How to construct a sequence of functions that are defined and continuous on $[0,1]$ and it converges to zero a.e. but on any interval it does not converge uniformly?
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what about $x^n$ ? – Theorem May 29 '13 at 09:50
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2See example 7.4, page 77, in Counterexamples in Analysis, Gelbaum and Olmsted. There, a sequence of continuous functions is constructed that converge to the function $f(x)=\cases{1/q,&$x=p/q$ in lowest terms, $p$ and $q$ integers, $q>0$\cr 0,&$x$ irrational}$ – David Mitra May 29 '13 at 09:58
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4$x^n\to 0$ uniformly on $[0,a]$ for $a<1$ – Norbert May 29 '13 at 09:58
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@David: with a couple more lines that could easily been an answer. – Willie Wong May 29 '13 at 10:04
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@DavidMitra:I've found it, thanks – jintok May 29 '13 at 10:06
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See example 7.4, page 77, in Counterexamples in Analysis, Gelbaum and Olmsted (the nice diagram on page 79 gives the essential idea of the construction).
There, a sequence of continuous functions is constructed that converges to the function $f(x)=\cases{1/q,&$x=p/q$ in lowest terms, $p$ and $q$ integers with $q>0$\cr 0,&$x$ irrational}$.
That the convergence is not uniform on any interval follows from the fact that $f$ is discontinuous at every rational $x$, and the fact that a uniform limit of continuous functions is continuous.
David Mitra
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