# Compact set of positive measure

• Nedeljko
In summary, the Lebesque measure of a compact set is greater than the measure of its interior. Convolution of the characteristic function of a compact set with itself is a nonnegative continuous function that is positive on an open interval - this is enough to show that the set U=\{x\in\mathbb{R}\,:\,f(x)>0\} is nonempty.f

#### Nedeljko

Let $$K\subseteq\mathbb R$$ is a compact set of positive Lebesque measure. Prove that the set $$K+K=\{a+b\,|\,a,b\in K\}$$ has nonempty interior.

What have you tried so far? Do you know of any links between the Lebesgue measure and the topology of R?

All open and all closed sets are measurable. For any measurable set $$E$$ and positive real $$\varepsilon$$ there are an closed set $$F\subseteq E$$ and an open set $$G\supseteq E$$ such that measure of $$G\setminus F$$ is less than $$\varepsilon$$.

The original problem is formulated for any measurable set $$E$$. 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.

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.

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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.

It's a theorem that if $f \in L^1(\mathbb{R})$ and $g \in L^\infty(\mathbb{R})$ then $f*g$ is continuous. (In fact, the same conclusion holds for $f \in L^p(\mathbb{R})$ and $g \in L^q(\mathbb{R})$, 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.

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.

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I proved thjat $$f$$ is continuous an that the open set $$U=\{x\in\mathbb{R}\,:\,f(x)>0\}$$ is the subset of $$K+K$$. But, how to prove that the set $$U$$ is nonempty? In fact, $$f(x)=m(K\cap(x-K))$$.

Note that if {f>0} was empty then the integral of f would be zero. Can you say what the integral of f is?

OK, application of Fubini's theorem gives $$\int_{\mathbb{R}}f\,dm=m(K)^2>0$$. Thanks.