# L^1 and almost everywhere convergence

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?

• 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.)
– Ian
Sep 24, 2022 at 23:11
• It is correct except that you forgot to mention the crucial hypothesis of independence in the first part. Borel-Cantelli requires independence. Sep 24, 2022 at 23:11
• @geetha290krm Right, fixed it. Thanks! Sep 24, 2022 at 23:16