Upper Sum Never Smaller than Lower Sum

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Theorem

Let $\left[{a \,.\,.\, b}\right]$ be a closed interval of the set $\R$ of real numbers.

Let $P = \left\{{x_0, x_1, x_2, \ldots, x_{n-1}, x_n}\right\}$ be a finite subdivision of $\left[{a \,.\,.\, b}\right]$.

Let $f: \R \to \R$ be a real function.

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

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

Let $U \left({P}\right)$ be the upper sum of $f \left({x}\right)$ on $\left[{a \,.\,.\, b}\right]$ belonging to the subdivision $P$.


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


Proof

For all $\nu \in 1, 2, \ldots, n$, let $\left[{x_{\nu - 1} \,.\,.\, x_{\nu}}\right]$ be a closed subinterval of $\left[{a \,.\,.\, b}\right]$.


As $f$ is bounded on $\left[{a \,.\,.\, b}\right]$, it is bounded on $\left[{x_{\nu - 1} \,.\,.\, x_{\nu}}\right]$.

So, let $m_\nu$ be the infimum and $M_\nu$ be the supremum of $f \left({x}\right)$ on the interval $\left[{x_{\nu - 1} \,.\,.\, x_{\nu}}\right]$.


By definition, $m_\nu \le M_\nu$.

So $m_{\nu} \left({x_{\nu} - x_{\nu - 1}}\right) \le M_{\nu} \left({x_{\nu} - x_{\nu - 1}}\right)$.

It follows directly that $\displaystyle \sum_{\nu \mathop = 1}^n m_{\nu} \left({x_{\nu} - x_{\nu - 1}}\right) \le \sum_{\nu \mathop = 1}^n M_{\nu} \left({x_{\nu} - x_{\nu - 1}}\right)$.

$\blacksquare$