Compact set of positive measure

  • Thread starter Nedeljko
  • Start date
  • #1
40
0

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
morphism
Science Advisor
Homework Helper
2,015
4
What have you tried so far? Do you know of any links between the Lebesgue measure and the topology of R?
 
  • #3
40
0
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].
 
  • #4
40
0
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.
 
  • #5
morphism
Science Advisor
Homework Helper
2,015
4
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:
  • #6
40
0
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.
 
  • #7
morphism
Science Advisor
Homework Helper
2,015
4
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.
 
  • #8
gel
533
5
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:
  • #9
40
0
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].
 
  • #10
gel
533
5
Note that if {f>0} was empty then the integral of f would be zero. Can you say what the integral of f is?
 
  • #11
40
0
OK, application of Fubini's theorem gives [tex]\int_{\mathbb{R}}f\,dm=m(K)^2>0[/tex]. Thanks.
 

Related Threads for: Compact set of positive measure

  • Last Post
Replies
1
Views
5K
Replies
6
Views
4K
Replies
5
Views
2K
Replies
8
Views
5K
  • Last Post
Replies
3
Views
3K
Replies
4
Views
15K
Replies
3
Views
4K
Replies
6
Views
2K
Top