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I'm thinking about the basic types of convergence for sequences of functions: convergence in measure, almost uniform convergence, convergence in Lp and point wise almost everywhere convergence. I'm looking for examples of sequences of functions that converge in one or more of these ways, but fail for others. I keep seeing the same examples over and over and I'd like to think about some new ones. Here are the examples I've seen:

$f_n=\chi_{[n, n+1]}$

$f_n=\chi_{A_{n}}$ where $A_1 = [0,1]$, $A_2 = [0,1/2]$, $A_3 = [1/2,1]$, $A_4 = [0,1/4]$, $A_5 = [1/4,1/2]$, $A_6 = [1/2,3/4]$, $A_7 = [3/4,1]$, $A_8 = [0,1/8]$ ...

$f_n= n\chi_{[1/n,2/n]}$

These are a great set of examples since they let you give a counterexample for the relations between the types of convergence when needed, but I would like to know of some more.

(EDIT: I added a bit to the 2nd example to make it more clear.)

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The sequence $(f_n)$ given by $f_n=n^{-1}\chi_{[-n,n]}$ converges uniformly, but not in $L^1$.

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  • $\begingroup$ Thanks. That one works like the 3rd one, but it is not on a finite measure space. $\endgroup$
    – S. Donovan
    Dec 30 '09 at 22:53
  • $\begingroup$ It's not like the third one, which only converges almost uniformly, not uniformly. $\endgroup$ Jul 1 '11 at 9:11
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When it comes to complex analytic functions on open subsets of $\mathbb{C}$, it is hard to come up with examples of pointwise convergent sequences that do not converge uniformly on compact sets. That is partly because it doesn't take much for a family of analytic functions to be normal. There is much more to be said about the matter, and for an exposition including examples I recommend this survey by Davidson.

[Edit: This post would be better if it included a description of such an example. I may add one when I have more time.]

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On $[0,1]$: $ f_n = a_n\chi_{[\alpha n,\alpha n + \varepsilon n^{-2}]\ {\rm mod}\ 1 } $ with $\alpha$ irrational, and $a_n = 1 $ or $ a_n = n^2 $.

This is, of course, also similar ...

Transferred from my comments below, and corrected (TeX was not shown, so I did not see that some of the code did not work):

Well, it was rather late (in my timezone). So I only typed in the $f_n$.

$ \int |f_n -0| $ is either $ = \varepsilon n^{-2} \to 0 $ or $ = \varepsilon \to \varepsilon $, respectively.

$ \int \bigcup_{n\ge N} \lbrace x | \chi_{[\alpha n,\alpha n + \varepsilon n^{-2}]\ {\rm mod}\ 1 } \ne 0 \rbrace \le \sum_{n\ge N}\varepsilon n^{-2} \to 0 $, i.e., $ f_n \to 0 $ almost everywhere.

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  • $\begingroup$ What is this an example of? $\endgroup$ Dec 31 '09 at 2:15
  • $\begingroup$ It's like the 2nd one when a_n = n^2 ? $\endgroup$
    – S. Donovan
    Dec 31 '09 at 3:12
  • $\begingroup$ It's not like the 2nd one, as it doesn't converge pointwise. $\endgroup$ Dec 31 '09 at 3:34
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    $\begingroup$ But the 2nd one won't converge pointwise. $\endgroup$
    – S. Donovan
    Dec 31 '09 at 5:04
  • $\begingroup$ Well, it was rather late (in my timezone). So I only typed in the $f_n$. $ \int f_n -0 $ is $ = \varepsilon n^{-2} $ or $ = \varepsilon $, respectively. $ \int \bigcup_{n\ge N} \{ x | \chi_{[\alpha n,\alpha n + \varepsilon n^{-2}]\ {\rm mod}\ 1 } \not= 0 \} \le 1 - \sum{n\ge N}\varepsilon n^{-2}$ \to 1 $ $\endgroup$ Dec 31 '09 at 10:43

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