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I proved a). It's easy because if $y>x$ then $$f(y)=\sum \limits_{n: x_n<y} c_n=\sum \limits_{n: x_n<x} c_n+\sum \limits_{n: x\leqslant x_n<y} c_n\geqslant f(x).$$ But how to prove b) i.e. $f(x_n+)-f(x_n-)=c_n.$

My sketch: Taking some $x_j\in E$. We know that $f(x_j+)$ is a $\lim \limits_{n\to\infty}{f(t_n)}$ for any $t_n\in (x_j,b)$ such that $t_n\to x_j$ as $n\to \infty$. Hence $$f(t_n)=\sum\limits_{k:x_k<t_n}c_k=(x_1+\cdots+x_j)+\sum\limits_{k>j:x_k<t_n}c_k. $$ Intuitively I know that the last sum tends to $0$ if $n\to \infty$ but I can't prove this strictly.

Can anyone help how to prove it?

RFZ
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2 Answers2

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So we have to prove that $c_n=f(x_n+)-f(x_n-)=\lim_{x\to x_n+}f(x)-\lim_{x\to x_n-}f(x)$. Note that if we show $c_n=\lim_{\epsilon\to 0+}f(x_n+\epsilon)-f(x_n-\epsilon)$ then we get our desired equality. But $f(x_n+\epsilon)-f(x_n-\epsilon)=\sum_{x_n-\epsilon\le x_m<x_n+\epsilon}c_m\ge c_n$ and that given $\delta>0$ since $\sum_mc_m$ converges we can always choose an appropiate $\epsilon'$ that removes a finite number of $c_m$ different from $c_n$ from the sum $\sum_{x_n-\epsilon\le x_m<x_n+\epsilon}c_m$ so that $\sum_{x_n-\epsilon'\le x_m<x_n+\epsilon'}c_m-c_n<\delta$, any other $\epsilon<\epsilon'$ will also have this property, we thus have proved $c_n=\lim_{\epsilon\to 0+}f(x_n+\epsilon)-f(x_n-\epsilon)$ and so the desired equality.

I'm gonna expand a little bit here: If we show that $c_n=\lim_{\epsilon\to 0+}f(x_n+\epsilon)-f(x_n-\epsilon)$ then we get \begin{align*} f(x_n+)-f(x_n-)&=\lim_{\epsilon\to 0+}f(x_n+\epsilon)-\lim_{\epsilon\to o+}f(x_n-\epsilon) \\ &=\lim_{\epsilon\to 0+}f(x_n+\epsilon)-f(x_n-\epsilon) \\ &=c_n \end{align*} as desired, the monotonicty has nothing to do. Remember that $\sum_mc_m=\lim_{k\to\infty}\sum_{m=1}^kc_m$. By the defition of limit for each $\delta$ you can always pick a $N$ so that for $k\ge N$ one has $\sum_mc_m-\sum_{m=1}^kc_m<\delta$. Now given a $\delta$ pick $\epsilon'$ so small that all of those $c_1,...,c_N$ (except of course $c_n$ that will always be in the sum no matter what) get removed in the sum $\sum_{x_n-\epsilon'\le x_m<x_n+\epsilon'}c_m$ so that $\sum_{x_n-\epsilon'\le x_m<x_n+\epsilon'}c_m-c_n<\delta$. Note also that $0\le\sum_{x_n-\epsilon'\le x_m<x_n+\epsilon'}c_m-c_n$. If $\epsilon<\epsilon'$ then the same inequalities also holds, we thus have proved $c_n=\lim_{\epsilon\to 0+}f(x_n+\epsilon)-f(x_n-\epsilon)$

Zero
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  • May be $с_n=f(x_n+)-f(x_n-)$. Right? – RFZ Sep 15 '15 at 04:27
  • What do you mean? – Zero Sep 15 '15 at 04:31
  • you wrote "So we have to prove that $c_n=f(x_n)-f(x_n-)$". I am asking maybe $c_n=f(x_n+)-f(x_n-)$" – RFZ Sep 15 '15 at 04:36
  • You are right. Is the proof too little expanded to be accepted? – Zero Sep 15 '15 at 04:40
  • I also have one question. $f(x_n+)-f(x_n-)=\lim \limits_{\varepsilon \to 0+}f(x_n+\varepsilon)-\lim \limits_{\varepsilon \to 0+}f(x_n-\varepsilon)$ Then you wrote that the last equality is $\lim \limits_{\varepsilon \to 0+}(f(x_n+\varepsilon)-f(x_n-\varepsilon))$. – RFZ Sep 15 '15 at 05:37
  • It's legal because in monotonic function $f(x+)$ and $f(x-)$ are exists. Am I right? – RFZ Sep 15 '15 at 05:43
  • Why your $\varepsilon′$ does not appear nowhere? Also I can't understand how you made this sum $- c_n$ less than $\delta$. P.S. I think that your proof is nice but it's written quite briefly. – RFZ Sep 15 '15 at 06:46
  • Can you explain me? – RFZ Sep 15 '15 at 13:27
  • Ok I expanded a little bit. Also something more or less similar is done in http://math.stackexchange.com/questions/1195127/remark-4-31-in-baby-rudin-how-to-verify-these-points?rq=1 – Zero Sep 15 '15 at 16:11
  • why after deleting terms $c_1, \cdots, c_N$ our series has form $\sum_{x_n-\epsilon'\le x_m<x_n+\epsilon'}c_m$? What kind she had before deleting? Only this moment I can't understand (maybe it's because it has complicated form). Also thanks for you edited post! – RFZ Sep 15 '15 at 16:35
  • After deleting those terms the series may not have the form $\sum_{x_n-\epsilon'\le x_m<x_n+\epsilon'}c_m$ for some $\epsilon'$, instead we pick the $\epsilon'$ so that in $\sum_{x_n-\epsilon'\le x_m<x_n+\epsilon'}c_m$ the terms $c_1,...c_N$ (except maybe $c_n$) aren't present but there may be even more terms not present (here is actually used the fact that the $c_m$'s are positive so that it doesn't matter that we delete some other $c_m$'s). Quite explicitely, we could take $\epsilon'=\min{|x_n-x_m|:m\in{1,2,...N}, m\neq n}$ – Zero Sep 15 '15 at 17:45
  • I think it's really hard for me :) From what series you deleted these terms? – RFZ Sep 16 '15 at 03:30
  • I think you should just forget about series and define for any $I$ (countable or otherwise) and $(a_i){i\in I}$ nonnegative reals $\sum{i\in I}a_i=\sup_{I_0\subset I \text{finite}}\sum_{i\in I_0}a_i$. It's much better to think of it this way, the problem with series is that you have to order them, here there is no order and is equivalent to the usual summation in series because the $a_i$'s are nonnegative – Zero Sep 16 '15 at 23:33
  • Why $\sum \limits_{i\in I}a_i=\sup ...$ What is this? What's relation between this and our problem? – RFZ Sep 17 '15 at 05:39
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First note that for $x<y$ $$f(y)-f(x) = \sum_{n:x\le x_n< y}c_n.$$ Now, if we fix $y$, and let $x\uparrow y$. Clearly, for any $N$, we can find $x=x(N)$ close to enough $y$ so that $x_1,\dots,x_N\not\in [x,y)$. Therefore $$\lim_{x\to y-}f(y)-f(x)\le \lim_{N\to\infty}\sum_{k>N}c_k=0.$$ Similarly, if we fix $x$ and let $y\downarrow x$, there are two cases,

  1. If $x=x_n$, then for any $N$, we can find $y=y(N)$ close to $x_n$ enough so that $\{x_1,\dots, x_N\} \cap [x_n,y)=\{x_n\}$. So $$c_n\le \lim_{y\to x_n+} f(y)-f(x_n)\le c_n+\lim_{N\to \infty}\sum_{k>N}c_k=c_n.$$

  2. If $x\ne x_n$ for any $n$, then the argument is the same as in the case of left limit.

Quang Hoang
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