# Does $\lim_{n\rightarrow\infty}\int_X|f_n-f|d\mu=0$ imply $\lim_{n\rightarrow\infty}|f_n-f|=0$?

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

• @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) – Clement C. Sep 26 '15 at 22:29
• 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 $f$. – 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:

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

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.

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