Will the Lebesgue integral of a real valued function always be a Riemann sum? If we have a real valued integral that is Lebesgue integrable but not Riemann integrable, can the value of the Lebesgue integral be given by a Riemann sum by choosing appropriate points in the partitioning process?
Example:
The indicator function for the irrationals: 
$$f(x)=\begin{cases} 1 &\mbox{if } x\not\in \mathbb{Q}\\
                     0 &\mbox{if } x \in \mathbb{Q} \end{cases}$$
has Lebesgue integral $1$ on $[0,1]$. This is the same value we would get if we formed the Riemann sum using irrational endpoints.  
 A: There are (at least) two ways to interpret your question. The first
is that you are asking if the value of the integral is actually given
by a Riemann sum. This is not the case, even for Riemann integrable
functions. Consider for example the function
$$
f:\left[0,1\right]\to\mathbb{R},x\mapsto\begin{cases}
1, & x\in\left[0,\frac{1}{\sqrt{2}}\right],\\
0, & \text{else}.
\end{cases}
$$
Then the integral $\int f\,{\rm d}x=\frac{1}{\sqrt{2}}$ is irrational,
but since $f$ only attains rational values, any Riemann sum
$$
\sum_{i=1}^{n}\frac{1}{n}f\left(\xi_{i}\right)\in\mathbb{Q}
$$
can never equal $\int f\,{\rm d}x$.
The second interpretation of the question is that you are asking if
there always exists a sequence $\left(\xi_{i}^{\left(n\right)}\right)_{n\in\mathbb{N},i\in\left\{ 1,\dots,n\right\} }$
such that we have $\xi_{i}^{\left(n\right)}\in\left[\frac{i-1}{n},\frac{i}{n}\right]$
and
$$
\int_{0}^{1}f\,{\rm d}x=\lim_{n\to\infty}\sum_{i=1}^{n}\frac{f\left(\xi_{i}^{\left(n\right)}\right)}{n}.
$$
Here, I am assuming (without loss of generality) that $f$ is defined
on $\left[0,1\right]$.
This is indeed true. For the proof, I will use the following well-known fact: If we define
$$
f_{n}:=\sum_{i=1}^{n}\left[\oint_{\frac{i-1}{n}}^{\frac{i}{n}}f\left(y\right)\,{\rm d}y\cdot\chi_{\left(\frac{i-1}{n},\frac{i}{n}\right)}\right],
$$
then $f_{n}\to f$ in $L^{1}$. Here, I am writing $\oint_A f \, dx := \frac{1}{\lambda(A)} \int_A f(x) \,dx$ for the mean value of $f$ over $A$, since $\fint$ does not seem to work.
The proof of this goes essentially
as follows: Define the operators $\Phi_{n}:L^{1}\to L^{1},f\mapsto f_{n}$.
This is easily seen to be bounded with $\left\Vert \Phi_{n}\right\Vert \leq1$
and for $f\in C\left(\left[0,1\right]\right)$ (which is a dense subset
of $L^{1}$), the convergence $f_{n}\to f$ is immediate. Now apply
a density argument.
For the proof of the actual claim, I will assume $f$ to be bounded, although this additional
assumption can probably(!) be removed. Note that we then also have
$\left\Vert f_{n}\right\Vert _{\infty}\leq\left\Vert f\right\Vert _{\infty}$
for all $n\in\mathbb{N}$.
Now, let $\varepsilon_{n}:=\left\Vert f_{n}-f\right\Vert _{L^{1}}$
and note $\varepsilon_{n}\to0$. To avoid trivialities, we will assume
$\varepsilon_{n}>0$ for all $n$. If this does not hold, we have
$f=f_{n}$ almost everywhere, so that $f$ is Riemann integrable.
In this case, the claim is trivial.
Now, the idea is to divide the intervals $\left(\frac{i-1}{n},\frac{i}{n}\right)$
for each $n\in\mathbb{N}$ into the "good" and the "bad" ones.
The set of (indices of) bad intervals will be
$$
I:=\left\{ i\in\left\{ 1,\dots,n\right\} \,\mid\,\forall x\in\left(\frac{i-1}{n},\frac{i}{n}\right):\:\left|f_{n}\left(x\right)-f\left(x\right)\right|\geq\sqrt{\varepsilon_{n}}\right\} .
$$
Note that we have
\begin{eqnarray*}
\varepsilon_{n} & > & \left\Vert f_{n}-f\right\Vert _{L^{1}}\\
 & \geq & \sum_{i\in I}\int_{\left(i-1\right)/n}^{i/n}\left|f_{n}\left(x\right)-f\left(x\right)\right|\,{\rm d}x\\
 & \geq & \sum_{i\in I}\frac{1}{n}\cdot\sqrt{\varepsilon_{n}}\\
 & = & \frac{\sqrt{\varepsilon_{n}}}{n}\cdot\left|I\right|
\end{eqnarray*}
and thus $\frac{\left|I\right|}{n}\leq\sqrt{\varepsilon_{n}}$.
Now, by definition of $I$ there is for each $i\in I^{c}=\left\{ 1,\dots,n\right\} \setminus I$
some $\xi_{i}\in\left(\frac{i-1}{n},\frac{i}{n}\right)$ with $\left|f_{n}\left(\xi_{i}\right)-f\left(\xi_{i}\right)\right|<\sqrt{\varepsilon_{n}}$.
For each $i\in I$, choose an arbitrary $\xi_{i}\in\left(\frac{i-1}{n},\frac{i}{n}\right)$.
Note that $f_{n}$ is constant on each of the intervals $\left(\frac{i-1}{n},\frac{i}{n}\right)$,
so that we have
$$
f_{n}\left(\xi_{i}\right)=f_{n}\left(x\right)\qquad\forall x\in\left(\frac{i-1}{n},\frac{i}{n}\right)
$$
and thus
$$
\int_{\left(i-1\right)/n}^{i/n}f\left(x\right)\,{\rm d}x=f_{n}\left(\xi_{i}\right)\cdot\int_{\left(i-1\right)/n}^{i/n}\,{\rm d}x=\frac{f_{n}\left(\xi_{i}\right)}{n}
$$
for all $i\in\left\{ 1,\dots,n\right\} $. This finally yields
\begin{eqnarray*}
\left|\int_{0}^{1}f\left(x\right)\,{\rm d}x-\sum_{i=1}^{n}\frac{f\left(\xi_{i}\right)}{n}\right| & \leq & \left|\int_{0}^{1}f\left(x\right)\,{\rm d}x-\int_{0}^{1}f_{n}\left(x\right)\,{\rm d}x\right|+\left|\int_{0}^{1}f_{n}\left(x\right)\,{\rm d}x-\sum_{i=1}^{n}\frac{f\left(\xi_{i}\right)}{n}\right|\\
 & \leq & \left\Vert f-f_{n}\right\Vert _{L^{1}}+\left|\sum_{i=1}^{n}\left[\int_{\left(i-1\right)/n}^{i/n}f_{n}\left(x\right)\,{\rm d}x-\frac{f\left(\xi_{i}\right)}{n}\right]\right|\\
 & \leq & \varepsilon_{n}+\left|\sum_{i=1}^{n}\frac{f_{n}\left(\xi_{i}\right)-f\left(\xi_{i}\right)}{n}\right|\\
 & \leq & \varepsilon_{n}+\sum_{i\in I}\frac{\left|f_{n}\left(\xi_{i}\right)\right|+\left|f\left(\xi_{i}\right)\right|}{n}+\sum_{i\in I^{c}}\frac{\left|f_{n}\left(\xi_{i}\right)-f\left(\xi_{i}\right)\right|}{n}\\
 & \leq & \varepsilon_{n}+2\left\Vert f\right\Vert _{\infty}\cdot\frac{\left|I\right|}{n}+\sum_{i\in I^{c}}\frac{\sqrt{\varepsilon_{n}}}{n}\\
 & \leq & \varepsilon_{n}+2\left\Vert f\right\Vert _{\infty}\cdot\sqrt{\varepsilon_{n}}+\frac{\left|I^{c}\right|}{n}\cdot\sqrt{\varepsilon_{n}}\\
 & \leq & \varepsilon+\sqrt{\varepsilon}\cdot\left(1+2\left\Vert f\right\Vert _{\infty}\right)\\
 & \xrightarrow[n\to\infty]{} & 0.
\end{eqnarray*}
This completes the proof, but note once more that I assumed $f$ to
be bounded.
