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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.

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    $\begingroup$ Riemann integral only partitions in intervals. That means that a minimum point within one interval of the partition will drive the reactangle's area down for all the interval. This means that the supremum of the lower sums could be finite but the upper sums being somewhere else and even going to infinity. On the other hand, the Lebesgue integral's simple functions are rectangles in which the bases are not just intervals. Therefore, they can go around those points. $\endgroup$
    – user547557
    Commented Apr 4, 2018 at 2:02
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    $\begingroup$ Are you aware that functions can be approximated by simple functions? $\endgroup$
    – IAmNoOne
    Commented Apr 4, 2018 at 2:27

2 Answers 2

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

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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.

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  • $\begingroup$ Thank you for your answer. Just to see if I understand. Your first equation tells why we only need sup for Lebesgue integral, because it implies the inf.? $\endgroup$ Commented Apr 4, 2018 at 2:08
  • $\begingroup$ Yes, if it were measurable, bounded, and on a bounded set. $\endgroup$
    – user284331
    Commented Apr 4, 2018 at 2:08
  • $\begingroup$ I accepted, as It answered my question. Can you point me in the right direction, as to how to generalize this? Lets say to the entire real line; it can be covered by a countable union of bounded intervals on which it holds. Is this OK? $\endgroup$ Commented Apr 4, 2018 at 2:11
  • $\begingroup$ The point is, for a proper Riemann integral, the domain of integral is a finite closed interval. If you want to say something about improper Riemann integral, it is another story. But you know that there is some function which is improper Riemann integrable but not Lebesgue integrable. $\endgroup$
    – user284331
    Commented Apr 4, 2018 at 2:13

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