# Refinements and the Mesh (Norm) of a Partition of a Closed Interval

Recall from the Partitions of a Closed Interval page that if $I = [a, b]$ where $a < b$, then a partition of this interval is a finite set $P = \{ a = x_0, x_1, ..., x_n = b \}$ such that:

(1)We said that the set of partitions on $[a, b]$ is $\mathscr{P} [a, b]$.

There are two more terms regarding partitions that we need to look at before we can move on to Riemann-Stieltjes Integrals. We define these terms below.

Definition: Let $I = [a, b]$ be a closed interval and let $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$. A partition $\bar{P} \in \mathscr{P}[a, b]$ is said to be a Refinement of $P$ if $P \subseteq \bar{P}$. If $\bar{P}$ is a refinement of $P$ we say that $\bar{P}$ is Finer than $P$. |

For example, consider the closed interval $I = [0, 1]$ and the partition $P = \left \{0, \frac{1}{2}, 1 \right \} \in \mathscr{P}[0, 1]$. Then $\bar{P} = \left \{ 0, \frac{1}{5}, \frac{1}{2}, 1 \right \} \in \mathscr{P}[0, 1]$ is such that $P \subset \bar{P}$, so $\bar{P}$ is a refinement of $P$ and $\hat{P}$ is finer than $P$.

Definition: Let $I = [a, b]$ be a closed interval and let $P = \{a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$. The Mesh or Norm of $P$ denoted $\| P \|$ is defined to be $\displaystyle{\| P \| = \max_{ k \in \{1, 2, ..., n \}} \{ \mid x_k - x_{k-1} \mid \}}$. |

Geometrically, the mesh of the partition $P = \{a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ is the length of the longest subinterval $[x_k, x_{k-1}]$, $\mid x_k - x_{k-1} \mid$.

For example, if $I = [0, 1]$ and $P = \left \{ 0, \frac{1}{8}, \frac{2}{3}, 1 \right \} \in \mathscr{P}[0, 1]$ then the mesh of $P$ is:

(2)We will now look at a rather simple theorem which tells us that if $\bar{P}$ is a refinement of $P$, then the mesh of $\bar{P}$ is less than or equal to the mesh of $P$.

Theorem 1: Let $[a, b]$ be a closed interval and let $P, \bar{P} \in \mathscr{P}[a, b]$ where $\bar{P}$ is a refinement of $P$. Then $\| \bar{P} \| \leq \| P \|$. |

**Proof:**Let $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$, and suppose that $\bar{P} \in \mathscr{P}[a, b]$ is a refinement of $P$. Then there exists a $c \in [a, b]$ such that $x_{k-1} \leq c \leq x_k$ for some $k \in \{1, 2, ..., n \}$. Then $\bar{P} = \{ a = x_0, x_1, ..., x_{k-1}, c, x_k, ..., x_n \}$. If $c = x_{k-1}$ or $c = x_k$ then $\| bar{P} \| = \| P \|$. Suppose that instead $P \subset \bar{P}$. Then the meshes of $P$ and $\bar{P}$ are:

- The only difference between the terms in $\| P \|$ and $\| \bar{P} \|$ is that $\mid x_k - x_{k-1} \mid$ in $\| P \|$ is replaced with the terms $\mid c - x_{k-1} \mid$ and $\mid x_k - c \mid$. By the triangle inequality we have that $\mid x_k - x_{k-1} \geq \mid c - x_{k-1} \mid$ and $\mid x_k - x_{k-1} \mid \geq x_k - c \mid$ since $c \in (x_{k-1}, x_k)$. Therefore $\| \bar{P} \| \leq \| P \|$. $\blacksquare$