Q/ Let $\{f_n\}$ be a sequence of measurable functions that converge to $f$ almost everywhere. Does $f_n$ converge in $L_1$? Justify your answer. How about the converse? (i.e if $\{f_n\}$ converges to $f$ in $L_1$, does it converge to $f$ a.e?) Either give a proof or counterexample).

For both cases if the corresponding result is not true, how about if we consider a subsequence?

A/ For the first part I am pretty sure its false for both the sequence and any subsequences by taking the counterexample;

Let our measure space be $(\mathbb{R},\lambda)$, $\;\;f_k=\frac{1}{k}$ for $x\in [0,k]$ and $0$ otherwise and let $f=0$

For the converse however I am not so sure, here is what I wrote as a proof;

Suppose $||f_n-f||_1=\int\,|f_n(x)-f(x)|\;d\mu \rightarrow 0$. This implies $|f_n(x)-f(x)|\rightarrow 0$ a.e, i.e $f_n\rightarrow f$ a.e

Does this even make sense? I don't really understand how the integral could tend to zero without whats inside tending to zero. I'm struggling in general to understand all the different forms of convergence and how they fit together, in the question do they mean pointwise convergence?


Regarding the converse of the second, the stuff in the integral doesn't all have to go to zero at the same time. So, imagine that you're converging to 0 again. You could have the first function be the indicator of $[0, 1)$, and the next two functions be $\chi_{[0, 1/2)}$ and $\chi_{[1/2, 1)}$. The next four functions would be the indicators of the four quarters that you can break $[0, 1)$ up into. The eight functions after that, the indicators for the eighths of the unit interval, and so on in that fashion. Clearly, these converge in $L_1$.

Now consider the sequence $\{ f_n(x) \}.$ For any $k$, there is some index $2^k \leq i < 2^{k+1}$ where $f_n(x) = 1$, so you clearly aren't getting pointwise convergence anywhere to $0$. As for the subsequence question, its harder, but I think you should put in some time working with it by yourself.

And yes, convergence a.e. is pointwise convergence almost everywhere.


Pointwsise convergence almost everywhere doesn't imply convergence in $L^{1}$ to the same $f$. Unless the sequence is dominated by an integrable function. See here.

And $L^{1}$ convergence doesn't imply convergence pointwise almost everywhere. However you can extract a subsequence that converge. As is donehere.

  • $\begingroup$ A counter example for what? Is my counterexample for the first part incorrect? $\endgroup$ – Someguy Mar 31 '15 at 12:46
  • $\begingroup$ In my opinion, this answer should be moved to the comments, since it only contains links. $\endgroup$ – Siminore Mar 31 '15 at 12:46
  • $\begingroup$ @Tom Hi, I couldn't follow what you were trying to prove. Are you saying that your sequence $ \{f_{k}\}$ converges to $0$ pointwise but doesn't converge in $L^{1}$? $\endgroup$ – yess Mar 31 '15 at 13:00
  • $\begingroup$ @yess yes that is what my first example is trying to show, the sequence I gave converges to 0 pointwise but the integral of $|f_k-0|$ over the whole space is 1 for every k $\endgroup$ – Someguy Mar 31 '15 at 13:05
  • $\begingroup$ @Tom I agree with your proof. $\endgroup$ – yess Mar 31 '15 at 13:12

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