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Suppose $f:[0,1]\rightarrow\mathbb{R}$ is a bounded function satisfying: for each $c\in [0,1]$ there exist the limits $\lim_{x\rightarrow c^+}f(x)$ and $\lim_{x\rightarrow c^-}f(x)$. Is true that the set of discontinuity of $f$ is countable?

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  • $\begingroup$ The same holds if we replace "there exists the limit $$\lim_{x\to c}f(x)$$ I'm wondering if I can adapt the proof I have from this to this statement. It is very similar to what copper.hat had written. In fact, we consider the sets $$\mathcal L(\epsilon)=\{x:|f(x)-\lim_{y\to x}f(y)|>\epsilon\}$$ and argue using GEdgar's idea that each of these sets is finite. $\endgroup$ – Pedro Tamaroff Mar 4 '13 at 23:51
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Hopefully this more tedious proof is more illustrative...

In light of the assumptions on $f$, there are two ways that $f$ can fail to be continuous: (1) the left and right hand limits differ, or (2) the limits are equal, but the function value differs from the limit (thanks to @GEdgar for pointing this out).

Let $\epsilon>0$ and $\Delta_\epsilon = \{ x |\, |\lim_{y \downarrow x}f(y) - \lim_{y \uparrow x}f(y)| \geq \epsilon \}$. If $\Delta_\epsilon$ is not finite, then since $[0,1]$ is compact, there exists an accumulation point $\hat{x} \in [0,1]$. Let $c_+ = \lim_{y \downarrow \hat{x}}f(y), c_- = \lim_{y \uparrow \hat{x}}f(y)$. (Note that it is possible that $c_- = c_+$.) By assumption, there exists some $\delta>0$ such that for $y \in (\hat{x}-\delta,\hat{x})$, $|f(y)-c_-| < \frac{\epsilon}{4}$ and for $y \in (\hat{x}, \hat{x}+\delta)$, $|f(y)-c_+| < \frac{\epsilon}{4}$. However, this implies that $\hat{x}$ is an isolated point of $\Delta_\epsilon$, which is a contradiction. Hence $\Delta_\epsilon$ is finite.

If we let $\Delta = \cup_n \Delta_{\frac{1}{n}}$, we see that $\Delta $ is at most countable.

@GEdgar has pointed out an omission in my proof: It is possible that the two limits coincide at a point, but the function is still discontinuous at that point, ie, $\Delta$ is not the entire set of discontinuities.

Let $\Gamma_\epsilon = \{ x |\, \lim_{y \downarrow x}f(y) = \lim_{y \uparrow x}f(y), \ |\lim_{y \downarrow x}f(y)-f(x) | \geq \epsilon \}$. Suppose, as above, that $\Gamma_\epsilon$ is not finite, and let $\hat{x}$ be an accumulation point. Let $c_+, c_-$ be the limits as above. Again, there exists a $\delta>0$ such that if $y \in (\hat{x}-\delta,\hat{x})$, $|f(y)-c_-| < \frac{\epsilon}{4}$, and similarly, if $y \in (\hat{x}, \hat{x}+\delta)$, $|f(y)-c_+| < \frac{\epsilon}{4}$. Consequently, $\hat{x}$ is isolated, hence a contradiction, and $\Gamma_\epsilon$ is finite.

If we let $\Gamma= \cup_n \Gamma_{\frac{1}{n}}$, we see that $\Gamma$ is at most countable.

Since the set of discontinuities is $\Delta \cup \Gamma$, we see that the set of discontinuities is at most countable.

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  • $\begingroup$ Discontinuity could also be $c_+ = c_- \ne f(c)$. $\endgroup$ – GEdgar Mar 4 '13 at 21:52
  • $\begingroup$ @GEdgar: I think it is fixed now, thanks! $\endgroup$ – copper.hat Mar 4 '13 at 23:59
  • $\begingroup$ Sorry to ask @copper.hat, but why $\hat{x}\in\Delta_\epsilon$ and if $\hat{x}\in\Delta_\epsilon$, why $c_-$ could be equal to $c_+$? $\endgroup$ – Tomás Mar 5 '13 at 11:15
  • $\begingroup$ @Tomás: It is not necessary that $\hat{x} \in \Delta_\epsilon$, all that is necessary is that in any neighborhood of $\hat{x}$, you can find points of $\Delta_\epsilon$. (In my earliest version, I incorrectly implied that $\hat{x} \in \Delta_\epsilon$.) $\endgroup$ – copper.hat Mar 5 '13 at 16:04
  • $\begingroup$ Oh, Ok. I am asking because this claim still contained in your proof. $\endgroup$ – Tomás Mar 5 '13 at 16:31
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Or try to go directly.

Assume the set $D$ of discontinuities is uncountable, select a condensation point of $D$ (a point where every neighborhood is $D$ is uncountable), then show that a one sided limits fails to exist there. I suggest the Cauchy criterion for nonexistence of a limit...

** more complete version **

If $f$ is discontinuous at a point $c$, then there is $\epsilon > 0$ so that for every $\delta>0$ there exists $x$ with $|x-c|<\delta$ and $|f(x)-f(c)|>\epsilon$. Let $D_\epsilon$ be the set of points satisfying this for a given $\epsilon$. The set of all discontinuities is $$ D = \bigcup_{\epsilon>0} D_\epsilon = \bigcup_{n=1}^\infty D_{1/n} $$ Suppose $D$ is uncountable. Then $D_{1/n}$ is uncountable for some $n$. Fix such an $n$. An uncountable set in $\mathbb R$ has an accuulation point, say $c$ is an accumulation point of $D_{1/n}$. Either every right-neighborhood of $c$ meets $D_{1/n}$ or every left-neighborhood of $c$ meets $D_{1/n}$.

Suppose every left-neighborhood of $c$ meets $D_{1/n}$; we will show that the left limit of $f$ does not exist at $c$. (The case for right-neighborhoods is similar, then the right limit does not exist.) Indeed, for any $\delta>0$, the interval $(c-\delta, c)$ meets $D_{1/n}$. Let $x \in (c-\delta,c)\cap D_{1/n}$. There is $\delta'>0$ so that $(x-\delta',x+\delta') \subset (c-\delta,c)$. By the definition of $D_{1/n}$, there is $y$ so that $|x-y|<\delta'$ and $|f(x)-f(y)|>1/n$. Summary:

For every $\delta>0$ there exists $x,y$ so that $c-\delta < x < c$, $c-\delta< y < c$, but $|f(x)-f(y)|>1/n$.

Thus the Cauchy criterion for existence of the left-hand limit $$ \lim_{t \to c^-} f(t) $$ fails.

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Yes as $f$ must be a regulated function and hence only has countable many discontinuities.

A regulated function is a function which has a right and a left hand limit. This is equivalent to that there is a sequence of step functions converging uniformly towards the function. As a step function only has finite many discontinuies, the regulated function can only have countable many discontinuies.

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  • $\begingroup$ Sorry, but what is a regular function? $\endgroup$ – Tomás Mar 4 '13 at 18:37
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    $\begingroup$ @Tomás: Regulated function $\endgroup$ – Martin Mar 4 '13 at 18:38
  • $\begingroup$ What is a regulated function? $\endgroup$ – copper.hat Mar 4 '13 at 18:39
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    $\begingroup$ Except that is not in your answer, it is buried in a comment. And it it still requires proof that you can approximate such functions in such a way by step functions. $\endgroup$ – Thomas Andrews Mar 4 '13 at 20:20
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    $\begingroup$ @ThomasAndrews this proof can be read in every analysis 1 script, the op asked whether it is true or not, I gave him the information he needs to google it and look it up. He didn't ask for a full proof $\endgroup$ – Dominic Michaelis Mar 4 '13 at 20:30
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If you consider the indicator function of the Cantor $1/3$-set $\chi_C(x)$ in $[0,1]$ then this function has uncountably many discontinuities since the Cantor set is uncountable. So, you need extra assumptions on $f(x)$, such as restricting the discontinuities only to jump-type.

Look also the answers of these questions:

How to show that a set of discontinuous points of an increasing function is at most countable

Theorems about functions with uncountable number of discontinuities

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