[過去ログ] ガロア第一論文と乗数イデアル他関連資料スレ6 (1002レス)
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266(3): 2024/01/29(月)07:58 ID:2Tor3z84(1/4) AAS
>>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
↓
省19
267(2): 2024/01/29(月)07:58 ID:2Tor3z84(2/4) AAS
つづき
If this set does not have zero Lebesgue measure, then by countable additivity of the measure there is at least one such n so that X1/n does not have a zero measure. Thus there is some positive number c such that every countable collection of open intervals covering X1/n has a total length of at least c. In particular this is also true for every such finite collection of intervals. This remains true also for X1/n less a finite number of points (as a finite number of points can always be covered by a finite collection of intervals with arbitrarily small total length).
For every partition of [a, b], consider the set of intervals whose interiors include points from X1/n. These interiors consist of a finite open cover of X1/n, possibly up to a finite number of points (which may fall on interval edges). Thus these intervals have a total length of at least c. Since in these points f has oscillation of at least 1/n, the infimum and supremum of f in each of these intervals differ by at least 1/n. Thus the upper and lower sums of f differ by at least c/n. Since this is true for every partition, f is not Riemann integrable.
We now prove the converse direction using the sets Xε defined above.[9] For every ε, Xε is compact, as it is bounded (by a and b) and closed:
For every series of points in Xε that is converging in [a, b], its limit is in Xε as well. This is because every neighborhood of the limit point is also a neighborhood of some point in Xε, and thus f has an oscillation of at least ε on it. Hence the limit point is in Xε.
Now, suppose that f is continuous almost everywhere. Then for every ε, Xε has zero Lebesgue measure. Therefore, there is a countable collections of open intervals in [a, b] which is an open cover of Xε, such that the sum over all their lengths is arbitrarily small. Since Xε is compact, there is a finite subcover – a finite collections of open intervals in [a, b] with arbitrarily small total length that together contain all points in Xε. We denote these intervals {I(ε)i}, for 1 ≤ i ≤ k, for some natural k.
省2
268(2): 2024/01/29(月)07:58 ID:2Tor3z84(3/4) AAS
つづき
We now show that for every ε > 0, there are upper and lower sums whose difference is less than ε, from which Riemann integrability follows. To this end, we construct a partition of [a, b] as follows:
Denote ε1 = ε / 2(b − a) and ε2 = ε / 2(M − m), where m and M are the infimum and supremum of f on [a, b]. Since we may choose intervals {I(ε1)i} with arbitrarily small total length, we choose them to have total length smaller than ε2.
Each of the intervals {J(ε1)i} has an empty intersection with Xε1, so each point in it has a neighborhood with oscillation smaller than ε1. These neighborhoods consist of an open cover of the interval, and since the interval is compact there is a finite subcover of them. This subcover is a finite collection of open intervals, which are subintervals of J(ε1)i (except for those that include an edge point, for which we only take their intersection with J(ε1)i). We take the edge points of the subintervals for all J(ε1)i − s, including the edge points of the intervals themselves, as our partition.
Thus the partition divides [a, b] to two kinds of intervals:
省7
269(2): 2024/01/29(月)08:07 ID:2Tor3z84(4/4) AAS
補足
>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.
”the Darboux integral definition of integrability”は、日wikipediaにも説明あるよ
もちろん、英wikipediaにも詳しい説明がある(下記)
(参考)
外部リンク:ja.wikipedia.org
省8
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