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Can anyone give an exam of Converge in measure, but not converge point-wise a.e.?

And also for the converse part, professor asks us to prove "pointwise a.e. implies converge in measure", but think about this function:

$$f_n(x)= \chi_{[n,\infty)}$$

It converge to $f(x)=0$ pointwise, but it seems that the difference measure between $f(x)$ and $f_n(x)$ is always infinity.

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    $\begingroup$ I guess there's a premise that the functions are integrable. $\endgroup$ Sep 25, 2013 at 17:43
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    $\begingroup$ Look at Counterexamples in en.wikipedia.org/wiki/Convergence_in_measure. $\endgroup$
    – copper.hat
    Sep 25, 2013 at 18:00
  • $\begingroup$ For the second part, your counter example is fine. $\endgroup$
    – leo
    Nov 2, 2013 at 23:12
  • $\begingroup$ Related $\endgroup$
    – leo
    Nov 2, 2013 at 23:15

2 Answers 2

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For the first part, consider the typewriter sequence (Example 4).

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for the first part :take $f_n$ on $[0,1]$ as $1_{E_{n}}$ and $E_{n}$are like this sequance
$E_{1}= [0,\frac{1}{2}) E_{2}= [\frac{1}{2},1]$
$E_{3}= [0,\frac{1}{3}) E_{4}= [\frac{1}{3},\frac{2}{3}] E_{5}= [\frac{2}{3},1]$
...

and for your other qustion the theorem is true whene $f_{n}$ has finite domain.

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