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Compact set of positive measure

  1. May 26, 2008 #1
    Let [tex]K\subseteq\mathbb R[/tex] is a compact set of positive Lebesque measure. Prove that the set [tex]K+K=\{a+b\,|\,a,b\in K\}[/tex] has nonempty interior.
     
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  3. May 26, 2008 #2

    morphism

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    What have you tried so far? Do you know of any links between the Lebesgue measure and the topology of R?
     
  4. May 27, 2008 #3
    All open and all closed sets are measurable. For any measurable set [tex]E[/tex] and positive real [tex]\varepsilon[/tex] there are an closed set [tex]F\subseteq E[/tex] and an open set [tex]G\supseteq E[/tex] such that measure of [tex]G\setminus F[/tex] is less than [tex]\varepsilon[/tex].
     
  5. May 27, 2008 #4
    The original problem is formulated for any measurable set [tex]E[/tex]. Using the fact that any measurable set of positive measure has a compact subset of positive measure, the general case is reduced to special case of compact sets.
     
  6. May 28, 2008 #5

    morphism

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    I thought I had an elementary solution, but I discovered some holes in my reasoning. Anyway, a quick and dirty way to solve the problem is to convolve the characteristic function of K with itself. You will get a nonnegative continuous function f whose integral is m(K)^2 > 0. Thus f must be positive on some open interval - but f is positive precisely on K+K.
     
    Last edited: May 28, 2008
  7. May 29, 2008 #6
    Are you sure that convolution is continuous? Why? Do you have a link? You state that the convolution is positive exactly on K+K, but K+K is compact and continuous function is positive on the open set.
     
  8. May 29, 2008 #7

    morphism

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    It's a theorem that if [itex]f \in L^1(\mathbb{R})[/itex] and [itex]g \in L^\infty(\mathbb{R})[/itex] then [itex]f*g[/itex] is continuous. (In fact, the same conclusion holds for [itex]f \in L^p(\mathbb{R})[/itex] and [itex]g \in L^q(\mathbb{R})[/itex], where p and q are convex conjugates.) You can find this in almost any text on harmonic analysis, e.g. it's on page 4 of Rudin's Fourier Analysis on Groups.

    The point of my post was the following: Since f is positive precisely on K+K (this statement actually requires a line of proof) and since f is positive on an open interval, then K+K must contain an open interval.
     
  9. May 29, 2008 #8

    gel

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    that's very neat.
    One thing though, f is positive on a subset of K+K (rather than precisely on K+K), which is exactly what was needed - and the condition that K is compact wasn't needed, only finite measure.

    ...and to show that convolution f*g is continuous is easy when g is continuous. The general case follows by taking limits.
     
    Last edited: May 29, 2008
  10. Jun 5, 2008 #9
    I proved thjat [tex]f[/tex] is continuous an that the open set [tex]U=\{x\in\mathbb{R}\,:\,f(x)>0\}[/tex] is the subset of [tex]K+K[/tex]. But, how to prove that the set [tex]U[/tex] is nonempty? In fact, [tex]f(x)=m(K\cap(x-K))[/tex].
     
  11. Jun 5, 2008 #10

    gel

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    Note that if {f>0} was empty then the integral of f would be zero. Can you say what the integral of f is?
     
  12. Jun 6, 2008 #11
    OK, application of Fubini's theorem gives [tex]\int_{\mathbb{R}}f\,dm=m(K)^2>0[/tex]. Thanks.
     
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