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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?

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    $\begingroup$ 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 $\endgroup$ Oct 24, 2018 at 20:42
  • $\begingroup$ @Hagen von Eitzen: Wouldn't the set of discontinuities of $\chi_D$ be $\text{bd}(D)$? $\endgroup$
    – quasi
    Oct 24, 2018 at 21:30
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    $\begingroup$ 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. $\endgroup$ Oct 24, 2018 at 21:40
  • $\begingroup$ @quasi If $\mu(D)$=0$, the interioir should be empty, so boundary=closure. $\endgroup$ Oct 24, 2018 at 21:40
  • $\begingroup$ @Hagen von Eitzen: Yes, of course. $\endgroup$
    – quasi
    Oct 24, 2018 at 21:41

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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 $\omega (f,x)\geq \frac{1}{k!% }- _{j=k+1}^{\infty }\frac{1}{j!}\geq \frac{1}{k!}% [1- _{j=k+1}^{\infty }\frac{1}{(k+1)(k+2)...(j)}]>\frac{1}{k!}% [1- _{j=k+1}^{\infty }\frac{1}{2^{j-k}}]=0.$

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