Currently I am reading Rick Durret's PTE book and I have trouble understanding a detail in the proof of the following theorem.

If $$ \ \ F_n \Rightarrow F_{\infty} \ \mbox{ then } \exists X_n \xrightarrow{a.s.} X_{\infty}$$ where $ X_n $ is has distribution $F_n$

Durret proves the theorem by construction i.e. he defines $Y_n(x) = \sup\{y:F(x)<y\}$ and then proves that $Y_n \xrightarrow{a.s.} Y$. While doing this he constructs an exceptional set as $\Omega_0 = \{ x : (a_x, b_x) = \emptyset \}$ where $ a_x = \sup\{y:F(y)<x\}$, $ b_x = \inf\{y:F(y)>x\}$.

The question is: when does $$(*) \ \ \ (a_x, b_x) \neq \emptyset$$ happen ?

If $x$ is a continuity point then $a_x = b_x$ . If $x$ is not a cont. point, then $x$ is a breakage point, i.e. a point where $\lim_{\epsilon \xrightarrow{-} 0} F(x+\epsilon) \neq F(x)$ since $F $ is right continuous, clearly $b_x = y_0$ where $F(y_0) = x$.

Also $\{y_0+\epsilon_n\}\subset \{y:F(y) < x\} $ where $\epsilon_n \xrightarrow{-} 0$, so $\sup\{y_0+\epsilon_n\} = y_0 = a_x$ and $(*)$ is not true where am I wrong ?

Then Durret by uses the fact that $(a_x, b_x)$'s that are countable, since they are disjoint and any of them contains a rational number to ignore $\Omega - \Omega_0$.

I also have quite a hard time understanding why the following part of the proof is valid: Durret denotes for convenience $Y_n$ as $F^{-1}_n$ .

$$(**) \ \ \ \liminf_{n \rightarrow \infty} F_n^{-1}(x) \geq F^{-1}(x)$$

Let $y<F^{-1}(x)$ be such that $F$ is cont. at $y$. Since $x \in \Omega_0$ then $F(y)<x$ and for sufficiently large $n$ $F_n(y) < x$, i.e., $F^{-1}_n(x) \geq y$. Since this holds for all $y$ satisfying the indicated restrictions, the result follows.

The text in bold is what I do not understand. We can build a sequence $\{y_n\}$ such that $y_n \rightarrow F^{-1}(x)$ but if $y_n = F^{-1}(x) + \epsilon_n$ where $\epsilon_n > 0$ then the restriction is not met, so all $y_n$'s should be less then $F^{-1}_n(x)$ , and if this is true I don see how this proofs $(**)$.



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