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If E is measurable then so it's each of it's translates

  1. Dec 2, 2007 #1
    I'm working with caratheodory's definition of measurability of sets as given in Royden. I'm trying to prove that given any measurable set E we have that E+y={x+y|where x is contained in E} is also measurable. I'm looking for a hint not the entire solution please.

    I think I've actually done this problem before in my real analysis class and i remember there being a 'special set' (the silver bullet) that relates E with E+y through operations allowed in sigma algebras...i could be wrong

    thanks for any help, again no solutions just hints please.
     
  2. jcsd
  3. Dec 2, 2007 #2
    actually I think it's simpler than I thought. We use the fact that lebesgue outer measure is translation invariant, that is m(A)=m(A+y) for any set A and any real number y. We also use two identities

    (E+y)^c=E^c+y (that is the compliment of the translate is the translate of the compliment) and
    [(A-y) intersect (E)]+y=A intersect (E+y)

    I will not prove these things as it is simple to do so.

    Let A be any subset of reals, put B=A-y, then since E is measurable
    mB=m(B intersect E)+m(B intersect E^c)
    since m is translation invariant
    mB=m([B intersect E]+y)+m([B intersect E^c]+y)
    Now using the first identity above we see that
    mB=m(A intersect (E+y))+m(B intersect (E^c+y))
    using the 2nd identity we have
    mB=m(A intersect (E+y))+m(B intersect (E+y)^c)
    and mB=mA because of trans invariance. QED
     
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