Converse to Froda: for every at most countable $E \subseteq \mathbb{R}$ there is a bounded nondecreasing $f : \mathbb{R} \to \mathbb{R}$ whose set of discontinuities is exactly $E$, every one of them a jump

If $E = \varnothing$, the constant function $0$ is nondecreasing, takes values in $[0,1]$, is continuous at every real, and has empty discontinuity set; all three claims hold vacuously for claim 3. Assume from here on that $E \ne \varnothing$ and fix a surjection $s : \mathbb{N} \to E$.

For every $n \in \mathbb{N}$, $\sum_{k<n} 1/2^{\,k+1} = 1 - 1/2^{\,n}$: each term is $1/2^{\,k+1} = 1/2^{\,k} - 1/2^{\,k+1}$, so the sum telescopes to $1/2^{\,0} - 1/2^{\,n} = 1 - 1/2^{\,n}$.

Define $a_{k}(x) := 1/2^{\,k+1}$ when $s(k) < x$ and $a_{k}(x) := 0$ otherwise, and note $0 \le a_{k}(x) \le 1/2^{\,k+1}$ for every $k$ and every real $x$.

For every real $\varepsilon > 0$ there is $n \in \mathbb{N}$ with $1/2^{\,n} < \varepsilon$: the partial sums $t_{n} := \sum_{k<n} 1/2^{\,k}$ converge to $2$, and $t_{n} = 2 - 2/2^{\,n}$ by the same telescoping as in step 1.2, so $|t_{n} - 2| = 2/2^{\,n} < \varepsilon$ for all large $n$, whence $1/2^{\,n} < \varepsilon/2 < \varepsilon$ for those $n$. Consequently the partial sums $1 - 1/2^{\,n}$ of $\sum_{k} 1/2^{\,k+1}$ have supremum $1$, so that series converges with sum $1$.

For every real $x$ the series $\sum_{k} a_{k}(x)$ converges and $0 \le f(x) \le 1$: its terms are nonnegative and its partial sums satisfy $\sum_{k<n} a_{k}(x) \le \sum_{k<n} 1/2^{\,k+1} = 1 - 1/2^{\,n} \le 1$, so they are bounded above by $1$ and the sum, being their supremum, lies in $[0,1]$.

Left continuity holds at every real $c$: given real $\varepsilon > 0$ take $n$ with $1/2^{\,n} < \varepsilon$; let $F := \{\, k < n : s(k) < c \,\}$; if $F = \varnothing$ put $x_{0} := c - 1$, and otherwise put $x_{0}$ to be a real with $\max\{s(k) : k \in F\} < x_{0} < c$, which exists because the maximum of the nonempty finite set $\{s(k) : k \in F\}$ is a real strictly below $c$.

Right continuity holds at every $c \notin E$: given real $\varepsilon > 0$ take $n$ with $1/2^{\,n} < \varepsilon$; since $c \notin E$ and $s$ has image $E$, no $k$ has $s(k) = c$, so every $k < n$ has $s(k) < c$ or $s(k) > c$. Let $G := \{\, k < n : s(k) > c \,\}$; if $G = \varnothing$ put $y_{0} := c + 1$, and otherwise put $y_{0}$ to be a real with $c < y_{0} < \min\{s(k) : k \in G\}$.

$f$ is nondecreasing: if $x \le y$ then $s(k) < x$ implies $s(k) < y$, so $a_{k}(x) \le a_{k}(y)$ for every $k$, hence $\sum_{k<n} a_{k}(x) \le \sum_{k<n} a_{k}(y)$ for every $n$, and taking suprema gives $f(x) \le f(y)$.

For all reals $x \le y$ and every $n \in \mathbb{N}$ with $a_{k}(x) = a_{k}(y)$ for every $k < n$, one has $f(y) - f(x) \le 1/2^{\,n}$: for $N \ge n$ the splitting $\sum_{k<N} a_{k}(y) = \sum_{k<n} a_{k}(y) + \sum_{k=n}^{N-1} a_{k}(y) \le \sum_{k<n} a_{k}(x) + \sum_{k=n}^{N-1} 1/2^{\,k+1}$ holds, the last sum being at most $\sum_{k \ge n} 1/2^{\,k+1} = 1 - (1 - 1/2^{\,n}) = 1/2^{\,n}$; so every partial sum of $\sum_{k} a_{k}(y)$ is at most $f(x) + 1/2^{\,n}$, and so is their supremum $f(y)$.

Let $c \in E$ and fix $k_{0}$ with $s(k_{0}) = c$. For every $y > c$ and every $N > k_{0}$ the finite sum $\sum_{k<N} a_{k}(y)$ exceeds $\sum_{k<N} a_{k}(c)$ by at least $1/2^{\,k_{0}+1}$, because the list $k \mapsto a_{k}(y) - a_{k}(c)$ has nonnegative entries, so the finite sum of its first $N$ entries is at least its entry at the index $k_{0}$, which is $a_{k_{0}}(y) - a_{k_{0}}(c) = 1/2^{\,k_{0}+1} - 0$. Hence $f(y) - 1/2^{\,k_{0}+1} \ge \sum_{k<N} a_{k}(c)$ for every $N$, the case $N \le k_{0}$ holding because the partial sums of a nonnegative series are nondecreasing; so $f(y) - 1/2^{\,k_{0}+1}$ is an upper bound of those partial sums and therefore at least their supremum $f(c)$.

With $x_{0}$ as in step 3.2 and any $x$ with $x_{0} < x \le c$: for $k < n$ with $s(k) < c$ we have $s(k) \le \max\{s(j) : j \in F\} < x_{0} < x$, so $a_{k}(x) = 1/2^{\,k+1} = a_{k}(c)$; and for $k < n$ with $s(k) \ge c \ge x$ we have $a_{k}(x) = 0 = a_{k}(c)$. So $a_{k}(x) = a_{k}(c)$ for every $k < n$, and step 4.2 applied to the pair $x \le c$ gives $0 \le f(c) - f(x) \le 1/2^{\,n} < \varepsilon$.

With $y_{0}$ as in step 3.3 and any $y$ with $c \le y < y_{0}$: for $k < n$ with $s(k) < c \le y$ we get $a_{k}(y) = 1/2^{\,k+1} = a_{k}(c)$, and for $k < n$ with $s(k) > c$ we have $s(k) \ge \min\{s(j) : j \in G\} > y_{0} > y$, so $a_{k}(y) = 0 = a_{k}(c)$. So $a_{k}(y) = a_{k}(c)$ for every $k < n$, and step 4.2 applied to the pair $c \le y$ gives $0 \le f(y) - f(c) \le 1/2^{\,n} < \varepsilon$.

So $f$ is discontinuous at $c$: for $\varepsilon := 1/2^{\,k_{0}+1} > 0$ and any real $\delta > 0$ the point $y := c + \delta/2$ satisfies $|y - c| < \delta$ and $|f(y) - f(c)| \ge \varepsilon$, so no $\delta$ witnesses the continuity condition at $c$.

Hence $f$ is continuous at every $c \notin E$: fix a real $\varepsilon > 0$, take $x_{0}$ as in step 3.2 and $y_{0}$ as in step 3.3 for that same $\varepsilon$, and put $\delta := \min\{c - x_{0}, y_{0} - c\} > 0$; then every real $x$ with $|x - c| < \delta$ satisfies $x_{0} < x < y_{0}$ and therefore $|f(x) - f(c)| < \varepsilon$, by step 5.1 when $x \le c$ and by step 5.2 when $x \ge c$.

Every point of $E$ is an interior point of the order-convex set $\mathbb{R}$, so both one-sided limits of $f$ exist there; step 5.1 gives $\lim_{x \to c^{-}} f(x) = f(c)$ and step 4.3 gives $\lim_{x \to c^{+}} f(x) \ge f(c) + 1/2^{\,k_{0}+1} > f(c)$. The two one-sided limits therefore differ, and the discontinuity at $c$ is a jump.

Claims 1, 2 and 3 hold for the function $f$ constructed in steps 1.1 and 2.1: claim 1 by steps 3.1 and 4.1, claim 2 by steps 6.1 and 5.3, and claim 3 by step 6.2.