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?

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 $E_{n}=%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
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\cap C_{n}^{0}$. 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 $\omega (f,x)\geq \frac{1}{k!%
}- _{j=k+1}^{\infty }\frac{1}{j!}\geq \frac{1}{k!}%
[1- _{j=k+1}^{\infty }\frac{1}{(k+1)(k+2)...(j)}]>\frac{1}{k!}%
[1- _{j=k+1}^{\infty }\frac{1}{2^{j-k}}]=0.$
