I am trying to get examples of sequences that converge in $L^1$ but not almost everywhere and vice versa. I know solutions to this are available on this website and elsewhere, but I think I have particularly simple examples that I did not see anywhere, and am wondering if maybe I am missing something. So here goes:

$L^1$ but not a.e.

Let $X_n \sim$ Bernoulli($1/n$) be independent and $X \equiv 0$. Then, $\mathbb{E}(X_n - X) = 1/n \to 0$ as $n \to \infty$. But by Borel-Cantelli $\limsup_{n \to \infty}X_n = 1$ almost surely. Here I saw most sources give the example of the typewriter sequence.

a.e. but not $L^1$

Let $\Omega = \mathbb{R}$ and $\lambda$ the Lebesgue measure. Let $f_n = 1_{[n, n +1]}$ and $f \equiv 0$. Then, $f_n \to f$ as $n \to \infty$ almost everywhere. But,

$$ \int |f_n - f|d\lambda = \int 1_{[n, n +1]} d\lambda = \lambda ([n, n+1]) = 1, $$ which obviously does not tend to $0$.

Are my solutions correct?

  • $\begingroup$ These are both pretty famous examples, actually. They're both correct, once you include independence in the first one. (Borel-Cantelli in that direction requires independence.) $\endgroup$
    – Ian
    Sep 24, 2022 at 23:11
  • 1
    $\begingroup$ It is correct except that you forgot to mention the crucial hypothesis of independence in the first part. Borel-Cantelli requires independence. $\endgroup$ Sep 24, 2022 at 23:11
  • $\begingroup$ @geetha290krm Right, fixed it. Thanks! $\endgroup$ Sep 24, 2022 at 23:16


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