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  "body": "\n## Proof\n\nWe use the oscillation of $f$ and Riemann's criterion, and compactness for the converse.\nLet $a \u003c b$ and let $f \\colon [a, b] \\to \\mathbb{R}$ be bounded. Fix $K \u003e 0$ with $|f(x)| \\le K$ for all $x \\in [a, b]$.\nA set $E \\subseteq \\mathbb{R}$ has measure zero if for each $\\varepsilon \u003e 0$ it is covered by countably many open intervals of total length less than $\\varepsilon$.\nWrite $U(P)$ and $L(P)$ for the upper and lower [[Darboux sums|0QRQVH0CRW]] of $f$ for a [[partition|069P20KYQT]] $P$, with $M_i$ and $m_i$ as in their definition.\n\n1. For a nonempty set $A \\subseteq [a, b]$, let $\\omega(A) = \\sup f(A) - \\inf f(A)$, using the [[supremum|0BTMQ04556]] and the [[infimum|04KFTA34C5]]. If $A \\subseteq B$, then $\\omega(A) \\le \\omega(B)$.\n   For $x, y \\in A$ we have $|f(x) - f(y)| \\le \\omega(A)$.\n2. For $x \\in [a, b]$ and $r \u003e 0$, let $A_r(x) = (x - r, x + r) \\cap [a, b]$. The [[oscillation|5XW4N8KY3Y]] of $f$ at $x$ is $\\omega(x) = \\inf_{r \u003e 0} \\omega(A_r(x)) \\ge 0$.\n3. We claim that $f$ is [[continuous at|0EEE2CZB2A]] $x$ if and only if $\\omega(x) = 0$. Suppose first that $f$ is continuous at $x$, and let $\\varepsilon \u003e 0$.\n   By the [[epsilon-delta form of continuity|0W8YHHGCJG]], there is $r \u003e 0$ with $|f(y) - f(x)| \u003c \\varepsilon / 2$ for $y \\in A_r(x)$. So $f(A_r(x)) \\subseteq [f(x) - \\varepsilon/2, f(x) + \\varepsilon/2]$ and $\\omega(x) \\le \\omega(A_r(x)) \\le \\varepsilon$. Hence $\\omega(x) = 0$.\n4. Conversely, let $\\omega(x) = 0$ and $\\varepsilon \u003e 0$. There is $r \u003e 0$ with $\\omega(A_r(x)) \u003c \\varepsilon$. By step 1, $|f(y) - f(x)| \u003c \\varepsilon$ for all $y \\in A_r(x)$. So $f$ is continuous at $x$.\n5. For $\\alpha \u003e 0$, let $D_\\alpha = \\{x \\in [a, b] : \\omega(x) \\ge \\alpha\\}$, and let $D$ be the set of points where $f$ is not continuous.\n   By steps 3 and 4 and the [[Archimedean property|0A3GWZ86AY]], $x \\in D$ if and only if $\\omega(x) \\ge 1/k$ for some integer $k \\ge 1$. So $D = \\bigcup_{k \\ge 1} D_{1/k}$, and $D_\\alpha \\subseteq D$ for each $\\alpha \u003e 0$.\n6. We claim that $D_\\alpha$ is [[closed|0G5AV9Q4RR]]. Let $x \\in [a, b] \\setminus D_\\alpha$. Then there is $r \u003e 0$ with $\\omega(A_r(x)) \u003c \\alpha$.\n   For $y \\in A_r(x)$, let $s = r - |y - x| \u003e 0$. Then $A_s(y) \\subseteq A_r(x)$, so $\\omega(y) \\le \\omega(A_s(y)) \\le \\omega(A_r(x)) \u003c \\alpha$. Thus $(x - r, x + r)$ does not meet $D_\\alpha$.\n7. Each point outside $[a, b]$ also has an open interval around it that misses $[a, b] \\supseteq D_\\alpha$. So the complement of $D_\\alpha$ in $\\mathbb{R}$ is open, and $D_\\alpha$ is closed. It is bounded, so it is compact by the [[Heine–Borel theorem|07PFYV65MB]].\n\nFirst assume that $f$ is [[Riemann integrable|0N1JQXZXHC]].\n\n8. Fix an integer $k \\ge 1$ and let $\\varepsilon \u003e 0$. By [[Riemann's criterion|017DH8WXTE]], there is a partition $P = (x_0, \\dots, x_n)$ with $U(P) - L(P) \u003c \\frac{\\varepsilon}{2k}$.\n9. Let $B$ be the set of indices $i$ such that the open interval $(x_{i-1}, x_i)$ contains a point $y$ of $D_{1/k}$. For such $i$ and $y$, some $A_r(y)$ lies in $[x_{i-1}, x_i]$.\n   So $M_i - m_i = \\omega([x_{i-1}, x_i]) \\ge \\omega(A_r(y)) \\ge \\omega(y) \\ge 1/k$.\n10. Every term $(M_i - m_i)(x_i - x_{i-1})$ of $U(P) - L(P)$ is at least $0$. So step 9 gives $\\frac{1}{k} \\sum_{i \\in B} (x_i - x_{i-1}) \\le U(P) - L(P) \u003c \\frac{\\varepsilon}{2k}$. Hence $\\sum_{i \\in B} (x_i - x_{i-1}) \u003c \\frac{\\varepsilon}{2}$.\n11. Each point of $D_{1/k}$ lies in some $(x_{i-1}, x_i)$ with $i \\in B$, or is a point $x_j$ of $P$. Let $\\eta = \\frac{\\varepsilon}{4(n + 1)}$.\n    The intervals $(x_{i-1}, x_i)$ for $i \\in B$ and $(x_j - \\eta, x_j + \\eta)$ for $j = 0, \\dots, n$ cover $D_{1/k}$. Their total length is less than $\\frac{\\varepsilon}{2} + 2(n + 1)\\eta = \\varepsilon$. So $D_{1/k}$ has measure zero.\n12. Now let $\\varepsilon \u003e 0$. For each $k \\ge 1$, step 11 gives countably many open intervals that cover $D_{1/k}$ with total length less than $\\varepsilon / 2^{k + 1}$.\n13. Together these intervals cover $D$ by step 5. They form a countable family, since a [[countable union of countable sets is countable|0Z8FM54P86]].\n14. Any finitely many of them lie in the families for $k \\le N$, for some $N$. So their total length is less than $\\sum_{k = 1}^{N} \\varepsilon / 2^{k + 1} \u003c \\varepsilon / 2$.\n    Hence the total length of the whole family is at most $\\varepsilon / 2 \u003c \\varepsilon$. So $D$ has measure zero.\n\nNow assume that $D$ has measure zero.\n\n15. Let $\\varepsilon \u003e 0$, and let $\\alpha = \\frac{\\varepsilon}{4(b - a)}$. By step 5, $D_\\alpha \\subseteq D$. So countably many open intervals of total length less than $\\frac{\\varepsilon}{4K}$ cover $D_\\alpha$.\n16. By step 7 and [[compactness via open covers|0P6PMJB4MA]], finitely many of them, say $G_1, \\dots, G_N$, already cover $D_\\alpha$. Their total length is still less than $\\frac{\\varepsilon}{4K}$. If $D_\\alpha$ is empty, take $N = 0$.\n17. Let $z \\in [a, b]$ lie in none of the $G_j$. Then $z \\notin D_\\alpha$, so $\\omega(z) \u003c \\alpha$. Choose $r_z \u003e 0$ with $\\omega(A_{r_z}(z)) \u003c \\alpha$, and let $H_z = (z - r_z / 2, z + r_z / 2)$.\n18. The open intervals $G_1, \\dots, G_N$ and $H_z$ cover $[a, b]$. By the Heine–Borel theorem $[a, b]$ is compact, so finitely many $H_{z_1}, \\dots, H_{z_m}$ together with $G_1, \\dots, G_N$ cover $[a, b]$.\n19. Let $P = (x_0, \\dots, x_n)$ be the partition whose points are $a$, $b$ and all endpoints of these $N + m$ intervals that lie in $(a, b)$.\n20. Fix $i$, and let $c$ be the midpoint of $[x_{i-1}, x_i]$. Choose an interval $V$ of the finite cover that contains $c$, and take $V = G_j$ for some $j$ when this is possible.\n21. We claim $(x_{i-1}, x_i) \\subseteq V$. Suppose that $y \\in (x_{i-1}, x_i)$ is not in $V$. Since $V$ is an interval containing $c$, one endpoint of $V$ lies between $c$ and $y$.\n    That endpoint lies in $(x_{i-1}, x_i)$, so it is a point of $P$ strictly inside a subinterval of $P$. This is impossible. So $[x_{i-1}, x_i]$ lies in the closure of $V$.\n22. Call $i$ of type 1 if $V = G_j$, and of type 2 if $V = H_z$. For type 2, $[x_{i-1}, x_i] \\subseteq [z - r_z/2, z + r_z/2] \\cap [a, b] \\subseteq A_{r_z}(z)$. So $M_i - m_i \\le \\omega(A_{r_z}(z)) \u003c \\alpha$ by step 1.\n23. For type 1, $M_i - m_i \\le 2K$. The type 1 subintervals with $V = G_j$ do not overlap and lie in the closure of $G_j$. So their total length is at most the length of $G_j$.\n24. Adding over $j$, the type 1 subintervals have total length less than $\\frac{\\varepsilon}{4K}$. The type 2 subintervals have total length at most $b - a$.\n25. Hence\n    $$U(P) - L(P) = \\sum_{i = 1}^{n} (M_i - m_i)(x_i - x_{i-1}) \\le 2K \\cdot \\frac{\\varepsilon}{4K} + \\alpha (b - a) = \\frac{3 \\varepsilon}{4} \u003c \\varepsilon.$$\n26. Such a partition exists for each $\\varepsilon \u003e 0$. By [[Riemann's criterion|017DH8WXTE]], $f$ is Riemann integrable on $[a, b]$.\n\nSteps 14 and 26 prove both directions.\n\n## Idea\n\nThe oscillation $\\omega(x)$ measures how much $f$ jumps near $x$, and $f$ is discontinuous exactly where it is positive. If $f$ is integrable, the subintervals where $f$ jumps by at least $1/k$ must have small total length, so each set $D_{1/k}$ is small. Conversely, the points with large oscillation form a compact set, so finitely many short intervals cover them. Elsewhere $f$ varies little, and the upper and lower sums come close together.\n",
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