渐近一致可积性与均值收敛等价性的反向证明求助

Hey there, let's work through the direction of the theorem you're missing—proving that if a sequence of random variables converges in $L^1$ (mean convergence), then it's asymptotically uniformly integrable. First, let's make sure we're on the same page with definitions and the exact statement.

Exact Theorem Statement #

Suppose ${X_n}$ is a sequence of random variables on a probability space $(\Omega, \mathcal{F}, P)$, and $X_n \xrightarrow{L^1} X$ (meaning $\mathbb{E}|X_n - X| \to 0$ as $n \to \infty$). Then ${X_n}$ is asymptotically uniformly integrable.

Key Definitions to Recall #

A sequence ${X_n}$ is asymptotically uniformly integrable if and only if two conditions hold:

• 1. Uniform Boundedness: $\sup_n \mathbb{E}|X_n| < \infty$
• 2. Uniform Absolute Continuity: For every $\epsilon > 0$, there exists a $\delta > 0$ such that for all measurable sets $A$ with $P(A) < \delta$, we have $\sup_n \mathbb{E}\left[|X_n| \cdot \mathbb{I}_A\right] < \epsilon$ (where $\mathbb{I}_A$ is the indicator function of set $A$).

Step-by-Step Proof #

Step 1: Prove Uniform Boundedness #

We use the triangle inequality for expectations:
$$\mathbb{E}|X_n| \leq \mathbb{E}|X_n - X| + \mathbb{E}|X|$$
Since $X_n \xrightarrow{L^1} X$, we know $\mathbb{E}|X_n - X| \to 0$. That means there exists some integer $N$ where for all $n > N$, $\mathbb{E}|X_n - X| < 1$. For these $n$, $\mathbb{E}|X_n| < 1 + \mathbb{E}|X|$.

For $n \leq N$, we have a finite set of values ${\mathbb{E}|X_1|, \mathbb{E}|X_2|, ..., \mathbb{E}|X_N|}$, which obviously has a finite upper bound. Let $M = \max\left{\mathbb{E}|X_1|, ..., \mathbb{E}|X_N|, 1 + \mathbb{E}|X|\right}$. Then $\sup_n \mathbb{E}|X_n| \leq M < \infty$, so condition 1 is satisfied.

Step 2: Prove Uniform Absolute Continuity #

Start by splitting the expectation using the triangle inequality again:
$$\mathbb{E}\left[|X_n| \cdot \mathbb{I}_A\right] \leq \mathbb{E}\left[|X_n - X| \cdot \mathbb{I}_A\right] + \mathbb{E}\left[|X| \cdot \mathbb{I}_A\right]$$
Let's fix an arbitrary $\epsilon > 0$ and handle each term separately:

1. Handle the $X$ term: Since $X \in L^1$ (because $X_n \xrightarrow{L^1} X$ implies $X$ is integrable), $X$ is absolutely continuous. That means there exists $\delta_1 > 0$ such that if $P(A) < \delta_1$, then $\mathbb{E}\left[|X| \cdot \mathbb{I}_A\right] < \epsilon/2$.
2. Handle the $X_n - X$ term for large $n$: Since $\mathbb{E}|X_n - X| \to 0$, there exists an integer $N$ where for all $n > N$, $\mathbb{E}|X_n - X| < \epsilon/2$. For any set $A$, $\mathbb{E}\left[|X_n - X| \cdot \mathbb{I}_A\right] \leq \mathbb{E}|X_n - X| < \epsilon/2$, regardless of $A$ (as long as $P(A) \leq 1$, which it always is).
3. Handle the finite number of small $n$: For each $n \leq N$, $X_n$ is in $L^1$, so each has its own absolute continuity constant $\delta_n > 0$ where $P(A) < \delta_n$ implies $\mathbb{E}\left[|X_n| \cdot \mathbb{I}_A\right] < \epsilon$.

Now take $\delta = \min\left{\delta_1, \delta_2, ..., \delta_N\right}$. For any set $A$ with $P(A) < \delta$:

• For $n > N$: $\mathbb{E}\left[|X_n| \cdot \mathbb{I}_A\right] \leq \epsilon/2 + \epsilon/2 = \epsilon$
• For $n \leq N$: $\mathbb{E}\left[|X_n| \cdot \mathbb{I}_A\right] < \epsilon$

This means $\sup_n \mathbb{E}\left[|X_n| \cdot \mathbb{I}_A\right] < \epsilon$, so condition 2 is satisfied.

Conclusion #

Both conditions for asymptotic uniform integrability are satisfied, so we've proven that $L^1$ convergence implies asymptotic uniform integrability.

内容的提问来源于stack exchange，提问作者TotalFailure