ( A+x_o) union( E intersection (A^c+x_o)) =< m ((E intersection A) union( E intersection A^c))

so m( (E intersection ( A + x_o) ) union ( E intersection ( A^c + x_o ) ) =
m(( E intersection A) union( E intersection A^c)

so m(E) = m ( (E intersection ( A + x_o) ) union ( E intersection ( A^c + x_o))
= m( ( E intersection A) union ( E intersection A^c))

Solution Summary

Lebesgue Measurable Sets are investigated. The solution is detailed and well presented.

$2.19