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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.

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Thanks in advance!

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  • $\begingroup$ When it says "Given $\epsilon > 0$, there exists $x' \in D, x'>x$ such that $$F(x) + \epsilon \geq F_D(x')$$" $\endgroup$ – afedder Dec 24 '14 at 10:51
  • $\begingroup$ The fact that there is guaranteed to be an $x' \in D$ such that $x' > x$ is a use of the density of $D$ in $\mathbb{R}$. $\endgroup$ – afedder Dec 24 '14 at 10:55
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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 $F(x)=F_D(x)$.

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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.

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    $\begingroup$ Thanks, this is a good answer. Have I understood you right if I say that $lim _{y \to x^+ , y \in D}F _D (y) $ makes no sense if $\exists \delta$: $\inf_{y \in D, y>x } |y-x |> \delta $? $\endgroup$ – Alexander Dec 25 '14 at 8:19
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    $\begingroup$ Yes! It makes no good sense. Formally, you could still interpret a limit expression like $\lim_{y\to x, y\in D}f(y)$, even if $D$ is not dense around $x$, following the usual definition of the limit. But then the value would not be unique, which is not what we want. So, it's better leave such limits undefined. $\endgroup$ – Blackbird Dec 25 '14 at 14:04
  • $\begingroup$ One more thing, how is it that that the limit wouldn't be unique? $\endgroup$ – Alexander Dec 26 '14 at 6:38
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    $\begingroup$ By the usual formal definition, $\lim_{y\to x, y\in D}f(y)$ is a number $c$ that satisfy the following: for every $r>0$, there is $s>0$ such that $|f(y)-c|<r$ for whichever $y\in D$ with $0<|y-x|<s$. Now if there is such $\delta>0$ that $D$ does not contain any point $y$ satisfying $0<|y-x|<\delta$, then any number $c$ would satisfy the definition, by simply choosing $s=\delta$, independently of $r$. $\endgroup$ – Blackbird Dec 26 '14 at 13:19

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