Upper Integral Never Smaller than Lower 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 integral of $f$ over $\closedint a b$ is less than or equal to the upper 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 integral be $L$, and the value of the upper integral be $U$.

, suppose $L > U$.

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

By the definitions of lower integral and upper 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 and upper 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, $L \le U$.