Discontinuities of Monotone Functions

Let $f$ be increasing on $[a, b]$ . Show that the left limit $f(x-) = \sup\{f(t) : t < x\}$ exists at every $x \in (a, b]$ , and the right limit $f(x+) = \inf\{f(t) : t > x\}$ exists at every $x \in [a, b)$ .

f(x-) = \sup\{f(t) : t < x\}, \qquad f(x+) = \inf\{f(t) : t > x\}

f(x-) \le f(x) \le f(x+) \quad \text{(left limit is left-continuous)}

At

x = q_1 = \frac{1}{2}

:

f(\frac{1}{2}-) = f(\frac{1}{2}) \approx 0.8373474478721619

, and the jump

f(\frac{1}{2}+) - f(\frac{1}{2}) = 2^{-1} = \frac{1}{2}

— the left limit equals the value; the rise is only from the right.

Show proof

For every $t < x$ , monotonicity gives $f(t) \le f(x)$ . So $\{f(t) : t < x\}$ is a nonempty bounded set with upper bound $f(x)$ , and by completeness its supremum $f(x-) := \sup\{f(t) : t < x\}$ exists as a real number.
This supremum is indeed the left limit. Given $\varepsilon > 0$ , the definition of the supremum yields $t_0 < x$ with $f(t_0) > f(x-) - \varepsilon$ . By monotonicity, $t_0 < t < x$ implies $f(t_0) \le f(t) \le f(x-) < f(x-) + \varepsilon$ , so $|f(t) - f(x-)| < \varepsilon$ — i.e. $\lim_{t \uparrow x} f(t) = f(x-)$ .
The right side is symmetric: $\{f(t) : t > x\}$ is bounded with lower bound $f(x)$ , so its infimum $f(x+) := \inf\{f(t) : t > x\}$ exists, and the same $\varepsilon$ argument gives $\lim_{t \downarrow x} f(t) = f(x+)$ . Hence a monotone function has both one-sided limits at every point.
Moreover $f(x-) \le f(x) \le f(x+)$ always holds: taking the supremum of $f(t) \le f(x)$ over $t < x$ gives $f(x-) \le f(x)$ , and similarly on the right. In the example, at $x = q_1 = \frac{1}{2}$ the left equality holds ( $f(\frac{1}{2}-) = f(\frac{1}{2}) \approx 0.8373474478721619$ ) while the right jumps by $\frac{1}{2}$ .

Show that a monotone function $f$ on $[a, b]$ has at most countably many discontinuities. Hint: show the jumps $j(x) = f(x+) - f(x-) \ge 0$ sum to at most $f(b) - f(a)$ , and that $\{x : j(x) > 1/n\}$ is finite with at most $n(f(b) - f(a))$ points.

j(x) := f(x+) - f(x-) \ge 0, \qquad \sum_{x} j(x) \le f(b) - f(a)

D = \bigcup_{n=1}^{\infty} \left\{x : j(x) > \frac{1}{n}\right\} \quad \text{countable union of finite sets}

n = 4

(threshold

\frac{1}{4}

): exactly one jump exceeds

\frac{1}{4}

—

k = 1

(size

\frac{1}{2}

) — within the bound

4(f(1) - f(0)) \approx 7.999999761581421

(

2^{-2} = \frac{1}{4}

is not

> \frac{1}{4}

).

Show proof

First, discontinuities can only be jumps. By problem 1, $f(x-)$ and $f(x+)$ always exist, so failure of $\lim_{t \to x} f(t) = f(x)$ forces $f(x-) < f(x+)$ — neither removable nor oscillatory discontinuities are possible; only a jump $j(x) = f(x+) - f(x-) > 0$ remains.
For finitely many points $x_1 < \cdots < x_m$ , the jump intervals $[f(x_i-), f(x_i+)]$ are disjoint (monotonicity orders them) and all lie inside $[f(a), f(b)]$ . Hence $\sum_{i=1}^{m} j(x_i) \le f(b) - f(a)$ , and since $m$ is arbitrary, the total of all jumps satisfies $\sum_x j(x) \le f(b) - f(a)$ .
Fix $n$ : if $j(x) > \frac{1}{n}$ at $m_n$ points, then $\sum j(x) \ge m_n \cdot \frac{1}{n}$ , so $m_n \le n(f(b) - f(a))$ — a finite set. But the discontinuity set $D = \{x : j(x) > 0\} = \bigcup_{n=1}^{\infty} \{x : j(x) > \frac{1}{n}\}$ is a countable union of finite sets, hence at most countable.
Numerical check ( $terms = 24$ , $n = 4$ ): the jumps are $2^{-k}$ ( $k = 1, \dots, 24$ ), and the only one exceeding $\frac{1}{4}$ is $2^{-1} = \frac{1}{2}$ ( $2^{-2} = \frac{1}{4}$ is not a strict exceedance). Against the bound $4(f(1) - f(0)) \approx 7.999999761581421$ , we get $1 \le 7.999\ldots$ — exactly as the theorem predicts.

Let $(q_k)$ enumerate the rationals of $(0, 1)$ and $g(x) = \sum_{k : q_k < x} 2^{-k}$ . Show that $g$ is increasing on $[0, 1]$ , discontinuous exactly at the rationals, and continuous at every irrational.

g(x) = \sum_{k : q_k < x} 2^{-k}, \qquad g(q_m+) - g(q_m) = 2^{-m}

\text{if } x \notin \mathbb{Q} \text{ and } N \text{ large, } \sum_{k > N} 2^{-k} < \varepsilon \text{ and } \delta = \operatorname{dist}(x, \{q_1, \dots, q_N\}) > 0

At

x = q_3 = \frac{2}{3}

:

g(\frac{2}{3}) \approx 0.841285169124603

, jumping up by

2^{-3} = \frac{1}{8} = 0.125

—

g(\frac{2}{3}+) \approx 0.966285169124603

.

Show proof

The series always converges: $0 \le \sum_{k : q_k < x} 2^{-k} \le \sum_{k=1}^{\infty} 2^{-k} = 1$ , so $g(x)$ is well defined with values in $[0, 1]$ . If $x < y$ then $\{k : q_k < x\} \subseteq \{k : q_k < y\}$ , so the sum only gains terms and $g(x) \le g(y)$ — $g$ is increasing.
Let $x = q_m$ . The sum defining $g(q_m)$ excludes $k = m$ (the condition $q_k < q_m$ leaves out $q_m$ itself), but as $t \downarrow q_m$ , eventually $q_m < t$ for all $t > q_m$ , adding $2^{-m}$ . Hence $g(q_m+) = g(q_m) + 2^{-m}$ — $g$ jumps by $\frac{1}{2^m}$ at $q_m$ , so it is discontinuous there.
Now let $x$ be irrational. Given $\varepsilon > 0$ , pick $N$ with $\sum_{k > N} 2^{-k} < \varepsilon$ (possible since the tail tends to 0). Since $x$ is irrational, $\delta := \min_{1 \le k \le N} |x - q_k| > 0$ . If $|t - x| < \delta$ , none of $q_1, \dots, q_N$ lies strictly between $x$ and $t$ , so the difference $g(t) - g(x)$ comes only from the tail: $|g(t) - g(x)| < \varepsilon$ . Hence $g$ is continuous at $x$ .
So $g$ is discontinuous at every rational and continuous at every irrational — a concrete example of a monotone function discontinuous on a dense countable set (and, by problem 2, on no larger a set than dense). Numerically, at $q_3 = \frac{2}{3}$ : $g(\frac{2}{3}) \approx 0.841285169124603$ , $g(\frac{2}{3}+) \approx 0.966285169124603$ — the gap $0.125 = 2^{-3}$ matches the caption's jump.

Discontinuities of Monotone Functions

Examples & Exercises