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I am trying to prove the following statement, but I have been stuck for a while and I am looking for an hint about how to prove it:

"Suppose $B\subset\mathbb{R}$ and $f:B\to\mathbb{R}$ is an increasing function. Prove that $f$ is continuous at every element of $B$ except for a countable subset of $B$."

What I have tried to do:

Since I wasn't able to tackle the original statement I tried by assuming that $B$ is also a Borel set so as to exploit in some way the fact that $f$ must then be a Borel measurable function. Then if $D:=\{x\in B: f\text{ is not continuous at }x\}$ and $C$ is a Borel set we have that $f^{-1}(C)$ is a Borel set and $f^{-1}(C)=(f^{-1}(C)\cap D)\cup (f^{-1}(C)\cap B\setminus D)$ and at this point I tried to go on by contradiction, assuming that $D$ is uncountable and tried to show that $(f^{-1}(C)\cap D)\cup (f^{-1}(C)\cap B\setminus D)$ is not a Borel set but I haven't succeeded so I would greatly appreciate any hint or comment or explanation that could nudge me towards a more fruitful approach, thanks.

EDIT (Proof): Let $D:=\{x\in B:f\text{ is not continuous at }x\}$: since $f$ is monotone (increasing) it can only have jump discontinuities so if $d\in D$ then $I_d:=(\lim\limits_{x \to d^-,\ x\in B\\}f(x),\lim\limits_{x \to d^+,\ x\in B}f(x))=(\sup_{x<d,\ x\in B}f(x),\inf_{x>d,\ x\in B} f(x))$ is a non-empty interval in $\mathbb{R}$ hence there must be a rational number $q_d$ in it which we can choose and since $I_d\cap I_{d'}=\emptyset$ for $d\neq d'$* this rational number is also unique so we can define an injective function $g:D\to\mathbb{Q}, g(d):=q_d$. Thus $\#(D)\leq \#(\mathbb{Q})=\#(\mathbb{N})$ ie the set of points at which $f$ is discontinuous is (at most) countable, as desired. $\square$

(*) Suppose $I_d\cap I_{d'}\neq\emptyset$ for some $d\neq d'$ (we can suppose wlog that $d<d'$): then there must be $y\in I_d\cap I_{d'}$ so $\sup_{x<d,\ x\in B}f(x)<y<\inf_{x>d,\ x\in B}f(x)$ and $\sup_{x<d',\ x\in B}f(x)<y<\inf_{x>d',\ x\in B}f(x)$ which implies (since $d\leq d'\Rightarrow \inf_{x>d,\ x\in B}f(x)\leq \sup_{x<d', x\in B}f(x)$) that $y<y$, contradiction.

lorenzo
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2 Answers2

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If $f$ is not continuous at $x\in B$, then $\inf\{\,f(t)\mid t\in B, t>x\,\}>\sup\{\,f(t)\mid t\in B, t<x\,\}$ (including the possibility $\inf=+\infty$ or $\sup=-\infty$). Pick a rational between this $\inf $ and $\sup$ and thus obtain an injective map from the set of discontinuities to $\Bbb Q$.

Hagen von Eitzen
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    Suppose $f$ is not continuous at $x\in B$ and $x'\in B$, then, if $B$ is not an interval, the intervals $(\sup{,f(t)\mid t\in B, t<x,}, \inf{,f(t)\mid t\in B, t>x,})$ and $(\sup{,f(t)\mid t\in B, t<x',}, \inf{,f(t)\mid t\in B, t>x',})$ may overlap. See comments under the question. – Ramiro Mar 21 '23 at 11:42
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Let us assume $B = (a, b).$ If $x < y < z$ then $f(x) \leq f(y) \leq f(z)$ and every discontinuity must be a jump discontinuity. If $\mathrm{J}_n$ is the set of jumps with size $> \dfrac{1}{n},$ then $\mathrm{J}_n$ is finite and the proof follows. Q.E.D.

William M.
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