Results 1 to 2 of 2

Math Help - Measurability/Dilations

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    75
    Thanks
    1

    Measurability/Dilations

    Hello. I wasn't going to ask this question on here, since I'm sure I've been over-thinking and just need to take a minute to reevaluate my solution attempt, but I need to get this done ASAP.

    Problem Statement: Let \delta = (\delta_{1}, ..., \delta_{d}), \delta_{i} > 0. Prove that if E \subset R^{d} is measurable, then so is \delta E where \delta E =: {(\delta_{1}x_{1}, ..., \delta_{d}x_{d}) : (x_{1}, ..., x_{d}) \in E} and m(\delta E) = \delta_{1}\delta_{2}...\delta_{d}m(E).

    Attempt at solution: The first thing I proved is that the statement is true for open and closed cubes; this is straight forward. Then we assume E \subset R^{d} is measurable, and so \forall \epsilon > 0 \exists O \subset R^{d}  s.t.  E  \subset O  and  m(O-E) < \epsilon.

    Now I'm not sure exactly what to do (in a succinct manner) to prove the result; I've basically scribbled down like 3 pages of scratchwork trying to make everything workout, but I keep getting the wrong inequalities, or I'm not quite sure what I'm doing is legit (very very new to measure theory here).

    Obviously if the above is satisfied, then we also have \delta E \subset \delta O and m(\delta O -\delta E) < \epsilon. This is clear from the definitions involved. So I guess this proves that \delta E is measurable. (Maybe this requires a more rigorous justification?).

    From here, I need to prove that the equality holds, but I'm not really sure how to do this in a rigorous fashion (I've attempted a few time son scratchwork, but I can't get a succinct proof to fall out). I was trying to use the above epsilon-inequality, since I can work with the term m(\delta O) since I have the cube-decomposition lemma for open sets in R^{d} and the proof that the theorem holds for cubes, but I can't work with m(\delta E) in any tangible way (or can I?).

    By the way, it is correct to interpret m(O-E) < \epsilon as m(O) - m(E) < \epsilon?

    Finally, here was my other attempt at the solution, which doesn't really seem rigorous:

    m(\delta E) = \inf m(\delta O) = \delta_{1}...\delta_{d} \inf m(O) = \delta_{1}...\delta{d} m(E). The first \inf is of course with respect to coverings of \delta E per the definition of Lebesgue measure (its agreement with the outer measure), and the second is for coverings of just E.

    Any help would be appreciated,

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2009
    Posts
    75
    Thanks
    1

    Re: Measurability/Dilations

    Thought about this problem a little more carefully, and I believe I have a proper proof.

    Suppose that E \subset R^{d} is a Lebesgue measureable set. Then \forall \epsilon > 0 \exists O_{\epsilon} s.t. \mu (O_{\epsilon} - E) < \epsilon which is to say  \mu (O_{\epsilon}) - \mu (E) < \epsilon and since \epsilon is arbitrary, we must have that E = \inf_{O \supset E} \mu (O). Note that O \subset R^{d} is an open set such that  O \supset E (but becomes equal to E in the limit represented by the infimum!).

    Correspondingly, we have by the above definitions: \mu (\vec{\delta} O_{\epsilon} - \vec{\delta} E) < \epsilon^{\prime} = \epsilon \cdot max(\delta_{1}, ..., \delta_{d}). But the \delta_{max} term is finite, and so \epsilon^{\prime} can \rightarrow 0. It follows \vec{\delta} E is measurable.

    To prove its measure is as we expect, we make the following computation:

    \begin{align*} \mu(\vec{\delta} E) &= \inf_{O \supset \vec{\delta} E} \mu (O) \\ &= \inf_{O \supset E} \mu (\vec{\delta} O) \\ &= \inf_{\{Q_{j}\} \supset E} \mu (\vec{\delta} \bigcup\limits_{j=1}^{\infty} Q_{j}) \\ &= \inf_{\{Q_{j}\} \supset E} \mu (\bigcup\limits_{j=1}^{\infty} \vec{\delta} Q_{j}) \\ &= \inf_{\{Q_{j}\} \supset E} \sum\limits_{j=1}^{\infty} |\vec{\delta} Q_{j}| \\ &= \inf_{\{Q_{j}\} \supset E} \sum\limits_{j=1}^{\infty}  \prod\limits_{i=1}^{d} (\delta_{i} b_{i} - \delta_{i} a_{i}) \\  &= \inf_{\{Q_{j}\} \supset E}  \delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d} \sum\limits_{j=1}^{\infty}|Q_{j}|  \\ &=\delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d} \inf_{\{Q_{j}\}  \supset E}\sum\limits_{j=1}^{\infty} \mu (Q_{j}) \\  &=\delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d} \inf_{O \supset E}  \mu (O) \\ &=\delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d}  \, m^{*}(E) \\ &= \delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d}\, \mu(E). \end{align*}.

    The justifications at various steps follow from the related definitions (dilation sets), hypothesis of Lebesgue measure (i.e. being able to pass to open epsilon-close covers, and thus to the limit represented as an inf), and the open set decomposition theorem (i.e. being able to then pass from open coverings to countable coverings of cubes).

    Please let me know if there are any flaws in my argument, and how they can be fixed! I'm still a little uneasy about the level of rigor in my solution, particularly in passing from the set in question, to open coverings, to countable unions of cubes and the use of the infimum along the way.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Applying dilations and reflection help
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 9th 2010, 04:17 AM
  2. transformations, dilations help
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 11th 2009, 04:45 PM
  3. another dilations question
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 14th 2008, 02:22 AM
  4. simple question about dilations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 12th 2008, 11:44 PM
  5. Dilations Help
    Posted in the Geometry Forum
    Replies: 2
    Last Post: December 18th 2007, 02:40 PM

/mathhelpforum @mathhelpforum