Discontinuities of Monotone Functions

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 $\frac{1}{2}$.

Reduction and endpoints: if $f$ is decreasing, $-f$ is increasing with the same discontinuity set and jump sizes $(-f)(x+) - (-f)(x-) = f(x-) - f(x+) = |f(x+) - f(x-)|$; the conclusion for $-f$ (bound $(-f)(b) - (-f)(a) = |f(b) - f(a)|$) is then the conclusion for $f$. So assume $f$ is increasing from here, with bound $f(b) - f(a)$.

At the endpoints only the inward one-sided limit exists, so read $f(a-) := f(a)$ and $f(b+) := f(b)$: the endpoint jumps are $j(a) = f(a+) - f(a)$ and $j(b) = f(b) - f(b-)$, and $j(x) \ge 0$ at every point with this convention. 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.

Take finitely many points $x_1 < \cdots < x_m$. If $x_i < x_{i+1}$ then for any $t$ with $x_i < t < x_{i+1}$, monotonicity gives $f(x_i+) = \inf_{s > x_i} f(s) \le f(t) \le \sup_{s < x_{i+1}} f(s) = f(x_{i+1}-)$. Hence $f(x_i+) \le f(x_{i+1}-)$ — the interiors of the jump intervals $[f(x_i-), f(x_i+)]$ and $[f(x_{i+1}-), f(x_{i+1}+)]$ are pairwise disjoint (they may share an endpoint).

The endpoint positions also hold: for $a < t < x_1$, $f(a+) \le f(t) \le f(x_1-)$, so $f(a+) \le f(x_1-)$, and likewise $f(x_m+) \le f(b-)$. Chaining everything: $f(a+) \le f(x_1-) \le f(x_1+) \le f(x_2-) \le \cdots \le f(x_m+) \le f(b-)$. The jump intervals are therefore disjoint subintervals of $[f(a+), f(b-)]$, and the sum of the lengths of disjoint subintervals is at most the length of the containing interval — telescoping along this chain gives the inequality below.

Every jump interval lies in $[f(a+), f(b-)] \subseteq [f(a), f(b)]$, so $\sum_{i=1}^{m} j(x_i) = \sum_{i=1}^{m} \big(f(x_i+) - f(x_i-)\big) \le f(b-) - f(a+) \le f(b) - f(a)$. Since $m$ is arbitrary, the sum over 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.

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). For $t > q_m$, $g(t) - g(q_m) = 2^{-m} + \sum_{k \ne m:\, q_m < q_k < t} 2^{-k}$: the $2^{-m}$ term switches on, plus contributions from other rationals in $(q_m, t)$.

The extra sum tends to $0$ as $t \downarrow q_m$: given $\varepsilon > 0$, choose $K$ with $\sum_{k > K} 2^{-k} < \varepsilon$; the finitely many $q_k$ with $k \le K$, $k \ne m$, stay a positive distance from $q_m$, so for $\delta > 0$ smaller than all those distances and $q_m < t < q_m + \delta$, no $k \le K$ with $k \ne m$ has $q_k \in (q_m, t)$, and the extra sum is at most $\sum_{k > K} 2^{-k} < \varepsilon$. Hence $g(t) \to g(q_m) + 2^{-m}$, i.e. $g(q_m+) = g(q_m) + 2^{-m}$ — $g$ 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.