Csdrhrt

As stated above, why can Lebesgue integral be defined only with supremum
$$
int_E f , dmu := supleft{ int_E s , dmu : 0 le s le f, s text{ simple } right}.
$$

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.

Assume we are dealing with a finite interval $E=[a,b]$ with bounded function $f$. One can prove that

begin{align*}
sup_{sleq f}~(L)int_{E}sdx=inf_{fleq 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 $f^{ast}(x)=lim_{deltarightarrow 0}sup_{U_{delta}(x)cap E}f$, $f_{ast}(x)=lim_{deltarightarrow 0}inf_{U_{delta}(x)cap E}f$, 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.

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

$$0 leqslant infleft{ int_E t , dmu : t geqslant f, t text{ simple } right} - supleft{ int_E s , dmu : 0 leqslant s leqslant f, s text{ simple } right} \ leqslant int_E t_epsilon , dmu - int_E s_epsilon , dmu = int_E (t_epsilon - s_epsilon) , dmu < epsilon , mu(E).$$

Since this is true for any $epsilon > 0$ we always have

$$infleft{ int_E t , dmu : t geqslant f, t text{ simple } right} = supleft{ int_E s , dmu : 0 leqslant s leqslant f, s text{ simple } right}, $$

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 $f = f^+ - f^-.$