Using Lebesgue measure test $f_{n}(x)=\chi_{[n-1,n]}$ for pointwise convergence, a.e convergence, convergence in measure and $L^{1}$ convergence

Is this correct?

1) Pointwise convergence

$\displaystyle\lim_{n\to\infty}\chi_{[n-1,n]}=0$ given $x\in\mathbb{R}$ and $n$ large $f_{n}(x)=0$ so $f_{n}\to 0$ pointwise.

2) Folland (Real analysis): pointwise convergence implies almost everywhere convergence, but it turns out that this sequence does not converge a.e... why?

3) If we fix $0<\varepsilon<1$ then $\mu(\{x:|f_{n}(x)|\ge \varepsilon\})=1$, and there is no convergence in measure.

4) No convergence in $L^{1}$ either as

$\displaystyle\int_{\mathbb{R}}\chi_{[n-1,n]}dm=1\not\to 0$

  • 1
    $\begingroup$ Re 2, why do you think it does not converge a.e.? See this math.stackexchange.com/a/225632/4583 $\endgroup$ – Ayman Hourieh Oct 17 '13 at 16:42
  • $\begingroup$ The sequence does converge almost everywhere. Everywhere indeed. Almost everywhere convergence does not imply convergence in measure. It may imply convergence in measure if the measure of the entire space is finite, I'm not sure whether it does, but here the space has infinite measure. $\endgroup$ – Daniel Fischer Oct 17 '13 at 16:44
  • $\begingroup$ @AymanHourieh: Well originally I thought it did converge a.e as well but then I saw a solution which said it did not converge a.e. So pointwise does imply convergence a.e right? $\endgroup$ – H.E Oct 17 '13 at 16:46
  • 1
    $\begingroup$ What is the question? $\endgroup$ – copper.hat Oct 17 '13 at 16:49
  • $\begingroup$ is my test for modes of convergence correct? and does pointwise convergence imply a.e convergence? $\endgroup$ – H.E Oct 17 '13 at 16:51

You got the idea. Here are some remarks:

  • Indeed, the sequence converges everywhere, hence in particular almost everywhere.
  • You proved that there cannot be convergence in $\mathbb L^1$/ in measure of the sequence $(f_n)$ to $0$. But we have to justify that since $f_n\to 0$ almost everywhere, the only candidate to be the limit is $0$. For $\mathbb L^1$, this requires the fact that we can extract an almost everywhere convergent subsequence, and also for convergence in measure after having restricted ourselves to finite measure components (here to compact intervals).

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.