Why convergence in Lp doesn't imply convergence almost surely?

Question

I've been searching for an hour to find why convergence in Lp doesn't imply almost sure convergence. Can somebody explain why?

EDIT: I've read that Lp is not sufficient nor necessary for almost sure convergence!! I don't know why?

Answer 1

The following example should illustrate why $L^p$ convergence doesn't imply almost sure convergence.

Suppose we have a set of simple functions $\{X_n\}$ defined on $([0,1],\mathcal{B}[(0,1)],\lambda)$—where $\lambda$ is the Lebesgue measure—in this way: $$X_1=1_{\left[0,\frac{1}{2}\right]}, \ X_2=1_{\left[\frac{1}{2},1\right]},\\ X_3=1_{\left[0,\frac{1}{3}\right]}, \ X_4=1_{\left[\frac{1}{3},\frac{2}{3}\right]}, \ X_5=1_{\left[\frac{2}{3},1\right]},$$ etc etc etc. For any $p>0$, $$\mathbb E(|X_1|^p)=\frac{1}{2}, \ E(|X_2|^p)=\frac{1}{2}, \\ E(|X_3|^p)=\frac{1}{3}, \ E(|X_4|^p)=\frac{1}{3},\ ...,\ E(|X_6|^p)=\frac{1}{4},\ ...$$ So $\mathbb E(|X_n|^p) \to 0$ and $X_n \stackrel{L^p}\rightarrow 0$. However, $X_n \not\to 0$ almost surely: for any $\omega \in [0,1]$, $X_n(\omega)=1$ for infinitely many n, and so $X_n \not\to 0$. $\>\square$

Let me know if you need further clarification.