Theorem – Lower sum <= upper sum

Contents
1. Theorem
2. Proof
3. Consequences of the theorem.

Theorem

Let $P$ and $Q$ be any two partitions of an interval $[a,b] \subset \mathbb{R}$ , and let $f$ be bounded on $[a,b]$ . Then, we have that

$L(f, Q) \leq U(f,P)$

Ie, the lower sum is always less than or equal to the upper sum.

Proof

Recall from this theorem that, for two partitions $S$ and $Q$ of $[a,b]$ , where $S$ is a refinement of $Q$ , we have,

$L(f, S) \leq L(f, Q) \leq U(f, Q) \leq U(f, S) \quad (1)$

If we define $S = A \cup B$ , then $S$ is a refinement of both $P$ and $Q$ , and so with $S = A \cup B$ , we can re-write $(1)$ as two different sets of inequalities,

$\begin{array}{cc} L(f, S) \leq L(f, P) \leq U(f, P) \leq U(f, S) \\\\ L(f, S) \leq L(f, Q) \leq U(f, Q) \leq U(f, S) \end{array}$

Hence, we have three cases. Firstly, if $U(f, Q) < U(f, P)$ , then $L(f,Q)\leq U(f, Q) < U(f, P)$ , so obviously $L(f, Q) \leq U(f,P)$ .

Secondly, if $U(f, Q) = U(f, P)$ , then it is obvious that $L(f, Q) \leq U(f, P)$ (this just follows from $(1)$ ).

Finally, if $U(f, Q) > U(f, P)$ , or $U(f, P) < U(f, Q)$ , then we have three inequalities

$\begin{array}{c} U(f, P) < U(f, Q) \\ L(f, P) \leq U(f, Q) \\ L(f, Q) + L(f, P) \leq U(f, Q) + U(f, P) \end{array}$ .

Adding the first two inequalities together gives,

$U(f, P) \geq 2\times U(f, Q) - L(f, P)$ . Secondly, re-writing the third inequality gives

$U(f, P) \geq L(f, P) + L(f, Q) - U(f, Q)$

Now, adding these two inequalities together, we obtain,

$\begin{array}{cc} 2\times U(f, P) &\geq 2\times U(f, Q) - L(f, P) + L(f, P) + L(f, Q) - U(f, Q) \\ &= U(f, Q) +L(f, Q) \end{array}$ .

But, as we know that $L(f, Q) \leq U(f, Q)$ , we have

$2\times U(f, P) \geq 2\times L(f, Q)$ , which of course gives $L(f, Q) \leq U(f, P)\text{.} ~\square$

Consequences of the theorem

As a consequence of this theorem, let us assume that for a function $f$ bounded on $[a,b]$ , that the upper integral of $f$ , $U(f)$ , is lower than the lower integral, $L(f)$ . Ie,

$U(f) < L(f)$

But then, this would imply that there exist partitions $P$ and $Q$ such that

$U(f, P) \leq L(f, Q) \quad(2)$ .

However, we have just proven the negation of $(2)$ , and so it follows that no such partitions exist.