Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a,b]$ is Riemann integrable if and only if its set of discontinuities has measure zero

For a real $\sigma > 0$ put $E_\sigma := \{\, x \in [a,b] : \omega_f(x) \ge \sigma \,\}$. By [L5], $E_\sigma \subseteq D$ for every $\sigma > 0$.

Each $E_\sigma$ has content zero, assuming $f$ integrable. Let $\sigma > 0$ and $\tau > 0$ be real. By [L3] fix a partition $P = (n,t)$ with $U(f,P) - L(f,P) < \sigma\tau \cdot 2^{-1}$. Let $B := \{\, i < n : (t_i,t_{i+1}) \cap E_\sigma \ne \varnothing \,\}$ and put $\chi_i := 1$ for $i \in B$ and $\chi_i := 0$ otherwise.

The exhaustion of $D$. Put $\sigma_k := 1/\iota(k+1)$ for $k \in \mathbb{N}$, a positive real by [L10]. Then $D = \bigcup_{k \in \mathbb{N}} E_{\sigma_k}$. For the inclusion from left to right, let $x \in D$, so $\omega_f(x) > 0$ by [L5] and $\omega_f(x)$ is a real by [L4]; if $\omega_f(x) \ge 1 = \sigma_0$ then $x \in E_{\sigma_0}$, and otherwise [L10] gives a natural $k+1 \ge 1$ with $1/\iota(k+1) < \omega_f(x)$, so $x \in E_{\sigma_k}$. For the reverse inclusion, $x \in E_{\sigma_k}$ gives $\omega_f(x) \ge \sigma_k > 0$, hence $x \in D$ by [L5].

The converse; this half of the proof is steps 2.3, 3.2, 4.2, 5.2, 6.2, 7.2, 8.1, 9.1, 10.1, 11.1, 12.1 and 13.1, and its symbols are its own. Assume $D$ null and let a real $\varepsilon > 0$ be given. Put $\sigma' := \varepsilon \cdot \bigl(2(b-a)\bigr)^{-1}$ and $\tau' := \varepsilon \cdot \bigl(8M_{+}\bigr)^{-1}$, both positive by [L14]. By step 1.1 and [L8], $E_{\sigma'} \subseteq D$ is null.

For $i \in B$ one has $M_i - m_i \ge \sigma$: fix $x \in (t_i,t_{i+1}) \cap E_\sigma$; since $(t_i,t_{i+1})$ is open there is a real $\rho > 0$ with $N_\rho(x) \subseteq (t_i,t_{i+1})$, so $[a,b] \cap N_\rho(x) \subseteq I_i$ and [L4] gives $\sigma \le \omega_f(x) \le \omega_f([a,b] \cap N_\rho(x)) \le \omega_f(I_i) = M_i - m_i$ by [L2].

$E_{\sigma'}$ is compact: by [L6] there is a closed $G$ with $E_{\sigma'} = [a,b] \cap G$, an intersection of two closed sets, hence closed; and $E_{\sigma'} \subseteq [a,b]$ is bounded. So [L7] applies.

Hence $\sigma\chi_i\Delta_i \le (M_i - m_i)\Delta_i$ for every $i < n$, the case $i \notin B$ because both $M_i - m_i \ge 0$ and $\Delta_i > 0$. Summing and using [L12] and [L2]: $\sigma \sum_{i<n}\chi_i\Delta_i \le U(f,P) - L(f,P) < \sigma\tau \cdot 2^{-1}$, so $\lambda := \sum_{i<n}\chi_i\Delta_i < \tau \cdot 2^{-1}$.

By [L9] applied to the compact null set $E_{\sigma'}$, it has content zero, so by [L8] there are $m \in \mathbb{N}$ and reals $c_0 \le e_0, \dots, c_m \le e_m$ with $E_{\sigma'} \subseteq \bigcup_{j \le m}[c_j,e_j]$ and $\sum_{j\le m}(e_j-c_j) \le \tau' \cdot 2^{-1}$. Put $\mu := \tau' \cdot \bigl(4(\iota(m)+1)\bigr)^{-1} > 0$ and $O_j := (c_j - \mu,\ e_j + \mu)$, an open interval containing $[c_j,e_j]$, of length $(e_j - c_j) + 2\mu$. Then $\sum_{j\le m}\bigl((e_j-c_j)+2\mu\bigr) \le \tau' \cdot 2^{-1} + 2\mu(\iota(m)+1) = \tau' \cdot 2^{-1} + \tau' \cdot 2^{-1} = \tau'$, by [L12] and [L10].

$E_\sigma$ is covered by the finite list of $2n+1$ closed intervals $[p_j,q_j]$, $j \le 2n$, defined by $[p_j,q_j] := [t_j,t_{j+1}]$ for $j < n$ with $j \in B$, $[p_j,q_j] := [a,a]$ for $j < n$ with $j \notin B$, and $[p_j,q_j] := [t_{j-n},t_{j-n}]$ for $n \le j \le 2n$: indeed a point of $E_\sigma$ lies in $[a,b]$, hence is one of $t_0,\dots,t_n$ or lies in some $(t_i,t_{i+1})$, and in the latter case $i \in B$. Its total length is $\sum_{j \le 2n}(q_j - p_j) = \sum_{i<n}\chi_i\Delta_i + 0 = \lambda < \tau$, by splitting the sum at $n$ ([L12]).

The family of good intervals. Let $\mathcal{W}$ be the set of all open intervals $(u,v)$ with $u < v$ such that either $(u,v) \subseteq O_j$ for some $j \le m$, or $\omega_f([a,b]\cap(u,v)) < \sigma'$. Every $x \in [a,b]$ lies in a member: if $x \in E_{\sigma'}$ then $x \in [c_j,e_j] \subseteq O_j$ for some $j \le m$ by step 4.2, and $O_j$ is itself a member; and if $x \notin E_{\sigma'}$ then $\omega_f(x) < \sigma'$, so by [L4] some real $\rho > 0$ has $\omega_f([a,b]\cap N_\rho(x)) < \sigma'$, and $N_\rho(x) = (x-\rho,x+\rho)$ is a member containing $x$.

As $\tau > 0$ was arbitrary, $E_\sigma$ has content zero by [L8], hence measure zero by [L9]; this used only that $f$ is integrable.

Cousin's construction: a partition each of whose subintervals lies in a member of $\mathcal{W}$. Let $S$ be the set of $y \in (a,b]$ such that some partition of $[a,y]$ has every subinterval contained in a member of $\mathcal{W}$. $S$ is nonempty: by step 5.2 fix $(\alpha,\beta) \in \mathcal{W}$ with $a \in (\alpha,\beta)$ and put $y_0 := \min\{(a+b)\cdot 2^{-1},\ (a+\beta)\cdot 2^{-1}\}$, so $a < y_0 \le b$, $y_0 < \beta$ and $\alpha < a$; the one-subinterval partition of $[a,y_0]$ has $[a,y_0] \subseteq (\alpha,\beta)$, so $y_0 \in S$. Also $S$ is bounded above by $b$, so $s := \sup S$ exists by [L13] and $a < y_0 \le s \le b$.

Integrability implies $D$ null. Assume $f$ integrable. By step 6.1 each $E_{\sigma_k}$ is null, and $k \mapsto E_{\sigma_k}$ is a sequence of subsets of $\mathbb{R}$, so [L11] applies and $\bigcup_k E_{\sigma_k} = D$ is null by step 2.2. This is the only use of [A1] in the proof.

$s = b$. By step 5.2 fix $(\alpha',\beta') \in \mathcal{W}$ with $s \in (\alpha',\beta')$. Since $\sup S = s > \alpha'$ there is $x \in S$ with $x > \alpha'$, and $x \le s$. Suppose $s < b$ and choose a real $y$ with $s < y < \min\{b, \beta'\}$, possible because $s < b$ and $s < \beta'$. Then $\alpha' < x \le s < y < \beta'$, so $[x,y] \subseteq (\alpha',\beta')$, and appending $y$ to a partition of $[a,x]$ witnessing $x \in S$ gives one for $[a,y]$ by [L1]; hence $y \in S$ with $y > \sup S$, which is impossible.

$b \in S$. By step 5.2 fix $(\alpha'',\beta'') \in \mathcal{W}$ with $b \in (\alpha'',\beta'')$. Since $\sup S = b > \alpha''$ there is $x \in S$ with $x > \alpha''$ and $x \le b$. If $x = b$ there is nothing to prove; otherwise $\alpha'' < x < b < \beta''$ gives $[x,b] \subseteq (\alpha'',\beta'')$, and appending $b$ as in step 7.2 puts $b$ in $S$. So there is a partition $P' = (n',t')$ of $[a,b]$, with subintervals $I'_i$ and lengths $\Delta'_i$ for $i < n'$, every subinterval of which lies in a member of $\mathcal{W}$.

Good and bad subintervals. Write $M'_i := \sup f[I'_i]$ and $m'_i := \inf f[I'_i]$. Call $i < n'$ good when $I'_i \subseteq W$ for some $W = (u,v) \in \mathcal{W}$ with $\omega_f([a,b]\cap(u,v)) < \sigma'$, and bad otherwise. For a good $i$, $I'_i \subseteq [a,b] \cap W$, so $M'_i - m'_i = \omega_f(I'_i) \le \omega_f([a,b]\cap W) < \sigma'$ by [L2] and [L4]. For a bad $i$, step 8.1 supplies a member containing $I'_i$, and it is not of the second kind, so $I'_i \subseteq O_j$ for some $j \le m$.

Bounding the bad lengths. For $j \le m$ put $J_j := \{\, i < n' : c_j - \mu \le t'_i \text{ and } t'_{i+1} \le e_j + \mu \,\}$ and $h^{j}_i := \Delta'_i$ for $i \in J_j$, $h^{j}_i := 0$ otherwise; a bad $i$ lies in some $J_j$ by step 9.1. Each $J_j$ consists of consecutive indices, since $i < i^{\dagger} < i^{\ddagger}$ with $i,i^{\ddagger} \in J_j$ gives $c_j - \mu \le t'_i \le t'_{i^{\dagger}}$ and $t'_{i^{\dagger}+1} \le t'_{i^{\ddagger}+1} \le e_j + \mu$.

$\sum_{i<n'}h^{j}_i \le (e_j - c_j) + 2\mu$ for each $j \le m$: the sum is $0$ when $J_j = \varnothing$; otherwise let $p := \min J_j$ and let $q$ be the least natural with $q > p$ and $q \notin J_j$, which exists by [L13] since $n' \notin J_j$, so that $J_j = \{\, i : p \le i < q \,\}$ by step 10.1. Splitting the sum at $p$ and at $q$ and discarding the vanishing outer parts, then telescoping ([L12]), gives $\sum_{i<n'}h^{j}_i = \sum_{i=p}^{q-1}\Delta'_i = t'_q - t'_p \le (e_j+\mu) - (c_j-\mu)$, using $p \in J_j$ and $q-1 \in J_j$.

Put $\beta_i := \Delta'_i$ for bad $i$ and $\beta_i := 0$ for good $i$. Then $\beta_i \le \sum_{j \le m}h^{j}_i$ pointwise by step 10.1, all terms being nonnegative, so by [L12] and step 11.1, $\sum_{i<n'}\beta_i \le \sum_{j\le m}\sum_{i<n'}h^{j}_i \le \sum_{j\le m}\bigl((e_j-c_j)+2\mu\bigr) \le \tau'$, the last step by step 4.2.

For every $i < n'$, $(M'_i-m'_i)\Delta'_i \le \sigma'\Delta'_i + 2M_{+}\beta_i$: for good $i$ by step 9.1 and $\beta_i \ge 0$, for bad $i$ by $M'_i - m'_i \le 2M_{+}$ from [L2] and $\beta_i = \Delta'_i$. Summing over $i < n'$ and using [L12], [L1] and step 12.1: $U(f,P')-L(f,P') \le \sigma'(b-a) + 2M_{+}\tau' = \varepsilon\cdot 2^{-1} + \varepsilon \cdot 4^{-1} < \varepsilon$.

The real $\varepsilon > 0$ of step 2.3 was arbitrary and step 13.1 produced a partition with $U - L < \varepsilon$, so $f$ is integrable by [L3]. With step 7.1 this proves both implications, and the criterion is established; the forward half is steps 1.1, 2.1, 2.2, 3.1, 4.1, 5.1, 6.1 and 7.1, working with $\sigma, \tau, P, B, \chi, \lambda$, and the converse half is the steps named in step 2.3, working with $\sigma', \tau', P', \mathcal{W}, S$.