Remark 4.31 from baby Rudin 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?

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 $c_n=\lim_{\epsilon\to 0+}f(x_n+\epsilon)-f(x_n-\epsilon)$

• May be $с_n=f(x_n+)-f(x_n-)$. Right? – ZFR Sep 15 '15 at 4:27
• What do you mean? – Zero Sep 15 '15 at 4:31
• you wrote "So we have to prove that $c_n=f(x_n)-f(x_n-)$". I am asking maybe $c_n=f(x_n+)-f(x_n-)$" – ZFR Sep 15 '15 at 4:36
• You are right. Is the proof too little expanded to be accepted? – Zero Sep 15 '15 at 4:40
• 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 $\lim \limits_{\varepsilon \to 0+}(f(x_n+\varepsilon)-f(x_n-\varepsilon))$. – ZFR Sep 15 '15 at 5:37

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.