Upper Sum Never Smaller than Lower Sum

Theorem
Let $\closedint a b$ be a closed interval of the set $\R$ of real numbers.

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

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

Let $f$ be bounded on $\closedint a b$.

Let $\map L P$ be the lower sum of $\map f x$ on $\closedint a b$ belonging to the subdivision $P$.

Let $\map U P$ be the upper sum of $\map f x$ on $\closedint a b$ belonging to the subdivision $P$.

Then $\map L P \le \map U P$.

Proof
For all $\nu \in 1, 2, \ldots, n$, let $\closedint {x_{\nu - 1} } {x_\nu}$ be a closed subinterval of $\closedint a b$.

As $f$ is bounded on $\closedint a b$, it is bounded on $\closedint {x_{\nu - 1} } {x_\nu}$.

So, let $m_\nu$ be the infimum and $M_\nu$ be the supremum of $\map f x$ on the interval $\closedint {x_{\nu - 1} } {x_\nu}$.

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

So $m_\nu \paren {x_\nu - x_{\nu - 1} } \le M_{\nu} \paren {x_\nu - x_{\nu - 1} }$.

It follows directly that $\ds \sum_{\nu \mathop = 1}^n m_\nu \paren {x_\nu - x_{\nu - 1} } \le \sum_{\nu \mathop = 1}^n M_\nu \paren {x_\nu - x_{\nu - 1} }$.