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Let $E\subset (a,b)$ be a countable subset of the open interval $(a,b)$. Let $E=\{x_n:n\in \mathbb N\}$. Let $\sum c_n$ be a convergent sequence such that $c_n\gt 0$ for all $n\in \mathbb N$.

Let $f:(a,b)\to \mathbb R$ be a real valued function defined as $f(x)= \sum_{x_n\lt x}c_n$. It is to be proven that:

  1. $f$ is monotonically increasing on $(a,b)$.
  2. $f$ is discontinuous on $E$.
  3. $f$ is continuous on $(a,b)\setminus E$

I tried to prove it like this:
Claim 1):$f$ is monotonically increasing on $(a,b)$.
Proof: For any $t\in (a,b)$, let $N_t=\{n: x_n\lt t\}$. For any $x,y\in (a,b)$ such that $x\lt y$, it follows that $N_x\subset N_y$ and therefore by definition of $f$, we have $f(x)\le f(y)$. This proves that $f$ is monotonically increasing.

Claim 2): $f$ is discontinuous on $E$.
Proof: Let $x_m\in E$ be an arbitrary point in $E$. Then we have
$\begin{align} f(x_m+)-f(x_m-)=&\inf\{f(t):x_m\lt t\lt b \}-\sup\{f(t):a\lt t\lt x_m\}\\=&\inf\{f(t):x_m\lt t\lt b \}+\inf\{-f(t):a\lt t\lt x_m\}\\=&\inf\{\sum_{x_n\lt t}c_n:x_m\lt t\lt b\}+\inf\{-\sum_{x_n\lt t}c_n:a\lt t\lt x_m\}\\=&\inf\{\sum_{x_m\le x_n\lt t} c_n: x_m\lt t\lt b\}=c_m \end{align}$.
Here I have used the result: For any two sets $A$ and $B,\inf (A+B)=\inf A+\inf B$. It follows that $f(x_m+)\ne f(x_m-)$ for any $x_m\in E$ and hence $f$ is discontinuous on $E$.

Claim 3)$f$ is continuous on $(a,b)\setminus E$
Proof: For any $x\in (a,b)\setminus E$, we have:
$f(x+)-f(x-)=\inf\{\sum_{x\lt x_n\lt t} c_n: x \lt t\lt b\}\tag1 $
Since $\sum c_n$ is convergent, given any $\epsilon\gt 0$, we can choose $N\in \mathbb N$ such that $\sum_{n=N+1}^\infty c_n\lt \epsilon$

Let's choose $\delta=\min \{|x-x_i|:i\in \{1,2,\cdots,N\}\}$ and therefore $(1)$ gives
$f(x+)-f(x-)=\inf\{\sum_{x\lt x_n\lt t} c_n: x \lt t\lt b\}= \inf\{\sum_{x\lt x_n\lt t} c_n: x \lt t\lt x+\delta\}\le \sum_{n=N+1}^\infty c_n\lt \epsilon$
Since $f$ is monotonically increasing, we have $f(x-)\le f(x)\le f(x+)$ and since $\epsilon $ is arbitrary, we have $f(x+)=f(x-)$ and therefore $f(x)=f(x-)=f(x+)$, which proves that $f$ is continuous at $x$. Since $x\in (a,b)\setminus E$ is arbitrary, it follows that $f$ is continuous on $(a,b)\setminus E$.

Is my proof correct? Thanks.

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    $\begingroup$ The proof of claim 1 is correct. For claim 2, your last equality is a bit fast : when you take the sum of the two sets $\left\{\sum_{x_{n}<t} c_{n}: x_{m}<t \leq b\right\}$ and $\left\{-\sum_{x_{n}<t} c_{n}: a \leq t<x_{m}\right\}$, the result is : $$\left\{\sum_{x_{n}<t} c_{n}-\sum_{x_{n'}<t'} c_{n'}: x_{m}<t \leq b, : a \leq t'<x_{m}\right\}$$ A little more work is required to finish the proof. $\endgroup$ – SolubleFish May 4 at 10:09
  • $\begingroup$ Your proof of claim (3) seems to be correct. Also, notice that you can adapt your proof of (2) to prove that $f(x+) - f(x-)= 0$ if $x$ is not one of the $x_m$., and vice versa, you can adapt your proof of (3) to prove (2). $\endgroup$ – SolubleFish May 4 at 10:10
  • $\begingroup$ @SolubleFish: Thanks a lot for reviewing my proof. I really appreciate that. Thank you! I thought that all terms in $B$ will get cancelled by elements in set $A$ and sums of only this form $\sum_{x_m\le x_n\lt t_n }$ will remain. But my proof (2) is correct. Right? $\endgroup$ – Koro May 4 at 10:16
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    $\begingroup$ The proof is correct, good job. I personally think too that the sup-inf thing in claim 2 was a little fast, but IMO the rest is very well written. Well done! Note that you produce, with this example, a function which is continuous exactly at some countable set. It's a very important and fruitful counterexample to know. $\endgroup$ – Teresa Lisbon May 7 at 20:36
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    $\begingroup$ (Correction : in the above post I meant : "a function which is discontinuous exactly at some countable set"). There is a precise characterization of sets which can be the exact set of discontinuity of any function. Precisely : a set can be the set of discontinuities of some real valued function if and only if it can be written as a countable union of closed sets. $\endgroup$ – Teresa Lisbon May 9 at 20:02
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For claim 2, the last equality is a bit shaky. To make the proof more detailed and rigorous : \begin{align*} f(x_m^+) - f(x_m^-) &= \inf \left\{\sum_{x_{n}<t} c_{n}: x_{m}<t < b\right\}+\inf \left\{-\sum_{x_{n}<t} c_{n}: a < t<x_{m}\right\} \\ &= \inf \left\{\sum_{x_{n}<t} c_{n}-\sum_{x_{n^{\prime}}<t^{\prime}} c_{n^{\prime}}: x_{m}<t < b,: a < t^{\prime}<x_{m}\right\} \\ &= \inf \left \{\sum_{t'\leqslant x_n < t} c_n : a < t' < x_m < t < b\right\} \\ &= \sum_{x_m \leqslant x_n \leqslant x_m} c_n \\ &= c_m \end{align*}

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  • $\begingroup$ This is very nice! +1. I like it. Thank you! Can you please elaborate more on second last line? $\endgroup$ – Koro May 4 at 10:46
  • $\begingroup$ Using this approach, the confusion that arises is that in $(t',t)$, there might be some another $x_p$ also for example if $E$ were dense. $\endgroup$ – Koro May 4 at 10:51
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    $\begingroup$ When you evaluate the last $\inf$, you find that the only term left in is $[x_m,x_m]$. Since the $(x_n)$ are distinct from each other, only $x_m$ is left. (The nice thing with this calculation is that you can replace $x_m$ with any $x$, and at the end you get $c_m$ if $x$ is equal to some $x_m$, or else $0$) $\endgroup$ – SolubleFish May 4 at 11:58
  • $\begingroup$ Thanks a lot :) $\endgroup$ – Koro May 4 at 12:02
  • $\begingroup$ One minor correction: The inequalities should be strict at endpoints $a$ and $b$. I have fixed the same in my post just now. $\endgroup$ – Koro May 5 at 3:55

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