Upper Sum Never Smaller than Lower Sum

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

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

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

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

Let $$L$$ be the lower sum of $f \left({x}\right)$ on $\left[{a \, . \, . \, b}\right]$ belonging to the subdivision $\left\{{x_0, x_1, x_2, \ldots, x_{n-1}, x_n}\right\}$.

Let $$U$$ be the upper sum of $f \left({x}\right)$ on $\left[{a \, . \, . \, b}\right]$ belonging to the subdivision $\left\{{x_0, x_1, x_2, \ldots, x_{n-1}, x_n}\right\}$.

Then $$L \le U$$.

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 always bounded on $$\left[{x_{\nu - 1} \,. \, . \, x_{\nu}}\right]$$

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 $$\sum_{\nu=1}^n m_{\nu} \left({x_{\nu} - x_{\nu - 1}}\right) \le \sum_{\nu=1}^n M_{\nu} \left({x_{\nu} - x_{\nu - 1}}\right)$$.