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Definition:

$f$ is Darboux Integrable $\iff \sup{\{L(f,P): P \text{ partition of } [a,b]\}}=\inf{\{U(f,P): P \text{ partition of } [a,b]\}}$

Let $f(x)=x^2$ and $P$ a partition of $[a,b]$.

Then how can $\sup{\{L(f,P): P \text{ partition of } [a,b]\}}$ and $\inf{\{U(f,P): P \text{ partition of } [a,b]\}}$ ever be equal if $P$ is a finite sequence of numbers from the interval $[a,b]$?

EDIT: Is it because they will never reach that value but they can take all values in between?

If so then if a function $g$ is integrable on $[a,b]$, then I can't say that $\exists P',P'': U(g,P')=L(g,P'')$ right?

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    $\begingroup$ The $\sup$ and $\inf$ is over all possible partitions. $\endgroup$ – Winther Jun 11 '17 at 13:38
  • $\begingroup$ ... which is an infinite set. $\endgroup$ – md2perpe Jun 11 '17 at 13:43
  • $\begingroup$ @Winther So if a function $g$ is integrable on $[a,b]$, then I can't say that $\exists P',P'': U(g,P')=L(g,P'')$ right? $\endgroup$ – ZeroPancakes Jun 11 '17 at 13:47
  • $\begingroup$ If $g$ is constant then you can, but not in general no just as $\inf_{n>0} \frac{1}{n} = 0$ and $\sup_{n>0} -\frac{1}{n} = 0$ does not mean that there is $n,m$ such that $\frac{1}{n} = - \frac{1}{m}$ (or $n$ such that $\frac{1}{n} = 0$). $\endgroup$ – Winther Jun 11 '17 at 13:50
  • $\begingroup$ @Winther Many thanks! $\endgroup$ – ZeroPancakes Jun 11 '17 at 13:51
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Given a partition $P$ (namely a finite set $ a=x_0 < x_1 < \dotsb < x_n=b$), it is easy to show that $U(f,P)=L(f,P)$ if and only if $f$ is constant on each interval of $P$: $$U(f,P)-L(f,P) = \sum_i \lvert x_i-x_{i-1} \rvert \Big( \sup_{[x_{i-1},x_i]} f - \inf_{[x_{i-1},x_i]} f \Big), $$ and $\sup_{A} f - \inf_{A} f \geq 0$, with equality if and only if $f$ is constant on $A$. Moreover, since $f(x_i)$ is considered in both $[x_{i-1},x_i]$ and $[x_i,x_{i+1}]$, we see that neighbouring intervals have to have the same constant value of $f$. Hence, by induction, $f$ is constant.

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