A bounded function on $[a,b]$ that is continuous except at finitely many points is Riemann integrable

Let a real $\varepsilon > 0$ be given. Put $\theta := \varepsilon \cdot \bigl(4(b-a)\bigr)^{-1}$ and $\eta := \varepsilon \cdot \bigl(16\,M_{+}(\iota(r)+1)\bigr)^{-1}$, both positive reals by [L10] and [L11].

Put $V := \bigcup_{k<r}\,(d_k - \eta,\ d_k + \eta)$, an open set by [L5], and $K := [a,b] \setminus V = [a,b] \cap (\mathbb{R}\setminus V)$, an intersection of two closed sets, hence closed by [L5], and bounded since $K \subseteq [a,b]$; so $K$ is compact by [L4].

$f$ is continuous at every point of $K$: a point $x \in K$ is not any $d_k$, since $d_k \in (d_k-\eta, d_k+\eta) \subseteq V$ and $K$ misses $V$. Hence the restriction $f|_K$ is continuous on $K$ by [L7].

By [L4] applied to $f|_K$ on the compact set $K$ with the value $\theta$, fix a real $\delta > 0$ such that $|f(x)-f(y)| < \theta$ for all $x,y \in K$ with $|x-y| < \delta$.

By [L10] fix a natural $N \ge 1$ with $1/\iota(N) < \delta \cdot (b-a)^{-1}$, so that the uniform partition $U_N$ has mesh $(b-a)/\iota(N) < \delta$ by [L1] and [L11]. Let $P = (n,t)$ be the partition obtained from $U_N$ by inserting, one after another, those of the $2r$ points $d_k - \eta$ and $d_k + \eta$ with $k < r$ that lie in $[a,b]$. By [L1] each insertion leaves the mesh no larger, so $\|P\| < \delta$, and $\operatorname{pts}(P)$ contains every one of those points that lies in $[a,b]$.

A dichotomy for each subinterval and each $k$. Fix $i < n$ and $k < r$, and write $c^{-} := d_k - \eta$ and $c^{+} := d_k + \eta$. Neither $c^{-}$ nor $c^{+}$ lies in the open interval $(t_i,t_{i+1})$: if such a point lies in $[a,b]$ it is a member of $\operatorname{pts}(P)$ by step 5.1, hence is some $t_j$ with $j \le n$, and no $t_j$ lies strictly between $t_i$ and $t_{i+1}$ by [L1]; and if it lies outside $[a,b]$ it is outside $[t_i,t_{i+1}]$ altogether.

Consequently either $(t_i,t_{i+1}) \cap (c^{-},c^{+}) = \varnothing$, or $I_i \subseteq [c^{-},c^{+}]$. Suppose the intersection contains a point $z$ and let $w \in (t_i,t_{i+1})$. If $w \le c^{-}$ then $c^{-}$ lies between $w$ and $z$, both in the order-convex set $(t_i,t_{i+1})$, so $c^{-} \in (t_i,t_{i+1})$, contradicting step 6.1; likewise $w \ge c^{+}$ is impossible. So $(t_i,t_{i+1}) \subseteq (c^{-},c^{+})$. Then $c^{-} < z < t_{i+1}$ with $c^{-} \notin (t_i,t_{i+1})$ forces $c^{-} \le t_i$, and symmetrically $c^{+} \ge t_{i+1}$, that is $I_i \subseteq [c^{-},c^{+}]$.

Call $i < n$ bad when $I_i \subseteq [d_k-\eta, d_k+\eta]$ for some $k < r$, and good otherwise. If $i$ is good then by step 7.1 the open interval $(t_i,t_{i+1})$ meets no $(d_k-\eta,d_k+\eta)$, hence $(t_i,t_{i+1}) \subseteq K$; since $K$ is closed, [L6] gives $I_i = [t_i,t_{i+1}] \subseteq K$.

For a good $i$: all $x,y \in I_i$ lie in $K$ and satisfy $|x-y| \le \Delta_i \le \|P\| < \delta$ by [L1] and [L11], so $|f(x)-f(y)| < \theta$ by step 4.1; therefore $\theta$ bounds the set whose supremum is $M_i - m_i$, and $M_i - m_i \le \theta$ by [L2].

Bounding the bad lengths. For $k < r$ put $J_k := \{\, i < n : d_k - \eta \le t_i \text{ and } t_{i+1} \le d_k + \eta \,\}$, so that $i$ is bad exactly when $i \in J_k$ for some $k < r$, and put $h^{k}_i := \Delta_i$ for $i \in J_k$ and $h^{k}_i := 0$ otherwise. Each $J_k$ is a set of consecutive indices: if $i < i' < i''$ with $i, i'' \in J_k$ then $d_k - \eta \le t_i \le t_{i'}$ and $t_{i'+1} \le t_{i''+1} \le d_k + \eta$, so $i' \in J_k$.

$\sum_{i<n} h^{k}_i \le 2\eta$ for each $k < r$. If $J_k = \varnothing$ the sum is $0$. Otherwise let $p := \min J_k$ and let $q$ be the least natural with $q > p$ and $q \notin J_k$, which exists by [L9] because $n \notin J_k$; by step 9.2 then $J_k = \{\, i : p \le i < q \,\}$, since an $i \in J_k$ with $i \ge q$ together with $p \in J_k$ and $p < q \le i$ would put $q \in J_k$. Splitting the sum at $p$ and at $q$ and discarding the vanishing outer parts ([L8]) gives $\sum_{i<n}h^{k}_i = \sum_{i=p}^{q-1}\Delta_i = t_q - t_p$ by telescoping, and $p \in J_k$, $q-1 \in J_k$ give $t_p \ge d_k-\eta$ and $t_q \le d_k+\eta$, whence $t_q - t_p \le 2\eta$.

Put $\beta_i := \Delta_i$ for $i$ bad and $\beta_i := 0$ for $i$ good. Then $\beta_i \le \sum_{k<r}h^{k}_i$ for every $i < n$, all terms being nonnegative and a bad $i$ lying in some $J_k$; so by [L8], $\sum_{i<n}\beta_i \le \sum_{i<n}\sum_{k<r}h^{k}_i = \sum_{k<r}\sum_{i<n}h^{k}_i \le \sum_{k<r}2\eta = 2\eta\,\iota(r)$, using step 10.1.

For every $i < n$ one has $(M_i - m_i)\Delta_i \le \theta\,\Delta_i + 2M_{+}\,\beta_i$: for good $i$ this is step 9.1 together with $\beta_i \ge 0$, and for bad $i$ it follows from $M_i - m_i \le 2M_{+}$ in [L2] and $\beta_i = \Delta_i$, together with $\theta\Delta_i \ge 0$.

Summing step 12.1 over $i < n$ and using [L8], [L1] and step 11.1: $U(f,P) - L(f,P) \le \theta\sum_{i<n}\Delta_i + 2M_{+}\sum_{i<n}\beta_i \le \theta(b-a) + 4M_{+}\eta\,\iota(r) \le \varepsilon\cdot 4^{-1} + \varepsilon\cdot 4^{-1} < \varepsilon$, the last estimate because $4M_{+}\eta\,\iota(r) \le 4M_{+}\eta(\iota(r)+1) = \varepsilon\cdot 4^{-1}$ by step 1.1.

The real $\varepsilon > 0$ of step 1.1 was arbitrary and step 13.1 produced a partition $P$ with $U(f,P) - L(f,P) < \varepsilon$, so [L3] applies and $f$ is Riemann integrable on $[a,b]$.