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 $\map U {f, P}$ and $\map U {f, Q}$ denote the upper sum of $f$ with respect to $P$ and $Q$ respectively.

Proof
Write:


 * $P = \set {x_0, x_1, \ldots, x_k}$

and:


 * $Q = \set {y_0, y_1, \ldots, y_l}$

where:


 * $a = x_0 < x_1 < \ldots < x_k = b$

and:


 * $a = y_0 < y_1 < \ldots < y_l = b$

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

Set:


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

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

Also set:


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

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

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

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


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

We can see that:


 * $a_1 = 0$

and:


 * $b_k = l$

We also clearly have:


 * $b_{i - 1} = a_i$ for each $1 \le i \le k$.

Note that:


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

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

So:


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

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

So, from Supremum of Subset, we have:


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

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

That is:


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

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

We can then write:


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

for each $1 \le i \le k$, giving: