Lebesgue measurable but not Riemann integrable

Question

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?

Answer 1

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.