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

Let a real $\epsilon >0$ be given. Put $\theta :=\epsilon \cdot (4(b-a){)}^{-1}$ and $\eta :=\epsilon \cdot (16M_{+}(\iota (r)+1){)}^{-1}$, 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 $f{\mathrm{\mid }}_K$ is continuous on $K$ by [L7].

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

By [L10] fix a natural $N\ge 1$ with $1\mathrm{/}\iota (N)<\delta \cdot (b-a{)}^{-1}$, so that the uniform partition $U_N$ has mesh $(b-a)\mathrm{/}\iota (N)<\delta$ 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 $\mathrm{\parallel }P\mathrm{\parallel }<\delta$, and $\mathrm{pts}(P)$ 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 $\mathrm{pts}(P)$ 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 $(t_i,t_{i+1})\cap (c^{-},c^{+})=\mathrm{\varnothing }$, 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 $c^{-}\mathrm{\notin }(t_i,t_{i+1})$ 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 $\mathrm{\mid }x-y\mathrm{\mid }\le {\mathrm{\Delta }}_i\le \mathrm{\parallel }P\mathrm{\parallel }<\delta$ by [L1] and [L11], so $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<\theta$ 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 $h_i^k:={\mathrm{\Delta }}_i$ for $i\in J_k$ and $h_i^k:=0$ otherwise. Each $J_k$ is a set of consecutive indices: if $i<i^{\mathrm{\prime }}<i^{\mathrm{\prime }\mathrm{\prime }}$ with $i,i^{\mathrm{\prime }\mathrm{\prime }}\in J_k$ then $d_k-\eta \le t_i\le t_{i^{\mathrm{\prime }}}$ and $t_{i^{\mathrm{\prime }}+1}\le t_{i^{\mathrm{\prime }\mathrm{\prime }}+1}\le d_k+\eta$, so $i^{\mathrm{\prime }}\in J_k$.

${\sum }_{i<n}{h}_i^k\le 2\eta$ for each $k<r$. If $J_k=\mathrm{\varnothing }$ the sum is $0$. Otherwise let $p:=\min J_k$ and let $q$ be the least natural with $q>p$ and $q\mathrm{\notin }{J}_k$, which exists by [L9] because $n\mathrm{\notin }{J}_k$; 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 ${\sum }_{i<n}{h}_i^k={\sum }_{i=p}^{q-1}{\mathrm{\Delta }}_i=t_q-t_p$ 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 ${\beta }_i:={\mathrm{\Delta }}_i$ for $i$ bad and ${\beta }_i:=0$ for $i$ good. Then ${\beta }_i\le {\sum }_{k<r}{h}_i^k$ for every $i<n$, all terms being nonnegative and a bad $i$ lying in some $J_k$; so by [L8], ${\sum }_{i<n}{\beta }_i\le {\sum }_{i<n}{\sum }_{k<r}{h}_i^k={\sum }_{k<r}{\sum }_{i<n}{h}_i^k\le {\sum }_{k<r}2\eta =2\eta \iota (r)$, using step 10.1.

For every $i<n$ one has $(M_i-m_i){\mathrm{\Delta }}_i\le \theta {\mathrm{\Delta }}_i+2M_{+}{\beta }_i$: 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 ${\beta }_i={\mathrm{\Delta }}_i$, together with $\theta {\mathrm{\Delta }}_i\ge 0$.

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

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