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Measurable sets

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data
    If [itex]A[/itex] is [itex]\lambda ^* [/itex]-measurable and [itex]x\in \mathbb{R} ^n[/itex]
    then [itex]x+A[/itex] is [itex]\lambda ^* [/itex]-measurable.

    My attempt at the proof is below, but i feel like it is not a correct proof.

    2. Relevant equations
    [itex]\lambda ^* [/itex] is the lebesgue outer measure

    3. The attempt at a solution
    let [itex]A[/itex] be a [itex]\lambda ^* [/itex]-measurable set, and let [itex]x\in \mathbb{R} ^n[/itex] and let [itex]S[/itex] be the entire space.
    Then [itex]\forall T\subset S[/itex], [itex]\lambda ^* (T) = \lambda ^* (T\cap A)+\lambda ^* (T\cap A^c )[/itex]
    Lesbesgue outer measure is translation invariant,
    so, [itex]\lambda ^* (T-x) = \lambda ^* ((T-x)\cap A) + \lambda ^* ((T-x)\cap A^c)[/itex]
    =\lambda ^* (T\cap (A+x)) + \lambda ^* (T\cap (A^c +x))
    = \lambda ^* (T\cap (A+x)) + \lambda ^* (T\cap (A+x)^c)
    = \lambda ^* (T)[/itex]

    so, [itex]x+A[/itex] is [itex]\lambda ^* [/itex]-measurable
  2. jcsd
  3. Oct 1, 2011 #2
    Using Caratheodory's Criterion is the correct way of doing this, but all you've concluded is what you knew to begin with: the outer measure is translation invariant.

    You know that A is measurable right? And to show that x+A is measurable you must show that
    [itex] \lambda^*(T \cap (x+A)) + \lambda^*(T \cap (x+A)^c ) = \lambda^*(T) [/itex]
    for every set [itex] T \subseteq \mathbb R^n [/itex]. So try starting with [itex] \lambda^*(T \cap (x+A)) + \lambda^*(T \cap (x+A)^c ) [/itex] and see if you can get the [itex] \lambda^*(T) [/itex]

    Edit: Your proof has the right ideas, but not the right set-up.
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