Supremum of Lower Sums Never Greater than Upper Sum

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Theorem

Let $\closedint a b$ be a closed real interval .

Let $f$ be a bounded real function defined on $\closedint a b$.

Let $S$ be a finite subdivision of $\closedint a b$.

Let $\map U S$ be the upper sum of $f$ on $\closedint a b$ with respect to $S$.

Let $\map L P$ be the lower sum of $f$ on $\closedint a b$ with respect to a finite subdivision $P$.


Then:

$\sup_P \map L P \le \map U S$


Proof

From Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, $\map U S$ is an upper bound for the real set:

$T = \leftset {\map L P: P}$ is a finite subdivision of $\rightset {\closedint a b}$

Since $\sup_P \map L P$ is the supremum of $T$:

$\sup_P \map L P \le \map U S$

Hence the result.

$\blacksquare$