Upper Sum of Refinement

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

Let $P$ be a finite subdivision of $\closedint a b$.

Let $Q$ be a refinement of $P$.

Then:


 * $\map U {f, P} \le \map U {f, Q}$

where $\map U {f, P}$ denotes the upper sum of $f$ with respect to $P$.

Proof
Write:


 * $P = \set {x_1, x_2, \ldots, x_k}$

and:


 * $Q = \set {y_1, y_2, \ldots, y_l}$

where:


 * $a = x_1 < x_2 < \ldots < x_k = b$

and:


 * $a = y_1 < y_2 < \ldots < y_l = b$

Since $P \subseteq Q$, we have $k \le l$ from Cardinality of Subset of Finite Set.

Set:


 * $M_i = \sup \set {\map f x : x \in \closedint {x_i} {x_{i + 1} } }$

for each $1 \le i \le k - 1$.

Also set:


 * ${\tilde M}_j = \sup \set {\map f x : x \in \closedint {y_j} {y_{j + 1} } }$

for each $1 \le j \le l - 1$.

Consider a pair of elements $\tuple {x_i, x_{i + 1} }$ in $P$.

Since $P \subseteq Q$, there exists $a_i, b_i$ such that:


 * $\tuple {x_i, x_{i + 1} } = \tuple {y_{a_i}, y_{b_i} }$

Note that:


 * $\closedint {y_j} {y_{j + 1} } \subseteq \closedint {x_i} {x_{i + 1} }$

for all $a_i \le j \le b_i - 1$.

So:


 * $\set {\map f x : x \in \closedint {y_j} {y_{j + 1} } } \subseteq \set {\map f x : x \in \closedint {x_i} {x_{i + 1} } }$

for all $a_i \le j \le b_i - 1$.

So, from Supremum of Subset, we have:


 * $\sup \set {\map f x : x \in \closedint {y_j} {y_{j + 1} } } \le \sup \set {\map f x : x \in \closedint {x_i} {x_{i + 1} } }$

for all $a_i \le j \le b_i - 1$.

That is:


 * ${\tilde M}_j \le M_i$

for each $\tuple {i, j}$ with $a_i \le j \le b_i - 1$

We can then write:


 * $\ds x_{i + 1} - x_i = \sum_{j \mathop = a_i}^{b_i - 1} \paren {y_{j + 1} - y_j}$

giving: