The set of points where a function $f:[a,b]\to\mathbb R$ is discontinuous is Lebesgue measurable

Question

Prove that the set of points where a function $f:[a,b]\to\mathbb R$ is discontinuous is Lebesgue measurable.

Lebesgue measure of set $A$ means that for any set $S\in\mathbb R,$ $m^*(S)=m^*(A\cap S)+m^*(A^c\cap S)$.

I think I should start with the definition of the negation of continuous function,

$\exists\epsilon>0,\forall\delta>0,\exists x\in D_f:\vert x-x_0\vert<\delta$ and $\vert f(x)-f(x_0)\vert\ge\epsilon$

and then should I see that $(x_0+\delta,x_0-\delta)$ is Lebesgue measure ?

Am I correct?

You should probably start by getting the negation correct. :) And, of course any open interval is Lebesgue-measurable, so that's certainly not the point. — Ted Shifrin, Mar 19 '19 at 23:49
See appendix M10 of Patrick Billingsley's "Convergence of Probability Measures" for a proof. — John, Mar 19 '19 at 23:58
See also: How to show that the set of points of continuity is a $G_{\delta}$. — Martin Sleziak, Mar 20 '19 at 06:12
@MartinSleziak Thank you. What's $G_\delta$ ? And how is this related to the solution of my exercise? — , Mar 20 '19 at 21:27
Wikipedia: Gδ set: A countable intersection of open sets. Every Borel measurable set is Lebesgue measurable. (In short, if you know that countable intersection of measurable sets is measurable and that every open set is measurable, you immediately get that any $G_\delta$-set is measurable. — Martin Sleziak, Mar 21 '19 at 04:30

Kavi Rama Murthy · Accepted Answer · 2019-03-20T05:13:48.333

2

Hint: if nothing is known about $f$ except that it maps $\mathbb R$ into itself you can proceed as follows: let $\omega(x) =\lim_{\delta \to 0} \sup \{|f(x)-f(y)|: |y-x|\ \leq \delta \}$. You can verify that $\{x:\omega (x) <t\}$ is open for each real number $t$. Hence $\omega$ is a Borel measurable function. Now verify that $f$ is continuous at $x$ iff $\omega (x)=0$. So the points of discontinuity form a Borel set. No need to worry about Lebesgue measurability.

edited Mar 20 '19 at 05:13

answered Mar 19 '19 at 23:57

Kavi Rama Murthy

311,013

Is the $a$ from ${x:\omega (x) <a}$ the $a$ of the domain ? – Mar 20 '19 at 02:45
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@Isa Sorry, I have changed it to $t$. – Kavi Rama Murthy Mar 20 '19 at 05:14
thank you Kavi Rama Murthy – May 09 '19 at 02:22
I wasn't able to show it following this approach. I instead followed Folland's approach in the exercise proving part (b) of Theorem 2.28. In particular I got stuck at showing ${x: \omega(x) < t}$ is open in $[a,b]$. How do you show that? – Simon SMN Nov 14 '22 at 17:38

The set of points where a function $f:[a,b]\to\mathbb R$ is discontinuous is Lebesgue measurable

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