弱收敛导出几乎必然收敛定理的证明细节澄清

Hey there! I totally get stuck on these tiny, nitty-gritty details in Durrett's proofs too—they're often crucial but easy to gloss over. Let's break down this part of the Skorokhod representation theorem proof step by step, focusing on that exception set $\Omega_0$.

First, let's recap Durrett's setup to make sure we're on the same page:

• We start with $U \sim \text{Uniform}(0,1)$ (our "base" random variable on the probability space $([0,1], \mathcal{B}, \lambda)$, where $\lambda$ is Lebesgue measure).
• For each $n$, define the left inverse of the CDF $F_n$:
  $$Y_n(x) = \sup{ y : F_n(y) < x }$$
  A key fact here is that $Y_n$ has exactly distribution $F_n$—this is the standard inverse transform trick for generating random variables from a CDF.
• Our goal is to show $Y_n(x) \xrightarrow{\text{a.s.}} Y_{\infty}(x)$, where $Y_{\infty}(x) = \sup{ y : F_{\infty}(y) < x }$ (which has distribution $F_{\infty}$).

First, let's clarify the definitions of $a_x$ and $b_x$ as Durrett uses them:

• $a_x = \sup{ y : F_{\infty}(y) < x }$ (this is exactly $Y_{\infty}(x)$, the left inverse of $F_{\infty}$)
• $b_x = \inf{ y : F_{\infty}(y) > x }$

The interval $(a_x, b_x)$ represents all values $y$ where $F_{\infty}(y) = x$ (if such $y$ exist). So $\Omega_0 = { x : (a_x, b_x) = \emptyset }$ is the set of $x$ where there is no $y$ with $F_{\infty}(y) = x$—in other words, $x$ falls into a "gap" between consecutive values of $F_{\infty}$ (i.e., $x$ is not in the range of $F_{\infty}$).

The critical point here is that $\Omega_0$ has Lebesgue measure zero. Here's why:

• CDFs are non-decreasing and right-continuous, so they can have at most countably many jump points. Each jump point corresponds to a gap in the range of $F_{\infty}$ (the interval between $F_{\infty}(y^-)$ and $F_{\infty}(y^+)$ for the jump point $y$).
• These gaps are countable, and the total length of all gaps sums to $0$ (since $F_{\infty}$ is normalized to $F_{\infty}(-\infty)=0$ and $F_{\infty}(\infty)=1$—there's no "missing" total probability).

Durrett's strategy is to show that for all $x \notin \Omega_0$, $Y_n(x) \to Y_{\infty}(x)$. Here's the core logic:

1. For $x \notin \Omega_0$, $(a_x, b_x) \neq \emptyset$, so there exists some $y$ where $F_{\infty}(y) = x$.
2. Since $F_n \Rightarrow F_{\infty}$, for any continuous point $z$ of $F_{\infty}$, $F_n(z) \to F_{\infty}(z)$.
3. Take $z < Y_{\infty}(x)$: then $F_{\infty}(z) < x$, so for large enough $n$, $F_n(z) < x$, which implies $Y_n(x) \geq z$. Taking the $\liminf$, we get $\liminf Y_n(x) \geq Y_{\infty}(x)$.
4. Take $z > Y_{\infty}(x)$: then $F_{\infty}(z) > x$, so for large enough $n$, $F_n(z) > x$, which implies $Y_n(x) \leq z$. Taking the $\limsup$, we get $\limsup Y_n(x) \leq Y_{\infty}(x)$.
5. Combining these, $\liminf Y_n(x) = \limsup Y_n(x) = Y_{\infty}(x)$, so $Y_n(x) \to Y_{\infty}(x)$ for $x \notin \Omega_0$.

Since $\Omega_0$ has measure zero, this means $Y_n \xrightarrow{\text{a.s.}} Y_{\infty}$, which is exactly what we needed to prove.

• $\Omega_0$ is the set of $x$ falling into gaps in the range of $F_{\infty}$ (no $y$ exists with $F_{\infty}(y)=x$).
• It's a null set because these gaps are countable and have total length zero.
• For all $x$ not in $\Omega_0$, weak convergence of $F_n$ to $F_{\infty}$ guarantees $Y_n(x)$ converges to $Y_{\infty}(x)$.

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