1
$\begingroup$

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 $$\lim_{n\rightarrow\infty}|f_n-f|=0$$

$\endgroup$
  • $\begingroup$ @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
  • 1
    $\begingroup$ 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
5
$\begingroup$

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

$$\Bbb{1}_{[0,1/2]},\Bbb{1}_{[1/2,1]},\Bbb{1}_{[0,1/3]},...$$

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.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ @89085731 You can't keep changing the question. What is this for? What do you want? $\endgroup$ – user223391 Sep 27 '15 at 0:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.