Upper Darboux Integral Never Smaller than Lower Darboux Integral

Theorem

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

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

The lower Darboux integral of $f$ over $\closedint a b$ is less than or equal to the upper Darboux integral of $f$ over the same bounds.

That is:

$\ds \underline {\int_a^b} \map f x \rd x \le \overline {\int_a^b} \map f x \rd x$

Proof

Let the value of the lower Darboux integral be $L$, and the value of the upper Darboux integral be $U$.

Aiming for a contradiction, suppose, suppose $L > U$.

Then $\epsilon = \dfrac {L - U} 2$ is positive.

By the definitions of lower Darboux integral and upper Darboux integral:

$\ds \sup_P \map L P > \inf_P \map U P$

where $P$ ranges over all subdivisions of $\closedint a b$, and $\map L P$ and $\map U P$ are the lower Darboux sum and upper Darboux sum, respectively.

By the Characterizing Property of Supremum of Subset of Real Numbers, there exists are partition $P_L$ such that:

$\map L {P_L} > L - \epsilon = \dfrac {L + U} 2$

Likewise, by the Characterizing Property of Infimum of Subset of Real Numbers, there is a partition $P_U$ such that:

$\map U {P_U} < U + \epsilon = \dfrac {U + L} 2 $

Thus:

$\map U {P_U} < \map L {P_L}$

But, by Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, this is a contradiction.

Therefore, $L \le U$.

$\blacksquare$