Integrability of $\frac{1}{f}$

Table of Contents

• Swapping Argument
• The Problem

Swapping Argument

This first section isn't too important, but I figured it'd be good for you to see the precise statement of the swapping argument done in class. (I did this in office hours, but I really overcomplicated it and made a mistake proving it, but this version should be correct.)

Lemma

Let $I \subseteq \R$ be a (non-empty) interval and $\func{h}{I}{\R}$ be a function such that $h\p{x, y} = -h\p{y, x}$ for all $x, y \in I$. Then

$$\sup_{x, y \in I} h\p{x, y} = \sup_{x, y \in I} \abs{h\p{x, y}}.$$

Proof.

Since $h\p{x, y} \leq \abs{h\p{x, y}}$, we already have

$$\sup_{x, y \in I} h\p{x, y} \leq \sup_{x, y \in I} \abs{h\p{x, y}}.$$

For the reverse inequality, let $u, v \in I$. Then there are two cases: $h\p{u, v} \geq 0$ and $h\p{u, v} < 0$. In the first case,

$$\abs{h\p{u, v}} = h\p{u, v} \leq \sup_{x, y \in I} h\p{x, y}.$$

In the second case,

$$\abs{h\p{u, v}} = -h\p{u, v} = h\p{v, u} \leq \sup_{x, y \in I} h\p{x, y}.$$

Thus, we have established the inequality

$$\abs{h\p{u, v}} \leq \sup_{x, y \in I} h\p{x, y}$$

for all $u, v \in I$, which proves the lemma. $\square$

In the problem below, we will use this lemma twice: once on the function $h\p{x, y} = \frac{1}{f\p{x}} - \frac{1}{f\p{y}}$ and again on $h\p{x, y} = f\p{x} - f\p{y}$. The lemma will tell us

The Problem

Proposition

Assume $\func{f}{\br{a,b}}{\R}$ is a Riemann integrable function and that there exists $c > 0$ such that $\abs{f\p{x}} \geq c$ for all $x \in \br{a, b}$. Then $\frac{1}{f}$ is also Riemann integrable.

Scratchwork: Given $\epsilon > 0$, we need to find a partition $P$ of $\br{a, b}$ such that

The only real fact we have to work with is that $f$ is Riemann integrable, so we need to relate $U\p{\frac{1}{f}, P} - L\p{\frac{1}{f}, P}$ to $\p{f, P} - L\p{f, P}$. Let $P$ be an partition of $\br{a, b}$. Then

Let $I_k = \br{t_{k-1}, t_k}$. Our goal is to bound $M\p{\frac{1}{f}, I_k} - m\p{\frac{1}{f}, I_k}$ by (a constant independent of $k$ and $P$) times $M\p{f, I_k} - m\p{f, I_k}$.

This tells us how to pick our partition.

Proof.

Let $\epsilon > 0$. Since $f$ is Riemann integrable, there exists a partition $P$ of $\br{a, b}$ such that

$$U\p{f, P} - L\p{f, P} < c^2 \varepsilon.$$

By the calculation above,

$$\begin{aligned} U\p{\frac{1}{f}, P} - L\p{\frac{1}{f}, P} &= \sum_{k=1}^n \underbrace{\br{M\p{\frac{1}{f}, \br{t_{k-1}, t_k}} - m\p{\frac{1}{f}, \br{t_{k-1}, t_k}}}}_{\leq \frac{1}{c^2} \br{M\p{f, I_k} - m\p{f, I_k}}} \p{t_k - t_{k-1}} \\ &\leq \frac{1}{c^2} \sum_{k=1}^n \br{M\p{f, I_k} - m\p{f, I_k}} \p{t_k - t_{k-1}} \\ &= \frac{1}{c^2} \br{U\p{f, P} - L\p{f, P}} \\ &< \frac{1}{c^2} \cdot c^2\epsilon \\ &= \epsilon. \end{aligned}$$

$\square$