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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

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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.

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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.

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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.

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

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