A probability function is determined on a dense set- Where is density used in the following proof? A probability function is determined on a dense set- Where is density used in the following proof?
Consider the following theorem and proof from Resnick's book A probability path. I cannot really see where the assumption that $D $ is dense in $R $ is used. Can you enlighten me? Is it needed for $(8.2) $? I think there would be sufficient if there existed an $x' \in D$ such that $x' \ge x $ for it to hold.


Thanks in advance!
 A: You’re right: as long as $D$ contains arbitrarily large reals, $F$ is right continuous. However, $F$ need not extend $F_D$ even if $D$ is dense in $\Bbb R$, so the last sentence of the statement of the lemma is a non sequitur.
To see this, let $D=\Bbb R\setminus\{1\}$, and let $F_D(x)=0$ if $x\le 0$ and $F_D(x)=1$ if $0<x\in D$. Then 
$$F(0)=1\ne 0=F_D(0)\;,$$
and $F\upharpoonright D\ne F_D$.
If the real goal was to prove that if $F$ and $G$ are right continuous df’s that agree on a dense $D\subseteq\Bbb R$, then $F=G$, it could have been accomplished much more simply. Let $x\in\Bbb R\setminus D$; $D$ is dense, so there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $D$ converging to $x$ from the right. $F$ and $G$ are right continuous, and $F(x_n)=G(x_n)$ for $n\in\Bbb N$, so
$$F(x)=\lim_nF(x_n)=\lim_nG(x_n)=G(x)\;,$$
and therefore $F=G$.
Also, one can prove that if $D$ is dense in $\Bbb R$, and $F_D$ is right continuous on $D$ as well as satisfying the hypotheses of Lemma $8.1.1$, then $F$ is a right continuous df extending $F_D$: the extra hypothesis that $F_D$ is right continuous on $D$ allows us to show that if $x\in D$, then $F(x)=F_D(x)$.
A: The Lemma proves two things at the same time: I) a distribution function is uniquely determined by its restriction to a dense set, and II) a function $F_D$ defined on a dense set and having the listed properties extends to a distribution function.  Part (I) is the more important part, I suppose.
For (II) you are right that denseness of $D$ is not necessary; it is sufficient that $D$ has arbitrarily large elements.  Note that the second equality in (8.1) may not hold if $D$ is not dense, because you may not be able to let $y$ tend to $x$ from above within $D$.
The proof of (I) is by noting that any distribution function $F$ must satisfy (8.1), in particular, the equality
   $$F(x)=\lim_{\substack{y\downarrow x\\ y\in D}}F_D(y)\;,$$
where $F_D$ denotes the restriction of $F$ to a dense set $D$.  The denseness of $D$ guarantees that the second equality in (8.1) holds.
