2
$\begingroup$

Let $f:[a,b]\rightarrow\mathbb{R}$ be bounded, and let $D$ be its set of discontinuities. Then Lebesgue's criterion states that $f$ is Riemann-integrable if and only if $D$ has Lebesgue measure $0$.

My question is, for any subset $D$ of $[a,b]$ with Lebesgue measure $0$, does there exist a Riemann integrable function $f:[a,b]\rightarrow\mathbb{R}$ whose set of discontinuities is $D$? Would the characteristic function of $D$ suffice, or is more complicated than that?

$\endgroup$
  • 1
    $\begingroup$ The set of discontinuities of $\chi_D$ would be $\overline D$. For $D=\Bbb Q$, we can take $f(x)=\begin{cases}\frac 1n&x=\frac mn\\0&\text{else}\end{cases}$ instead $\endgroup$ – Hagen von Eitzen Oct 24 '18 at 20:42
  • $\begingroup$ @Hagen von Eitzen: Wouldn't the set of discontinuities of $\chi_D$ be $\text{bd}(D)$? $\endgroup$ – quasi Oct 24 '18 at 21:30
  • 2
    $\begingroup$ According to Wikipedia en.wikipedia.org/wiki/Classification_of_discontinuities the set of discontinuities is a Borel set. According to the answer here math.stackexchange.com/questions/1742137/… not every set of Lebesgue measure $0$ is Borel. $\endgroup$ – Sasha Kozachinskiy Oct 24 '18 at 21:40
  • $\begingroup$ @quasi If $\mu(D)$=0$, the interioir should be empty, so boundary=closure. $\endgroup$ – Hagen von Eitzen Oct 24 '18 at 21:40
  • $\begingroup$ @Hagen von Eitzen: Yes, of course. $\endgroup$ – quasi Oct 24 '18 at 21:41
2
$\begingroup$

It is well known that any $F_{\sigma}$ set is the set of discontinuities of some function. We can also make this function bounded. So any $F_{\sigma}$ set of measure $0$ is the set of discontinuities of a Riemann integrable function. As pointed out in the comments not every of measure $0$ is the set of discontinuities of a Riemann integarble function.

Proof of the fact that any $F_{\sigma}$ set is the set of discontinuities of a bounded function:

Let $A=\cap_{n=1}^{\infty }G_{n}$ with $G_{n}$ open and $% G_{n+1}\subset G_{n}$ for all $n$. Let $f_{n}=I_{C_{n}\backslash E_{n}\text{ }}$where $C_{n}=G_{n}^{c}$ and $E_{n}=% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \cap C_{n}^{0}$. Let $f= \sum_{n=1}^{\infty } \frac{1}{n!}f_{n}$. We claim that $f$ has the desired properties. First let $x\in A$. Then $% f_{n}(x)=0$ for all $n$. In fact, for each $n$, $f_{n}$ vanishes in a neighbourhood of $x$. Hence each $f_{n}$ is continuous at $x$. By uniform convergence of the series defining $f$ we see that $f$ is also continuous at $x$. Now let $x\in A^{c}.$ Let $k$ be the least positive integer such that $% x\in C_{k}$. If $x\in C_{k}^{0}$ then, in sufficiently small neighbourhoods of $x,$ $f_{k}$ take both the values $0$ and $1$ and so its oscillation at $% x $ is $1$. We claim that the oscillation of $f_{j}$ at $x$ is $0$ for each $% j<k:$ since $x\notin C_{j}$ it follows that points close to $x$ are all in $% C_{j}^{c}$ and hence $f_{j}$ vanishes at those points. Now $\omega (f,x)\geq \frac{1}{k!}\omega (f_{k},x)- _{j=k+1}^{\infty }\frac{1}{j!}$ since $\omega (f_{j},.)\leq 1$ everywhere. Thus $\omega (f,x)\geq \frac{1}{k!% }- _{j=k+1}^{\infty }\frac{1}{j!}\geq \frac{1}{k!}% [1- _{j=k+1}^{\infty }\frac{1}{(k+1)(k+2)...(j)}]>\frac{1}{k!}% [1- _{j=k+1}^{\infty }\frac{1}{2^{j-k}}]=0.$

$\endgroup$

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.