# Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions

## Theorem

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

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

Let $P$ and $Q$ be finite subdivisions of $\left[{a \,.\,.\, b}\right]$.

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

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

Then $L \left({P}\right) \le U \left({Q}\right)$.

## Proof

Let $P' = P \cup Q$.

We observe:

$P'$ is either equal to $P$ or finer than $P$
$P'$ is either equal to $Q$ or finer than $Q$

We find:

$L \left({P}\right) \le L \left({P'}\right)$ by the definition of lower sum and $P'$ refining $P$
$L \left({P'}\right) \le U \left({P'}\right)$ by Upper Sum Never Smaller than Lower Sum
$U \left({P'}\right) \le U \left({Q}\right)$ by the definition of upper sum and $P'$ refining $Q$

By combining these inequalities, we conclude:

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

$\blacksquare$