TITLE: Lebesgue measurable but not Riemann integrable
QUESTION [7 upvotes]: Every bounded function $f:[a,b]\to\mathbb R$, which is Riemann integrable, it is also Lebesgue integrable.
On the other hand 
$$
g(x)=\left\{
\begin{array}{lll}
1 & \text{if} & x\in\mathbb Q\cap[0,1], \\
0 & \text{if} & x\in [0,1]\setminus\mathbb Q,
\end{array}
\right. \tag{1}
$$
i.e., $g=\chi_{[0,1]\cap\mathbb Q}$, is Lebesgue integrable in $[0,1]$, but not Riemann integrable.
In fact, a bounded function $f:[a,b]\to\mathbb R$ is Riemann integrable if and only if it is almost everywhere continuous.
Now, the function $g=\chi_{[0,1]\cap\mathbb Q}$ is not Riemann integrable BUT is it  almost everywhere equal to $h\equiv 0$, which IS Riemann integrable.
My question is the following:
Is there a bounded function $f:[a,b]\to\mathbb R$, which is Lebesgue integrable, and it is not equal almost everywhere to a Riemann integrable function?

REPLY [6 votes]: The characteristic function of a "fat Cantor set" (a Cantor set with positive measure), is Lebesgue integrable, but is not equal almost everywhere to a Riemann integrable function, because every point of the fat Cantor set is a discontinuity of its characteristic function.