Q:

Riemann/Lebesgue integrability of a function

Let $f_n:[0,1]\to\mathbb{R}$ be a function defined as:
$$f_n(x)=\begin{cases}1\ \text{if}\ x=r_n\\0\ \text{otherwise}\end{cases}$$
where $r_n$ is an enumeration of rationals in $[0,1]$. Then, is the limit  function $f=\lim_{n\to\infty}f_n$ Riemann integrable/Lebesgue integrable? 
I think the function is Lebesgue integrable, as its absolute value is bounded, since 
$$f(x)=\begin{cases}1\ \text{if}\ x\in\mathbb{Q}\\0\ \text{otherwise}\end{cases}$$.
The function is easily seen to be non-Riemann integrable because of nowhere continuity.  Am I right?  Thanks beforehand. 

A:

$f$ is not Riemann integrable, that is true, but the reason is not of missing any continuity.  It is just that the upper Darboux sum will always be 1 and the lower Darboux sum will always be zero given any partition of $ [0,1]$ and therefore the function is not Riemann integrable.