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 $B:=\{i<n:(t_i,t_{i+1})\cap E_{\sigma }\mathrm{\ne }\mathrm{\varnothing }\}$ and put ${\chi }_i:=1$ for $i\in B$ and ${\chi }_i:=0$ otherwise.

The exhaustion of $D$. Put ${\sigma }_k:=1\mathrm{/}(k+1)$ 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 $1\mathrm{/}(k+1)<{\omega }_f(x)$, 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 $\epsilon >0$ be given. Put ${\sigma }^{\mathrm{\prime }}:=\epsilon \cdot (2(b-a){)}^{-1}$ and ${\tau }^{\mathrm{\prime }}:=\epsilon \cdot (8M_{+}{)}^{-1}$, both positive by [L14]. By step 1.1 and [L8], $E_{{\sigma }^{\mathrm{\prime }}}\subseteq D$ 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].

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

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

By [L9] applied to the compact null set $E_{{\sigma }^{\mathrm{\prime }}}$, 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 $E_{{\sigma }^{\mathrm{\prime }}}\subseteq {\bigcup }_{j\le m}[c_j,e_j]$ and ${\sum }_{j\le m}(e_j-c_j)\le {\tau }^{\mathrm{\prime }}\cdot 2^{-1}$. Put $\mu :={\tau }^{\mathrm{\prime }}\cdot (4(m+1){)}^{-1}>0$ 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 ${\sum }_{j\le m}((e_j-c_j)+2\mu )\le {\tau }^{\mathrm{\prime }}\cdot 2^{-1}+2\mu (m+1)={\tau }^{\mathrm{\prime }}\cdot 2^{-1}+{\tau }^{\mathrm{\prime }}\cdot 2^{-1}={\tau }^{\mathrm{\prime }}$, 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 $j\mathrm{\notin }B$, 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 ${\sum }_{j\le 2n}(q_j-p_j)={\sum }_{i<n}{\chi }_i{\mathrm{\Delta }}_i+0=\lambda <\tau$, 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 ${\omega }_f([a,b]\cap (u,v))<{\sigma }^{\mathrm{\prime }}$. Every $x\in [a,b]$ lies in a member: if $x\in E_{{\sigma }^{\mathrm{\prime }}}$ 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 $x\mathrm{\notin }{E}_{{\sigma }^{\mathrm{\prime }}}$ then ${\omega }_f(x)<{\sigma }^{\mathrm{\prime }}$, so by [L4] some real $\rho >0$ has ${\omega }_f([a,b]\cap N_{\rho }(x))<{\sigma }^{\mathrm{\prime }}$, 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 each $E_{{\sigma }_k}$ is compact by [L6] and [L7], so [L11] applies to this sequence of compact content-zero sets. Thus ${\bigcup }_k{E}_{{\sigma }_k}=D$ is null by step 2.2.

$s=b$. By step 5.2 fix $({\alpha }^{\mathrm{\prime }},{\beta }^{\mathrm{\prime }})\in \mathcal{W}$ with $s\in ({\alpha }^{\mathrm{\prime }},{\beta }^{\mathrm{\prime }})$. Since $\sup S=s>{\alpha }^{\mathrm{\prime }}$ there is $x\in S$ with $x>{\alpha }^{\mathrm{\prime }}$, and $x\le s$. Suppose $s<b$ and choose a real $y$ with $s<y<\min \{b,{\beta }^{\mathrm{\prime }}\}$, possible because $s<b$ and $s<{\beta }^{\mathrm{\prime }}$. Then ${\alpha }^{\mathrm{\prime }}<x\le s<y<{\beta }^{\mathrm{\prime }}$, so $[x,y]\subseteq ({\alpha }^{\mathrm{\prime }},{\beta }^{\mathrm{\prime }})$, 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 $({\alpha }^{\mathrm{\prime }\mathrm{\prime }},{\beta }^{\mathrm{\prime }\mathrm{\prime }})\in \mathcal{W}$ with $b\in ({\alpha }^{\mathrm{\prime }\mathrm{\prime }},{\beta }^{\mathrm{\prime }\mathrm{\prime }})$. Since $\sup S=b>{\alpha }^{\mathrm{\prime }\mathrm{\prime }}$ there is $x\in S$ with $x>{\alpha }^{\mathrm{\prime }\mathrm{\prime }}$ and $x\le b$. If $x=b$ there is nothing to prove; otherwise ${\alpha }^{\mathrm{\prime }\mathrm{\prime }}<x<b<{\beta }^{\mathrm{\prime }\mathrm{\prime }}$ gives $[x,b]\subseteq ({\alpha }^{\mathrm{\prime }\mathrm{\prime }},{\beta }^{\mathrm{\prime }\mathrm{\prime }})$, and appending $b$ as in step 7.2 puts $b$ in $S$. So there is a partition $P^{\mathrm{\prime }}=(n^{\mathrm{\prime }},t^{\mathrm{\prime }})$ of $[a,b]$, with subintervals $I_i^{\mathrm{\prime }}$ and lengths ${\mathrm{\Delta }}_i^{\mathrm{\prime }}$ for $i<n^{\mathrm{\prime }}$, every subinterval of which lies in a member of $\mathcal{W}$.

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

Bounding the bad lengths. For $j\le m$ put $J_j:=\{i<n^{\mathrm{\prime }}:c_j-\mu \le t_i^{\mathrm{\prime }}\text{ and }{t}_{i+1}^{\mathrm{\prime }}\le e_j+\mu \}$ and $h_i^j:={\mathrm{\Delta }}_i^{\mathrm{\prime }}$ for $i\in J_j$, $h_i^j:=0$ otherwise; a bad $i$ lies in some $J_j$ by step 9.1. Each $J_j$ consists of consecutive indices, since $i<i^{†}<i^{‡}$ with $i,i^{‡}\in J_j$ gives $c_j-\mu \le t_i^{\mathrm{\prime }}\le t_{i^{†}}^{\mathrm{\prime }}$ and $t_{i^{†}+1}^{\mathrm{\prime }}\le t_{i^{‡}+1}^{\mathrm{\prime }}\le e_j+\mu$.

${\sum }_{i<n^{\mathrm{\prime }}}{h}_i^j\le (e_j-c_j)+2\mu$ for each $j\le m$: the sum is $0$ when $J_j=\mathrm{\varnothing }$; otherwise let $p:=\min J_j$ and let $q$ be the least natural with $q>p$ and $q\mathrm{\notin }{J}_j$, which exists by [L13] since $n^{\mathrm{\prime }}\mathrm{\notin }{J}_j$, 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 ${\sum }_{i<n^{\mathrm{\prime }}}{h}_i^j={\sum }_{i=p}^{q-1}{\mathrm{\Delta }}_i^{\mathrm{\prime }}=t_q^{\mathrm{\prime }}-t_p^{\mathrm{\prime }}\le (e_j+\mu )-(c_j-\mu )$, using $p\in J_j$ and $q-1\in J_j$.

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

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

The real $\epsilon >0$ of step 2.3 was arbitrary and step 13.1 produced a partition with $U-L<\epsilon$, 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 ${\sigma }^{\mathrm{\prime }},{\tau }^{\mathrm{\prime }},P^{\mathrm{\prime }},\mathcal{W},S$.