Proving discontinuity and continuity of a real valued function on an open interval.

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.

• 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. – SolubleFish May 4 at 10:09
• 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). – SolubleFish May 4 at 10:10
• @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? – Koro May 4 at 10:16
• 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. – Teresa Lisbon May 7 at 20:36
• (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. – Teresa Lisbon May 9 at 20:02

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}
• 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. – Koro May 4 at 10:51
• 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 $0$) – SolubleFish May 4 at 11:58
• One minor correction: The inequalities should be strict at endpoints $a$ and $b$. I have fixed the same in my post just now. – Koro May 5 at 3:55