Convergence Almost Surely does not imply convergence of moments

Last updated Nov 1, 2022

Let $(\Omega, \mathcal{M}, \mathbb{P})$ be a Probability Space. There exists a sequence of Random Variables $(X_n){n=1}^{\infty}$ on $\Omega$ s.t. $X{n} \to X$ but $\mathbb{E}|X_{n}| \not\to \mathbb{E}|X|$.

We construct the example. Let $([0,1], \mathcal{B}([0,1]), \lambda)$ be our probability space. Suppose $X_{n}$ is defined as $$X_{n} = 0*1_{x \leq 1 - \frac{1}{n}} + 7n1_{x > 1 - \frac{1}{n}}$$

Then $X_{n} \to 0$ almost surely. To see this, observe that for $s \in [0, 1)$, there exists $N \in \mathbb{N}$ s.t. $1 - \frac{1}{N} > s$ (by the Archimedean property of the Natural Numbers). For all $n \geq N$, we have $\frac{1}{n} \leq \frac{1}{N}$ so $$X_{n}(s) = 0*1_{x \leq 1 - \frac{1}{n}}(s) + 7n1_{x > 1 - \frac{1}{n}}(s) = 0$$

Thus $\mathbb{P}[X_{n} \to 0] = \lambda([0, 1)) = 1 \checkmark$ .

On the other hand, observe that

$$\mathbb{E}|X_{n}| = 0*\lambda([0, 1 - \frac{1}{n}]) + 7n \lambda\left(\left(1-\frac{1}{n}, 1\right)\right) = 0 + \frac{7n}{n} = 7$$

While

$$\mathbb{E}|X| = 0$$

Since $7 \not\to 0$ we have that the Moments do not converge. $\blacksquare$

1. In this example, all random variables are non-negative, so the construction also shows that almost sure convergence does not imply expectation converges.
2. Any form of convergence implied by almost sure (Convergence in Distribution, Convergence in Probability) will also have this counterexample.
3. This is an example where Fatou’s Lemma is strict.
4. This example also works as a general example for generic Measure Spaces.

1. Adapted from Does Convergence in Distribution Imply Convergence of Expectation

• Lebesgue Measure
• Real Numbers
• Closed Interval