Oscillation of a Bounded Function on a Set

# Oscillation of a Bounded Function on a Set

We will soon look at a very nice criterion for determining whether a bounded function $f$ on a closed interval $[a, b]$ is Riemann integrable or not. In order to prove this particular result (which we will see soon), we will first need to define the oscillation of a function on a set.

 Definition: Let $f$ be a bounded function on the interval $[a, b]$, and let $T \subseteq [a, b]$. Then the Oscillation of $f$ on $T$ denoted $\Omega_f (T)$ is defined as $\Omega_f (T) = \sup \{ f(x) - f(y) : x, y \in T \}$.

We should note that for a bounded function $f$ on $[a, b]$ and $T \subseteq [a, b]$ that $\Omega_f (T)$ always exists. To show this, consider the difference for $x, y \in T$:

(1)
\begin{align} \quad f(x) - f(y) \end{align}

Since $f$ is a bounded function we have that for all $x \in T$ that $\mid f(x) \mid < M$ where $M \in \mathbb{R}$, $M > 0$. Furthermore, $\mid f(y) \mid < M$. So:

(2)
\begin{align} \quad f(x) - f(y) \leq \mid f(x) - f(y) \mid \leq \mid f(x) \mid + \mid f(y) \mid \leq M + M = 2M \end{align}

So the set $\{ f(x) - f(y) : x, y \in T \}$ is bounded. Furthermore, this set is nonempty and by The Completeness Property of the Real Numbers we have that $\sup \{ f(x) - f(y) : x, y \in [a, b] \}$ exists in $\mathbb{R}$.

For example, if $f$ is bounded on $[a, b]$ and $c \in [a, b]$ where $T = \{ c \} \subset [a, b]$, then the oscillation of $f$ on $T$ is $0$ since:

(3)
\begin{align} \quad \Omega_f(T) = \sup \{ f(x) - f(y) : x, y \in \{ c \} \} = \sup \{ 0 \} = 0 \end{align}

Of course, if $T$ is a more complicated subset of $[a, b]$ then $\Omega_f (T)$ is generally more difficult to compute. None the less, we are not so much as interested in computing the oscillation of a function on a subset $T \subseteq [a, b]$ as we are with the general results. We first prove that the oscillation of $f$ on $T$ is always nonnegative.

 Proposition 1: Let $f$ be a bounded function on $[a, b]$ and let $T \subseteq [a, b]$. Then $\Omega_f(T) \geq 0$.
• Proof: Let $T \subseteq [a, b]$ and consider the oscillation of $f$ on $T$:
(4)
\begin{align} \quad \Omega_f (T) = \sup \{ f(x) - f(y) : x, y \in T \} \end{align}
• Note that for each $x, y \in T$ that either $f(x) > f(y)$, $f(x) = f(y)$, or $f(x) < f(y)$.
• If $f(x) > f(y)$ then $f(x) - f(y) > 0$ and so $\Omega_f(T) > 0$.
• If $f(x) = f(y)$ then $f(x) - f(y) = 0$ so $\Omega_f(T) \geq 0$.
• Lastly, if $f(x) < f(y)$ then the difference $f(y) - f(x) \in \{ f(x) - f(y) : x, y \in T \}$ and $\Omega_f(T) > 0$.
• In all cases we see that $\Omega_f(T) \geq 0$. $\blacksquare$

We will now look at a similar definition of the oscillation of a bounded function $f$ on $[a, b]$ - this time not with respect to a subset $T$ of $[a, b]$ but with respect to a specific point.