Supremum of Lower Sums Never Greater than Upper Sum

From ProofWiki
Jump to navigation Jump to search


Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval .

Let $f$ be a bounded real function defined on $\left[{a \,.\,.\, b}\right]$.

Let $S$ be a finite subdivision of $\left[{a \,.\,.\, b}\right]$.

Let $U \left({S}\right)$ be the upper sum of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to $S$.

Let $L \left({P}\right)$ be the lower sum of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to a finite subdivision $P$.


$\sup_P L \left({P}\right) \le U \left({S}\right)$


Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions gives that $U \left({S}\right)$ is an upper bound for the real set {$L \left({P}\right)$: $P$ is a finite subdivision of $\left[{a \,.\,.\, b}\right]$}.

Since $\sup_P L \left({P}\right)$ is the least upper bound of the same set, we find:

$\sup_P L \left({P}\right) \le U \left({S}\right)$