I have to prove the following

THEOREM Let $f:[a,b]\to\Bbb R$ be such that $\lim\limits_{y\to x}f(y)$ exists for every $x\in[a,b]$. Then the set $\Delta\subset[a,b]$ where $f$ is discontinuous is countable.


PROPOSITION Let $f:[a,b]\to\Bbb R$ be such that $\lim\limits_{y\to x}f(y)$ exists for every $x\in[a,b]$. Define $g:[a,b]\to\Bbb R$ as $$g(x)=\lim_{y\to x}f(y)$$ Then $g$ is continuous.


Let $\alpha\in[a,b]$. Then $g(\alpha)=\lim_{y\to\alpha}f(y)$. Thus, for $\epsilon >0$ given there exists a $\delta>0$ such that $0<|y-\alpha|<\delta$ implies $|g(\alpha)-f(y)|<\epsilon/2$. This is $g(\alpha)-\epsilon/2<f(y)<g(\alpha)+\epsilon/2$. Let $x$ such that $0<|x-\alpha|<\delta$. Then

$$g(\alpha)-\epsilon/2\leq \lim_{y\to x}f(y)\leq g(\alpha)+\epsilon/2$$

But this means $|g(x)-g(\alpha)|\leq \epsilon/2<\epsilon$.$\;\blacktriangle$

Now we can move on:

PROOF Let $\epsilon >0$ and consider the set $$\Lambda(\epsilon)=\left\{x\in[a,b]:\left|\lim\limits_{y\to x}f(y)-f(x)\right|>\epsilon\right\}$$

The claim is that $\Lambda(\epsilon)$ is finite for every choice of $\epsilon$. Indeed, suppose $\Lambda(\epsilon)\subset[a,b]$ was not finite. Then, it has an accumulation point, $\lambda$, in $[a,b]$. By hypothesis $\lim\limits_{y\to \lambda}f(y)$ exists. Thus for this $\epsilon >0$ there exists a $\delta >0$ such that $$\tag 1 \left|f(y)-\lim_{x\to \lambda}f(x)\right|<\epsilon /2$$ whenever $0<|y-\lambda |<\delta$. Since $\lambda$ is an accumulation point, for each $\delta >0$ there exists a $w\in\Lambda(\epsilon)$ such that $0<|w-\lambda|<\delta$ and

$$\tag 2 \left|f(w)-\lim_{x\to w}f(x)\right|>\epsilon$$

And because of the previous proposition, for this $\epsilon>0$ there exists $\delta' >0$ such that for each $y$ with $0<|y-\lambda|<\delta'$ we have

$$\tag 3 \left|\lim_{x\to y}f(x)-\lim_{x\to \lambda}f(x)\right|<\epsilon/2$$

Let $\delta''=\min(\delta,\delta')$. We thus obtain from $(1)$,$(2)$ and $(3)$ a $w\in \Lambda(\epsilon)$ such that $0<|w-\lambda|<\delta''$ with

$$ \left|f(w)-\lim_{x\to \lambda}f(x)\right|<\epsilon /2$$

$$ \left|f(w)-\lim_{x\to w}f(x)\right|>\epsilon$$

$$ \left|\lim_{x\to w}f(x)-\lim_{x\to \lambda}f(x)\right|<\epsilon/2$$

which, by the triangle inequiality, is absurd. Then $\Lambda(\epsilon)$ is finite for each $\epsilon>0$. In particular, for each $n\in\Bbb N$, $\Lambda(1/n)$ is finite. But

$$\bigcup_{n\in\Bbb N}\Lambda(1/n)=\Delta$$ which is the countable union of finite sets, whence, it is at most countable.$\; \blacktriangle$

ADD I cannot understand Spivak's proof of this. He considers $[0,1]$ in his exercise (I had forgotten)

If the set of such $a$ was infinite, it would have an accumulation point in $[0,1]$. Call it $x$. For every $\delta >0$ there would be an $a$ such that $0<|x-a|<\delta/2$ and $$\tag 1 \lim_{y\to a}f(y)-f(a)|>\epsilon$$ Thus, there would exist an $a'$ such that $|a'-a|<\delta/2$ (from where $0<|x-a'|<\delta$) such that $$\tag 2 |f(a')-f(a)|>\epsilon$$ But since $\ell=\lim\limits_{y\to x }f(y)$ for some $\ell$, there exists a $\delta >0$ such that $$|f(y)-\ell|<\epsilon/2$$ whenever $0<|y-x|<\delta$. In particular, if $0<|x-a|<\delta$ and $0<|x-a'|<\delta$ then $\epsilon < |f(a')-f(a)|<|f(a')-\ell|+|f(a)-\ell|<\epsilon$ which is absurd.

My question is: where does he get $$\epsilon < |f(a')-f(a)|$$ from?

  • 1
    $\begingroup$ I would like to know an example of such $f$ with infinite $\Delta$. $\endgroup$ – Makoto Kato Nov 19 '12 at 20:18
  • 1
    $\begingroup$ @MakotoKato I don't know if this is too simple, but set $f(x)=0$ if $x\neq 1/n$ and $1/n$ otherwise. $\endgroup$ – Pedro Tamaroff Nov 19 '12 at 20:23
  • $\begingroup$ Isn't your first proposition redundant? You don't seem to use it in the proof of the theorem. $\endgroup$ – Rhys Nov 19 '12 at 20:41
  • $\begingroup$ @Rhys I use it to get $(3)$. $\endgroup$ – Pedro Tamaroff Nov 19 '12 at 20:43
  • $\begingroup$ In the start of the spivack's proof should not you write "if $a$ was uncountable". I think the difference between your proof and spivack's is the spivack does not prove why for that every $a$ you have $\tag 1 \lim_{y\to a}f(y)-f(a)|>\epsilon$ where $\epsilon$ is fixed for every $a$. $\endgroup$ – clark Nov 19 '12 at 21:36

In Spivak’s proof let $$L=\lim_{y\to a}f(a)\;,$$ and let

$$d=|L-f(a)|=\left|\lim_{y\to a}f(y)-f(a)\right|>\epsilon\;.$$

Then there is an $\eta>0$ such that $|L-f(y)|<\frac12(d-\epsilon)$ whenever $0<|y-a|<\eta$ and $y\in[0,1]$, and for all such $y$ we have


Now just pick $a'$ so that $0<|a'-a|<\min\left\{\frac{\delta}2,\eta\right\}$.

  • $\begingroup$ Oh, OK. And what about my proof? $\endgroup$ – Pedro Tamaroff Nov 19 '12 at 21:03
  • $\begingroup$ @Peter: It appears to be okay. $\endgroup$ – Brian M. Scott Nov 19 '12 at 21:09

In step 1, when he mentions $$|\lim_{y\rightarrow a}f(y) - f(a)| > \epsilon $$ $$\Rightarrow \liminf_{y\rightarrow a}f(y) > f(a) + \epsilon $$ $$\Rightarrow f(y) > f(a) + \epsilon$$ for infinitely many $y \in N_{\delta}(a)$. So let one of those $y=a'$. Thus we get $f(a') - f(a) > \epsilon$. Since both sides are non negative, your desired result follows.

  • $\begingroup$ Thanks. Could this be done without $\liminf$ considerations?. What about my proof? $\endgroup$ – Pedro Tamaroff Nov 19 '12 at 20:57
  • $\begingroup$ The proof appears fine. At least I didn't see any flaws in it. $\endgroup$ – Gautam Shenoy Nov 19 '12 at 21:05

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