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?
Best Answer
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.