Why can Lebesgue integral be defined only with supremum, but Riemanns needs both sup. and inf.? As stated above, why can Lebesgue integral be defined only with supremum
$$
 \int_E f \, d\mu := \sup\left\{ \int_E s \, d\mu : 0 \le s \le f, s \text{ simple } \right\}.
$$
but the Riemann integral needs to be expressed in terms of both the sup and inf?
What is the key difference? It seems important to me, but everywhere, I jut see the definition, without an explanation, why it is defined like this.
Please help me understand, I can not find an explanation anywhere.
 A: If $f$ is bounded and measurable on a set $E \subset \mathbb{R}$ then for any $\epsilon>0$ there are simple functions $s_\epsilon$ and $t_\epsilon$ such that $s_\epsilon \leqslant f \leqslant t_\epsilon$ and $t_\epsilon - s_\epsilon < \epsilon$.  This is not difficult to prove and $E$ need not be bounded.
If, in fact, $E$ is bounded, then by the basic properties of integrals of simple functions (monotone and linear) we have
$$0 \leqslant \inf\left\{ \int_E t \, d\mu :    t \geqslant f, t \text{ simple } \right\} - \sup\left\{ \int_E s \, d\mu : 0 \leqslant s \leqslant f, s \text{ simple } \right\} \\ \leqslant \int_E t_\epsilon \, d\mu - \int_E s_\epsilon \, d\mu  =  \int_E (t_\epsilon - s_\epsilon) \, d\mu  < \epsilon \, \mu(E).$$
Since this is true for any $\epsilon > 0$ we always have 
$$\inf\left\{ \int_E t \, d\mu : t \geqslant f, t \text{ simple } \right\} = \sup\left\{ \int_E s \, d\mu : 0 \leqslant s \leqslant f, s \text{ simple } \right\}, $$
and either the $\inf$ or $\sup$ defines the Lebesgue integral. The Riemann integral need not exist.
If, however, $f$ is nonnegative and either $f$ is unbounded or $\mu(E) = \infty$ or both, then the integral is defined as 
$$\int_E f = \sup \left\{\int_E g: 0 \leqslant g \leqslant f, \, g \text{ bounded, measurable, and of finite support} \right\}.$$
Such $\int_E g$ are defined in the first part and this can be extended in general using $f = f^+ - f^-.$
A: Assume we are dealing with a finite interval $E=[a,b]$ with bounded function $f$. One can prove that
\begin{align*}
\sup_{s\leq f}~(L)\int_{E}sdx=\inf_{f\leq t}~(L)\int_{E}tdx
\end{align*}
if $f$ is measurable, where $(L)$ stands for the Lebesgue integral, here $s,t$ run through all the simple functions.
For the upper Darboux integral $\overline{S}(f)$, one has 
\begin{align*}
(L)\int_{E}f^{\ast}dx=\overline{S}(f),
\end{align*}
and the lower Darboux integral $\underline{S}(f)$, one has
\begin{align*}
(L)\int_{E}f_{\ast}dx=\underline{S}(f),
\end{align*}
where $f^{\ast}(x)=\lim_{\delta\rightarrow 0}\sup_{U_{\delta}(x)\cap E}f$, $f_{\ast}(x)=\lim_{\delta\rightarrow 0}\inf_{U_{\delta}(x)\cap E}f$, where $U_{\delta}(x)$ stands for the deleted ball centered at $x$ with radius $\delta>0$.
If $f$ were Riemann integrable, then 
\begin{align*}
(R)\int_{E}fdx=\overline{S}(f)=\underline{S}(f)=(L)\int_{E}f^{\ast}dx=(L)\int_{E}f_{\ast}dx.
\end{align*}
One can see that for a bounded measurable function $f$ on a bounded set, one need no to define for the lower one for Lebesgue integral, but for Riemann integral, one needs to do so. Hope this helps.
