Suppose $(X,M,\mu)$ is a measurable space,$g_n,g,f_n,f\in L^1(X,R)$,if $|g_n(x)|\leq f_n(x)$ and $g_n\rightarrow g$ pointwise,if $$\lim_{n\rightarrow\infty}\int_X|f_n-f|d\mu=0,$$does that imply $$\lim_{n\rightarrow\infty}|f_n-f|=0$$

  • $\begingroup$ @avid19 I deleted my comment -- trying to make it more precise made it clear (for instance, as it is the OP doesn't specify the last limit it pointwise for $x$). What are the counterexamples for a.e. convergence? (I'm forgetting quite a lot) $\endgroup$ – Clement C. Sep 26 '15 at 22:29
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    $\begingroup$ The best you can do in general (without more knowledge about the $f_n$) is that a subsequence of the $f_n$ converges pointwise almost everywhere to $f$. $\endgroup$ – Cameron Williams Sep 26 '15 at 23:56

No. Let $X=[0,1]$ with Borel sets and standard Lebesgue measure. Consider the sequence of indicator functions:


This is a traveling bump and clearly $\int_{[0,1]} f_n(x) dx\to 0$, but each point goes back and forth between $0$ and $1$. So this converges NOWHERE but converges in $L^1$.

As a side note, $(X, M, \mu)$ is a measure space. A measurable space is without the actual measure.

Edit I don't understand your edit. Just let $g_n(x)=0$. $g_n(x)$ won't help anything.

  • $\begingroup$ would that have a.e. convergence? $\endgroup$ – 89085731 Sep 26 '15 at 23:51
  • $\begingroup$ @89085731 As I explicitly said, it converges nowhere. $\endgroup$ – user223391 Sep 26 '15 at 23:52
  • $\begingroup$ what if $|g_n(x)|<f_n(x)$ and $g_n\rightarrow g$,does that imply $f_n\rightarrow f$ $\endgroup$ – 89085731 Sep 27 '15 at 0:03
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    $\begingroup$ @89085731 Then that is blatantly false. Think about it for 10 seconds and I'm sure you'll find an example. $\endgroup$ – user223391 Sep 27 '15 at 0:09
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    $\begingroup$ @89085731 You can't keep changing the question. What is this for? What do you want? $\endgroup$ – user223391 Sep 27 '15 at 0:20

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