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Let $F:\mathbb{R}\rightarrow \mathbb{R}$ be increasing and let $G(x) = \lim_{y\rightarrow x^+}F(y) (= F(x^+)$. The set of points at which $F$ is discontinuous is countable.

My question is if say we have a point $x\in\mathbb{R}$ where $F$ is discontinuous then $$\lim_{y\rightarrow x^+} F(y) > \lim_{y\rightarrow x^-}F(y)$$ does there exist a injective map from the discontunities to subsets of $\mathbb{Q}$?

Ramiro
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Wolfy
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  • Hint: the interval from $F(x^-):=\lim_{y\to x^-}F(y)$ to $F(x^+):=\lim_{y \to x^+}F(y)$ is nonempty, hence contains a rational number. – kccu Jul 26 '16 at 19:22
  • I see, so we can say that there exists a rational number $p$ in between $$\lim_{y\rightarrow x^+} F(y) > \lim_{y\rightarrow x^-}F(y)$$ so can we say that $p$ came from an injective map of discontinuites to subsets of the rational numbers? – Wolfy Jul 26 '16 at 19:24
  • You can define the injective map using the fact that such a rational $p$ exists for each point of discontinuity $x$. It still remains to argue that this map is in fact injective. – kccu Jul 26 '16 at 19:25
  • @kccu ok so besides proving the map is injective from what I said in the second comment formal enough? As in does it sound ok? – Wolfy Jul 26 '16 at 19:54

1 Answers1

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@Wolfy, your intuition is correct. Here is a detailed proof following your intuition.

Let $F:\mathbb{R}\rightarrow \mathbb{R}$ be increasing. The set of points at which $F$ is discontinuous is countable.

Proof:

Since $F$ is increasing, we have that, for any $x\in\mathbb{R}$, the lateral limits $\lim_{y\rightarrow x^-}F(y)$ and $\lim_{y\rightarrow x^+} F(y)$ actually exit.

So, for any $x\in\mathbb{R}$ where $F$ is discontinuous, we have $$F(x^-)=\lim_{y\rightarrow x^-}F(y) < \lim_{y\rightarrow x^+} F(y)=F(x^+) $$ So open interval $(F(x^-), F(x^+))$ is non-empty.

Moreover , if $x_1,x_2\in\mathbb{R}$, $x_1<x_2$, we have $$F(x_1^+)=\lim_{y\rightarrow x_1^+}F(y) < \lim_{y\rightarrow x_2^-} F(y)=F(x_2^-) $$

So, if $x_1,x_2\in\mathbb{R}$, $x_1 \neq x_2$ and $F$ is discontinuous at $x_1$ and $x_2$, we have that $(F(x_1^-), F(x_1^+))$ and $(F(x_2^-), F(x_2^+))$ are disjoint.

Let $D=\{ x\in\mathbb{R} : F \textrm{ is discontinuous at } x \}$.

Let us define $\psi: D \rightarrow \mathbb{Q}$, where for each $x \in D$, we choose a rational in $(F(x^-), F(x^+))$. Its is possible because, for each $x\in D$, $(F(x^-), F(x^+))$ is a non-empty open interval.

Moreover, $\psi$ is injetive. In fact, if $x_1,x_2\in D$ and $x_1 \neq x_2$, $(F(x_1^-), F(x_1^+))$ and $(F(x_2^-), F(x_2^+))$ are disjoint, so $\psi(x_1)\neq \psi(x_2)$.

So $\psi$ is an injective map from $D$ into $\mathbb{Q}$.

Ramiro
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  • quick question about $F$ in section 3.5. Throughout the section is $F$ always defined as $F:\mathbb{R}\rightarrow \mathbb{R}$? – Wolfy Jul 27 '16 at 00:49
  • @Wolfy , Thoughout the section 3.5, $F$ is a function from $\mathbb{R}$ into $\mathbb{C}$, EXCEPT where explictly said it is into $\mathbb{R}$ (that is, real-valued). – Ramiro Jul 27 '16 at 03:08
  • I see so for the 3.28 lemma, just because $F\in BV$ then we can't say that $F$ is increasing unless $F:\mathbb{R}\rightarrow \mathbb{R}$; this is from Theorem 3.27b.) – Wolfy Jul 27 '16 at 03:13
  • @Wolfy , Yes. You can for instance in Theorem 3.27a.) that $F\in BV$ means in general that $F$ is complex-valued. For Lemma 3.28, $F$ is complex-valued. – Ramiro Jul 27 '16 at 03:22
  • ahh ok, having some trouble with lemma 3.28, I thought $F$ would be increasing \ – Wolfy Jul 27 '16 at 04:01
  • @Ramiro, when you define $\phi:D\to\mathbb Q$, you use the axiom of choice, is there another way to do it? thanks – Framate Jun 27 '17 at 05:37
  • @Miguemate There are different way to prove the result, but as far as I know all of then depends (directly or indirectly) on the axiom of choice. – Ramiro Jun 28 '17 at 05:16