# Is every measure $0$ set a set of discontinuities of a Riemann integrable function?

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?

• 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 – Hagen von Eitzen Oct 24 '18 at 20:42
• @Hagen von Eitzen: Wouldn't the set of discontinuities of $\chi_D$ be $\text{bd}(D)$? – quasi Oct 24 '18 at 21:30
• 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. – Sasha Kozachinskiy Oct 24 '18 at 21:40
• @quasi If $\mu(D)$=0\$, the interioir should be empty, so boundary=closure. – Hagen von Eitzen Oct 24 '18 at 21:40
• @Hagen von Eitzen: Yes, of course. – quasi Oct 24 '18 at 21:41

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 $$% j since $$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.$$