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I am working a problem on this book which asks to prove or disprove that if $f_n \rightarrow f$ uniformly on $E_1, E_2, E_3, \dots,$ then $f_n \rightarrow f$ uniformly on $\cup_{n=1}^\infty E_n$.

Two ideas come to mind: 1) If $E_n$ is decreasing, then surely the above statement holds. 2) If a finite union was put in place of the countable union, i.e. $\cup_{n=1}^k E_n$ then the proof for uniform convergence would be straightforward seeing that we can take the maximum of a set with a finite number of elements.

The second idea gives me the intuition that the statement above is not necessarily true.

However I am finding it difficult to find counterexamples. I know that $f_n(x)=x^n$ is not uniformly convergent on $[0,1]$ but that it is uniformly convergent on $[0,\sigma]$ where $0\leq\sigma<1$. I cannot think of sets whose countable union is $[0,1]$, unfortunately. Any hints on how to go about constructing one, if possible?

Anyone has thoughts about how I should proceed?

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Consider $f_n(x):=x^n$ as you did and $E_k:=[0,1-k^{-1}]$.

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  • $\begingroup$ Yes I considered that but isn't it that $\cup_{n=1}^\infty E_n = [0,1)$ in that case? $\endgroup$ – Tomas Jorovic Aug 22 '13 at 14:48
  • $\begingroup$ Still the supremum $f_n$ is constant one. Sorry for my other answer. I misreas that question $\endgroup$ – Marc Palm Aug 22 '13 at 14:51
  • $\begingroup$ Oh I see what you mean. Indeed there's a difference between $[0,\sigma], \sigma<1$ and $[0,1)$. thank you!!! $\endgroup$ – Tomas Jorovic Aug 22 '13 at 14:53
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Your idea with $f_n=x^n$ is right. Use these as your sets : $E_0 = \left\{1\right\}$ and $E_k=\left[0,1-k^{-1}\right]$.

Just as you said, $f_n\rightarrow0$ on all $E_k$, for $k\geq1$, and on $E_0$, we have that $f_n(1)=1 \,\,\forall n$. Furthermore, the union of all $E_k, k\geq0$ is the unit interval, on which $f_n$ does not converge uniformly.

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Another idea is that power series are only locally uniformly convergent. Think of the $e$-function. The approximating sequence $e_n(x):=\sum_{j=0}^n x^n/j!$ converges on any bounded set uniformly. Hence, taking choosing the sets $E_l=(-\infty,l)$ and the using the previous considerations you find a counter example, since the series does not converge uniformly on $\mathbb{R}$

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