Monday, February 11, 2013 CC-BY-NC
Introduction to Riemann sums

Maintainer: admin

Definition: Let $I$ be a closed and bounded interval. A partition of $I$ is a finite, ordered set

$$ P = \{ x_0, x_1, \ldots, x_n \} $$

of points in $I$ such that $x_0 = a < x_1 < x_2 < \ldots < x_n = b$.

Remark: A partition divides your interval $I$ into non-overlapping intervals $[x_0, x_1]$, $[x_1, x_2]$ etc, the union of which is the original interval.

Definition: Given an interval $I = [a,b]$, a partition $P$ of $I$ and $f : I \to \mathbb{R}$ bounded on $I$, we define a Riemann Sum of $f$ corresponding to $P$ to be any expression of the form

$$ \sum_{i=1}^n f(x_i^*)(x_1 - x_{i-1}), \quad \text{where } x_i^* \in [x_{i-1}, x_i] $$

Definition: Let $f : I \to \mathbb{R}$ be bounded on $I = [a,b]$, and let $P$ be a partition of $I$. Let $m_k = \inf\{ f(x) : x \in [x_{k-1}, x_k] \}$ and $M_k = \sup\{ f(x) : x \in [x_{k-1}, x_k] \}$. We define the lower sum of $f$ corresponding to $P$ as

$$ L_f(P) = \sum_{k=1}^n m_k(x_k - x_{k-1}) $$

We define the upper some $U_f(P)$ similarly, with $M_k$.

Lemma: Let $f : I \to \mathbb{R}$ be bounded on $I = [a,b]$ and let $P$ be any partition of $I$. Then $L_f(P) \leq U_f(P)$

Proof: Let $P = \{ x_0, x_1, \ldots, x_n \}$. It follows that $m_k \leq M_k$ for all $k$ from 1 to $n$. Now, since $x_k - x_{k-1} \geq 0$, $m_k(x_k - x_{k-1}) \leq M_k(x_k - x_{k-1})$ and thus for the sums

$$ \sum_{k=1}^n m_k(x_k - x_{k-1}) \leq \sum_{k=1}^n M_k(x_k - x_{k-1}) $$

i.e. $L_f(P) \leq U_f(P)$. $\blacksquare$

Definition: Let $I = [a,b]$ and $P = \{x_0, x_1, \ldots, x_n\}$ be a partition of $I$. A partition $Q = \{y_0, y_1, \ldots, y_m\}$ is said to be a refinement of $P$ if $P \subseteq Q$, i.e. if every point of $P$ is also a point of $Q$. In other words, $Q$ is a refinement of $P$ $\iff$ you can adjoin a finite number of elements to $P$ to get $Q$.

Lemma: If $f : I \to \mathbb{R}$ is bounded, $P$ is a partition of $I$ and $Q$ is a refinement of $P$, then

$$ L_f(P) \leq L_f(Q) \quad \text{and} \quad U_f(Q) \leq U_f(P) $$

Proof: Observe that by a previous lemma, $L_f(Q) \leq U_f(Q)$. Let $P = \{x_0, x_1, \ldots, x_n\}$ be a partition of $I$. Let $z \in I$ be such that $z \not\in P$. We can create a new partition $P' = P \cup \{z\}$, a refinement of $P$. Since $P$ decomposes $I$ into a finite number of non-overlapping intervals, there exists some $k$ such that $1 \leq k \leq n$ for which $z \in [x_{k-1}, x_k]$. $z$ bisects this interval into $[x_{k-1}, z]$ and $[z, x_k]$ which are both subsets of $[x_{k-1}, x_k]$. Thus $\inf\{ f(x) : x \in [x_{k-1}, x_k]\} \leq \inf\{ f(x) : x \in [x_{k-1}, z]\} = m_k'$ and similarly $\inf\{ f(x) : x \in [x_{k-1}, x_k]\} \leq \inf\{ f(x) : x \in [z, x_k]\} = m_k''$. Thus:

$\begin{align} m_k(x_k - x_{k-1}) &= m_k(x_k - z + z - x_{k-1}) \\ &= m_k(x_k - z) + m_k(z - x_{k-1}) \\ &\leq m_k''(x_k - z) + m_k'(z - x_{k-1}) \end{align}$

so $L_f(P) \leq L_f(P')$. A similar process works for the upper sum. As we noted earlier, if $Q$ is a refinement of $P$ then $Q$ can be obtained by adding a finite number of points to $P$, one at a time. As our process holds for any of these steps, our proof is complete. $\blacksquare$