Almost Sure Convergence does not imply Lp Convergence

Last updated Nov 6, 2022

Suppose $(\Omega, \mathcal{M}, \mathbb{P})$ is a Probability Space and $(X_{n})$ is a Sequence of Random Variables on $\Omega$. Suppose $X_{n} \to X$ almost surely. Then, it does not follow that $X_{n} \to X$ in $L^{p}$ for any $p \geq 1$.

We prove by constructing a Counterexample. Suppose $(X_{n})$ is a Sequence of Random Variables defined on Probability Space $([0,1], \mathcal{B}([0,1]), \lambda)$ and for $x \in [0,1]$

$$X_{n}(x) = \begin{cases} n^{2} & x \in [0, \frac{1}{n}) \\ 0 & x \in [\frac{1}{n}, 1] \end{cases}$$ Then for $x \in (0, 1]$, if $n \geq \lceil{\frac{1}{x}}\rceil$, we have that $X_{n}(x) = 0$. Thus on $(0, 1]$, $X_{n} \to 0$ pointwise. Since $(0, 1]^{C} = {0}$ is a Null Set, we have that $X_{n} \to 0$ almost surely.

Now suppose $X_{n} \overset{L^{p}}{\to} X$ for some $p \geq 1$. Recall that Lp Spaces are Banach so $X \in L^{p}$. Then, because Norms are Continuous:

$$\mathbb{E}(|X_{n}|^{p})^{\frac{1}{p}} \to \mathbb{E}(|X|^{p})^{\frac{1}{p}}$$ and therefore $$\mathbb{E}(|X_{n}|^{p}) \to \mathbb{E}(|X|^{p})$$ Now observe that, $$\begin{align*} \mathbb{E}(|X_{n}|^{p}) &= \frac{n^{2p}}{n}\\ &=n^{2p-1}\\ &\geq n^{1} &(\text{since }p \geq 1)\\ &\to \infty\\ \end{align*}$$ This contradicts $X_{n} \overset{L^{p}}{\to} X$ and therefore $X_{n}$ does not converge in Lp. $\blacksquare$

• Lp Convergence
• Borel Sigma Algebra
• Lebesgue Measure