Lower Sum of Refinement

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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:

$\map L {f, P} \le \map L {f, Q}$

where $\map L {f, P}$ and $\map L {f, Q}$ denotes the lower Darboux sum of $f$ with respect to $P$ and $Q$.


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:

$m_i = \inf \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$

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

Also set:

${\tilde m}_j = \inf \set {\map f x : x \in \closedint {y_{j - 1} } {y_j} }$

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 Infimum of Subset, we have:

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

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

That is:

$m_i \le {\tilde m}_j$

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:

\(\ds \map L {f, P}\) \(=\) \(\ds \sum_{i \mathop = 1}^k m_i \paren {x_i - x_{i - 1} }\) Definition of Lower Darboux Sum
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^k m_i \paren {\sum_{j \mathop = a_i + 1}^{b_i} \paren {y_j - y_{j - 1} } }\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^k \paren {\sum_{j \mathop = a_i + 1}^{b_i} m_i \paren {y_j - y_{j - 1} } }\)
\(\ds \) \(\le\) \(\ds \sum_{i \mathop = 1}^k \paren {\sum_{j \mathop = a_i + 1}^{b_i} {\tilde m}_j \paren {y_j - y_{j - 1} } }\) since $m_i \le {\tilde m}_j$ for each $\tuple {i, j}$ with $a_i + 1 \le j \le b_i$
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^l {\tilde m}_j \paren {y_j - y_{j - 1} }\)
\(\ds \) \(=\) \(\ds \map L {f, Q}\)

$\blacksquare$


Sources

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