>>261-263
>>>260 その指摘が不明確かと

やれやれ、日wikipediaに書いてあることを自慢して
ちょっとツッコミあると沈没か？ｗ

さて
リーマン可積分⇒微分可能でない点の集合が測度０
　↓
リーマン可積分⇒連続でない点の集合が測度０
ですな

"R^n の有界閉区間 I 上の有界関数 f: I → R に対し、
f が I 上リーマン可積分であることと、
f がほとんど至るところ連続であること（※）は同値
（※ f の不連続点全体の集合が零集合）"
だったね

これの証明は、下記の英wikipediaにある
（Integrability 可積分性条件 the Lebesgue-Vitali theorem な）
(参考)
https://en.wikipedia.org/wiki/Riemann_integral
Riemann integral

Integrability
A bounded function on a compact interval [a, b] is Riemann integrable if and only if it is continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). This is the Lebesgue-Vitali theorem (of characterization of the Riemann integrable functions). It has been proven independently by Giuseppe Vitali and by Henri Lebesgue in 1907, and uses the notion of measure zero, but makes use of neither Lebesgue's general measure or integral.
The integrability condition can be proven in various ways,[4][5][6][7] one of which is sketched below.

Proof
The proof is easiest using the Darboux integral definition of integrability (formally, the Riemann condition for integrability) – a function is Riemann integrable if and only if the upper and lower sums can be made arbitrarily close by choosing an appropriate partition.
One direction can be proven using the oscillation definition of continuity:[8] For every positive ε, Let Xε be the set of points in [a, b] with oscillation of at least ε. Since every point where f is discontinuous has a positive oscillation and vice versa, the set of points in [a, b], where f is discontinuous is equal to the union over {X1/n} for all natural numbers n.

つづく