Countable Discontinuities of Monotonic Functions

Countable Discontinuities of Monotonic Functions

Recall from the Monotonic Functions page that $f$ is called a monotonic function on the interval $[a, b]$ if either $f$ is increasing or decreasing on $[a, b]$, and that for all $x, y \in [a, b]$ such that $x < y$, $f$ is increasing on $[a, b]$ if $f(x) \leq f(y)$ and decreasing on $[a, b]$ if $f(x) \geq f(y)$.

We will now look at a remarkable theorem which says that if $f$ is a monotonic function on an interval $[a, b]$ then $f$ has at most a countably infinite number of discontinuities on $[a, b]$. In order to prove this theorem, we will first need to look at the following lemma.

 Lemma 1: If $f$ is an increasing function and $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ is a partition of $[a, b]$ i.e., $a = x_0 < x_1 < ... < x_n = b$ then $\displaystyle{\sum_{k=1}^{n-1} [f(x_k^+) - f(x_k^-)] \leq f(b) - f(a)}$.
• Proof: Let $f$ be an increasing function on the interval $[a, b]$ and let $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ be a partition of $[a, b]$ so that then:
(1)
\begin{align} \quad a = x_0 < x_1 < ... < x_n = b \end{align}
• Let $y_1, y_2, ..., y_n \in [a, b]$ be points such that:
(2)
\begin{align} \quad a = x_0 < y_1 < x_1 < y_2 < ... < y_k < x_k < y_{k+1} < ... < y_n < x_n = b \end{align}
• In particular, for each $k \in \{ 1, 2, ..., n - 1 \}$ we have that $y_k < x_k < y_{k+1}$. Let $f(x_k^-) = \lim_{a \to x_k^-} f(a)$ and let $f(x_k^+) = \lim_{a \to x_k^+} f(a)$. Since $f$ is an increasing function we obtain the following inequality:
(3)
\begin{align} \quad f(y_k) \leq f(x_k^-) \leq f(x_k) \leq f(x_k^+) \leq f(y_{k+1}) \end{align}
• Therefore $f(y_{k+1}) - f(y_k) \geq f(x_k^+) - f(x_k^-)$ for each $k \in \{1, 2, ..., n -1 \}$. Summing these quantities from $k = 1$ to $k = n - 1$ yields:
(4)
\begin{align} \quad \sum_{k=1}^{n-1} [f(x_k^+) - f(x_k^-)] & \leq \sum_{k=1}^{n-1} [f(y_{k+1}) - f(y_k)] \\ \quad \sum_{k=1}^{n-1} [f(x_k^+) - f(x_k^-)] & \leq [f(y_2) - f(y_1)] + [f(y_3) - f(y_2)] + ... + [f(y_n) - f(y_{n-1})] \\ \quad \sum_{k=1}^{n-1} [f(x_k^+) - f(x_k^-)] & \leq f(y_n) - f(y_1) \end{align}
• Since $a = x_0 < y_1 < ... < y_n < x_n = b$ and $f$ is increasing, we have that $f(b) - f(a) \geq f(y_n) - f(y_1)$ and hence:
(5)
\begin{align} \quad \sum_{k=1}^{n-1} [f(x_k^+) - f(x_k^-)] \leq f(b) - f(a) \quad \blacksquare \end{align}
 Theorem 2: Let $f$ be a monotonic function on the interval $[a, b]$. Then $f$ has at most a countably infinite number of discontinuities.
• Proof: There are two cases to consider - both of which are analogous so we will only consider the case when $f$ is monotonically increasing. First let $f$ be an increasing function on the interval $[a, b]$. We note that a discontinuity occurs at $x \in [a, b]$ when $f(x^-) \neq f(x^+)$, and in particular since $f$ is an increasing function, a discontinuity occurs when $f(x^+) - f(x^-) > 0$ so there exists a natural number $m > 0$ such that:
(6)
\begin{align} \quad 0 < \frac{1}{m} < f(x^+) - f(x^-) \end{align}
• Now, for each natural number $m > 0$ define the set $S_m$ by:
(7)
\begin{align} \quad S_m = \left \{ x \in (a, b) : f(x^+) - f(x^-) > \frac{1}{m} \right \} \end{align}
• Let $x_1, x_2, ..., x_{n-1} \in S_m$ be such that $x_1 < x_2 < ... < x_{n-1}$. Then $x_1, x_2, ..., x_{n-1}$ are discontinuities of $f$ such that their jump $f(x_k^+) - f(x_k^-) > \frac{1}{m}$. Consider the sum of these jumps, $\displaystyle{\sum_{k=1}^{n-1} [f(x_k^+) - f(x_k^-)]}$. By Lemma 1 we have an upper bound for the sum of these discontinuity jumps:
(8)
\begin{align} \quad \sum_{k=1}^{n-1} [f(x_k^+) - f(x_k^-)] \leq f(b) - f(a) \end{align}
• Now since $f(x_k^+) - f(x_k^-) > \frac{1}{m}$ for each $k \in \{ 1, 2, ..., n-1 \}$ we have that:
(9)
\begin{align} \quad f(b) - f(a) \geq \sum_{k=1}^{n-1} [f(x_k^+) - f(x_k^-)] \geq \sum_{k=1}^{n-1} \frac{1}{m} = \frac{n - 1}{m} \end{align}
• Therefore we see that $\frac{n - 1}{m} \leq f(b) - f(a)$, so $n - 1 \leq m(f(b) - f(a))$ and $n \leq m(f(b) - f(a)) + 1$. So the number of discontinuities $n$ in $S_m$ is bounded above. Therefore $S_m$ must be a finite set of discontinuities, which of course, is countable. Now since for all discontinuities $x \in (a, b)$ there exists a natural number $m > 0$ such that $f(x^+) - f(x^-) > \frac{1}{m}$, we see that the set of all discontinuities of $f$ on $(a, b)$ is $\displaystyle{\bigcup_{m=1}^{\infty} S_m}$. Each $S_m$ is a finite (countable) set, so the union of infinitely many finite sets is countable, so at most $f$ has an infinitely countable number of discontinuities on $[a, b]$. $\blacksquare$