If $(X_n)$ is i.i.d. and $ \frac1n\sum\limits_{k=1}^{n} {X_k}\to Y$ almost surely then $X_1$ is integrable (converse of SLLN)

Question

Let $(\Omega,\mathcal F,P)$ be a finite measure space.

Let $X_n:\Omega \rightarrow \mathbb R$ be a sequence of iid r.v's

I need to prove that if: $ n^{-1}\sum _{k=1}^{n} {X_k} $ converges almost surely to $Y$ then all $X_k$ have expectation.

If I understand correctly then $X_k$ has expectations means $X_k$ is in $\mathcal L^1(\Omega)$.

And I know that on finite measure space Converging in expectations is converging in $\mathcal L^1(\Omega)$ and it's stronger than almost sure convergence.

And I know that from linearity of expectation even if one of the sequence is not in $\mathcal L^1(\Omega)$ then $Y$ is not in $\mathcal L^1(\Omega)$.

How do I continue?

Answer 1

The statement is actually the converse of the strong law of large numbers.

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of iid random variables and suppose that the sequence $S_n := \sum_{j=1}^n X_j$ satisfies $n^{-1} S_n \xrightarrow[]{n \to \infty} Y$ almost surely for some random variable $Y$. Then $$\mathbb{E}(|X_1|)<\infty.$$

Proof: Since $$\frac{X_n}{n} = \frac{S_n}{n} - \frac{n-1}{n} \frac{S_{n-1}}{n-1}$$ we find that $X_n/n$ converges to $0$ almost surely; in particular,

$$\mathbb{P} \left( \left| \frac{X_n}{n} \right| \geq 1 \, \, \text{infinitely often} \right)=0.$$

Applying the (converse) Borel-Cantelli lemma, we obtain

$$\sum_{n \geq 1} \mathbb{P}(|X_1| \geq n) = \sum_{n \geq 1} \mathbb{P} \left( \left| \frac{X_n}{n} \right| \geq 1 \right) < \infty.$$

As

$$\mathbb{E}(|X_1|) \leq 1 + \sum_{n \geq 1} \mathbb{P}(|X_1| \geq n)$$

(see e.g. this question for a proof of this inequality), this proves $\mathbb{E}(|X_1|)<\infty$.