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Is it true that if f continuous and goes to $0$ at $\infty$ then f is Lebesgue function?

I don't know how to prove it.

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No. Consider $f(x)=\cases{1,&$ -1$\cr 1/|x|, &\text{otherwise} }$.

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    +1, I was about to write down exactly the same example ;) - but reason for my comment: are there some similar obvious examples for $f\in C^1, C^2...$ or does the thesis actually hold, if $f$ is often enough differentiable?2012-12-08
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    @Fant You could "smooth out" the function above at $x=\pm 1$ so that it becomes infinitely differentiable. Or you can start with the function $$f(x)=\cases{ 1-\text{exp}\bigl[ -{1\over x^2}\text{exp}(-{1\over(1-x)^2})\bigr], &$0 This is infinitely differentiable, with derivative $0$ outside the interval $(0,1)$. Suitably scaling this function will give you an infinitely differentiable function with any desired constant value on $(-\infty,n)$ and any other desired smaller constant value on $[n+1,\infty)$.2012-12-08
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    @Fant You can "paste" a bunch of these together to obtain a counterexample that is infinitely differentiable.2012-12-08
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    "smoothin" out was my idea, but didn't see directly how. 2nd example looks nice, i will check this .. ty2012-12-08
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    But probably, someone more clever than myself can come up with a simpler counterexample...2012-12-08