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Discontinuities of regulated function

My question is why does the right limit $f(c+)$ not exist.

So far I have $|\lim_{y \to x_k+}{f(y)}- \lim_{y \to x_k-}{f(y)}|≥\frac{1}{n}$

$x_k$ is a sequence which strictly decreases to c.

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Paul
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  • There are many discussions on this topic. But I am as lazy as you and don't want to search through them. Here is a hint: If $\cal C$ is a class of functions all of whom have a finite right-hand limit at a point $c$ and $\hat{\cal C}$ is the collection of uniform limits of sequences from $\cal C$ then each function in the larger family $\hat{\cal C}$ also has a finite right-hand limit at the point $c$. – B. S. Thomson Nov 08 '15 at 20:17
  • I'm not familiar with any of the stuff u mentioned. And I couldn't find anything related to what I wanted – Paul Nov 08 '15 at 20:30
  • There are two definitions of regulated functions: (i) functions that have finite right-hand and left-hand limits at each point or (ii) uniform limits of step functions. You are not using Definition (i) or you wouldn't be asking about $f(c+)$ since you know it exists. You aren't confused about uniform limits since you must have already studied them and know that uniform limits preserve many properties. Since step functions have finite right-hand and left-hand limits at each point, so too do regulated functions. So explain your dilemma a bit better. – B. S. Thomson Nov 08 '15 at 20:48
  • How does the fact that Dn was taken to be the limit I posted suggest f(c+) doesn not exist? X_k is a sequence decreasing to c... – Paul Nov 08 '15 at 20:55
  • I am more confused here than you. You assume that $f$ is regulated? Right? So why are you fussing with $f(c+)$? It does exist! All regulated functions have this property. I am sure you have a genuine question, but I don't know what it is yet. – B. S. Thomson Nov 08 '15 at 21:05
  • Yes it exists but I need to prove it doesn't exist for a contradiction to the question on the link I provided. Here: http://math.stackexchange.com/questions/1507957/discontinuities-of-regulated-function#comment3071658_1508001 – Paul Nov 08 '15 at 21:06
  • Ok better. You say "My question is why does the right limit $f(c+0)$ not exist." But you are looking for discontinuities. You want to collect the points where $f(c+)\not =f(c-)$ or where $f(c+)=f(c-)\not= f(c)$. You are not trying to find a contradiction to the fact that $f(c+)$ exists. – B. S. Thomson Nov 08 '15 at 21:13
  • But in that thread I was using an argument which should give a contradiction. Now I am confused. – Paul Nov 08 '15 at 21:17
  • If you know any other approaches to the question please let me know – Paul Nov 08 '15 at 21:34

1 Answers1

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Ok, as stated the question was not immediately clear to me, but Paul has cleared it up since he has already tried to get help on it before.

Problem: Prove that set of points of discontinuity of a regulated function $f:[a,b]\to\mathbb{R}$ is countable. [But only using the fact that right and left hand limits exist!]

Asked and answered many times before, but let us take Paul's beginning of a proof and show how to proceed.

Let $$D_n= \{y\in (a ,b ): | f(y+) -f(y-)| > 1/n \}$$ for $n=1,2,3, \dots$. If this set is infinite then there is an accumulation point $\hat{y}$ in $[a,b]$.

If this point $\hat{y}\geq a$ find a $\delta>0$ so that $|f(\hat{y}+t) -f(\hat{y}+)|< 1/4n $ for $0<t<\delta$. In the interval $(\hat{y},\hat{y}+\delta)$ take any two points $u$, $v$ and check that $|f(u)-f(v)|< 1/2n$. That means that there are no points of $D_n$ in that interval.

Similarly if this point $\hat{y}\leq b$ find a $\delta>0$ so that $|f(\hat{y}-t) -f(\hat{y}-)|< 1/4n $ for $0<t<\delta$. In the interval $(\hat{y},\hat{y}-\delta)$, for the same reason, there are no points of $D_n$ here either.

This is a contradiction since $\hat{y}$ was selected as an accumulation point of $D_n$. Hence each set $D_n$ is finite.

[Note to Paul: we are contradicting the fact that $D_n$ is infinite, we are not contradicting the fact that the one-sided limits exist.]

There are more cases to consider here than just $|f(y+) -f(y-) |> 0$ but the same proof will work for all of them.

B. S. Thomson
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  • Thank you so much. Is this the only way to go about answering the question using limits? I have not really covered accumulation points or left/right limits too much besides their definition. – Paul Nov 08 '15 at 22:08
  • There are always lots of ways of answering any question. Just keep trying problems that use the Bolzano-Weierstrass theorem and work with limsups and liminfs etc. Definitions don't go very far without practice. [Try to learn chess by studying the rules but never playing a game.] – B. S. Thomson Nov 08 '15 at 22:14
  • One question where have you used the B-W theorem? – Paul Nov 08 '15 at 23:08
  • BW theorem was used at the point where we considered the possibility that $D_n$ is infinite. If so, then $D_n$ is an infinite bounded set inside the interval $[a,b]$ and, by the BW theorem, there must exist a point $c$ in that interval which is a point of accumulation of $D_n$. [The point $c$ doesn't have to be in $D_n$ itself.] – B. S. Thomson Nov 08 '15 at 23:25