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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ (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

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*}

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.