상위 노트: Study MOC
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ap-calc-bc· Type:math-concept· Stage:evergreenParent: Study MOC
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한 줄 요약
a bounded function is riemann integrable when every discontinuity of it is jump type. also, jump discontinuities are always countable, so the real condition is the type of discontinuity, not the number of them.
Definition 1.
Integrability f is integrable on [a, b] if the riemann sums go to one same value, no matter how we pick the sample points: the important part: one value for every choice of sample points. the dirichlet function will fail exactly at this part.
Definition 2.
Countable A mathematical set is countable if either it is finite or it can be put in one to one correspondence with the set of natural numbers.
Definition 3.
Jump discontinuity f has a jump discontinuity at a point p if both one-sided limits exist there, but they are not equal: the part that both limits must exist is easy to forget. if a one-sided limit does not even exist (dirichlet: at every point), that is not a jump. that one is called a discontinuity of the second kind.
Definition 4.
Oscillation on a set A: osc(f, A) = (sup of f on A) minus (inf of f on A). it means how much f can move inside A. at a point x: shrink an interval around x and watch the osc. the limit exists, because the osc only goes down when the interval shrinks. two facts:
- w(x) = 0 exactly when f is continuous at x
- at a jump point p, with left limit L- and right limit L+: w(p) is the biggest of L-, L+, f(p) minus the smallest of them. so w(p) is at least the jump size, which is bigger than 0.
Setup
integral interval [a, b], partition P: for each subinterval, so w_i is sup minus inf, not max minus min. for a discontinuous f, the max and min maybe do not exist. here U(P) is the biggest riemann sum I can make on P (push every sample point to where f is big), and L(P) is the smallest one.
| symbols | |
|---|---|
| Δt_i | t_i − t_(i−1) |
| |P| | the biggest Δt_i |
| W | osc(f, [a, b]), the total shake of f |
| M | the biggest |f(x)| |
정리
one fact about real numbers that I will trust if closed intervals are nested (I_1 ⊇ I_2 ⊇ I_3 ⊇ …) and their lengths go to 0, then there is exactly one point that lives inside all of them. I cannot prove this with AP calc tools. but it is basically saying the real number line has no hole, so I will just trust it. this fact appears every time I cut an interval in half again and again.
정리
Criterion (the main tool) bounded f is integrable ⟺ for every e > 0 there is a partition P with Gap(P) < e. meaning of Gap(P): it is exactly how much the riemann sum can still move when we re-pick the sample points on P. direction 1 (one line proof): if f is integrable with value I, then for a small enough mesh, every riemann sum sits between I − e/4 and I + e/4. so U(P) and L(P) sit there too, so Gap(P) is at most e/2. → so if the Gap stays above some fixed number for every partition, then f is NOT integrable. dirichlet will use this direction. direction 2 (Gap small ⟹ integrable): a fully careful proof needs one more argument that compares two different partitions. I did not write it here. I will trust it in this note, and maybe make a separate note for it.
Lemma 0. jump points are always countable
정리
for ANY function f on [a, b], the set J of jump discontinuities is countable. so a case like uncountable many jumps does not exist from the start.
증명
split J into up jumps (L- < L+) and down jumps (L- > L+). take an up jump point p, with left limit L- and right limit L+.
- pick a rational number q with L- < q < L+
- because the two one-sided limits exist, p has a left zone (p − d, p) where f < q, and a right zone (p, p + d) where f > q, for some small d > 0
- pick a rational u inside the left zone, and a rational v inside the right zone. now p gets the tag (q, u, v). three rational numbers.
claim: two different up jump points never get the same tag. suppose p < p’ share the same tag (q, u, v). then u < p < p’ < v. v sits inside the right zone of p, so f > q on the whole (p, v). u sits inside the left zone of p’, so f < q on the whole (u, p’). but the interval (p, p’) is not empty, and it sits inside both. so f > q and f < q at the same points. impossible. so different up jumps always get different rational triples. triples of rational numbers are countable (same zigzag counting idea as counting fractions), so the up jumps are countable. the down jumps work the same way, and a union of two countable sets is countable.
Main Theorem
정리
f is bounded on [a, b], and every discontinuity of f is a jump discontinuity. then f is integrable. (at the two endpoints only one side makes sense, so there the condition is: that one-sided limit exists.) working form: both one-sided limits of f exist at every point. this is the only thing the proof will use.
Case 0. no discontinuity at all (continuous f)
in AP calc we just accept that continuous means integrable. here is an actual proof, and its lemma becomes the engine for everything after.
정리
Lemma B (cutting in half) if w(x) < e for every x in a closed interval K, then K can be cut into finite many closed pieces, and every piece has osc ≤ e.
증명
let me call an interval good when such a finite cut exists. suppose K is not good. cut it in half. if both halves were good, joining the two cuts would make K good. so at least one half is not good. cut that half again. repeating this, I get nested not-good intervals I_1 ⊇ I_2 ⊇ … with lengths going to 0. by the trusted fact, they trap exactly one point x*. but w(x*) < e. the osc of shrinking intervals around x* goes down to w(x*), so some small interval around x* already has osc < e. for a big k, I_k sits inside that small interval, so osc(I_k) < e. then I_k is good with just one piece. contradiction.
증명
Case 0 finished continuous means w(x) = 0 at every point, so lemma B works for any e’. use e’ = e/(b − a) and take the finite cut as the partition P. then Gap(P) is at most e’ times the total length, which is e’(b − a) = e. done by the criterion.
Case 1. finite many jump discontinuities
plan: cover the k jump points with very small intervals, and run case 0 on the rest.
증명
jumps at p_1 < … < p_k, fix e > 0. (if f has no value at some p_j, give it any bounded value. see the repair log below.)
- pick h > 0 so small that the intervals [p_j − h, p_j + h] do not touch each other (cut them at a, b if needed), and 2khW ≤ e/2
- cost of the covers: each cover piece has osc at most W and length at most 2h, and there are k of them. so together they give at most 2khW ≤ e/2 to the Gap
- the leftover: finite many closed intervals with no jump inside, so f is continuous on each of them. case 0 cuts each one so that its part of the Gap is at most e/(2(b − a)) times its length. adding all of them: at most e/2
total Gap ≤ e.
Case 2. infinite many jump discontinuities (the general case)
countably many, even dense, anything is fine, as long as all of them are jumps. (and by lemma 0, countable is the only infinite option anyway.)
the problem: I cannot cover infinite many points with finite many small intervals. here is the key idea that saves everything:
big jumps cannot be infinite many. if they were, they must pile up somewhere, and at the piling point a one-sided limit breaks. so for each size e, only finite many points shake more than e. then the finite-case trick works again, at every scale.
정리
Lemma A (big shakes are only finite many) under the condition of the theorem, for every e > 0, the set D_e of points x with w(x) ≥ e is finite.
증명
suppose D_e has infinite many points. step 1 (they pile up). cut [a, b] in half. one half still contains infinite many points of D_e. cut that half again, and repeat. I get nested intervals with lengths going to 0, each containing infinite many points of D_e. by the trusted fact they trap one point c. every tiny interval around c contains some I_k, so it contains infinite many points of D_e. so I can pick points s_1, s_2, … in D_e with s_j ≠ c and s_j → c. infinite many of them are on the same side of c. say the left side (the right side case is the same). call them x_1, x_2, …, so x_m < c and x_m → c. step 2 (pull out two crowds of values). w(x_m) ≥ e means every interval around x_m has osc ≥ e, because the shrinking oscs go down TO w(x_m), so each of them is at least e. take the interval around x_m with radius r_m = (c − x_m)/2. the osc there is at least e, so I can pick u_m near the sup and v_m near the inf with because of the size of r_m, both u_m and v_m are still smaller than c, and u_m, v_m → c. step 3 (the left limit dies). if the left limit L of f at c existed, then f(u_m) → L and f(v_m) → L, so f(u_m) − f(v_m) → 0. but it stays above e/2. contradiction. so the left limit at c does not exist. then c is a discontinuity, but not a jump (a jump needs both one-sided limits). this breaks the condition of the theorem.
증명
Main theorem, finished fix e > 0 and let e0 = e/(2(b − a)). by lemma A, the set D_e0 is finite. call its points z_1, …, z_m. (if it is empty: just lemma B, done.)
- cover: put a small interval around each z_j. keep the radius smaller than half of the smallest gap between the z_j (so the covers never touch each other), and make the total length of the covers at most e/(2W). these pieces give at most W times e/(2W), which is e/2, to the Gap. also, because the covers do not touch each other and are small, no endpoint of a cover can be another z_i. so the leftover truly avoids every point of D_e0.
- leftover: finite many closed intervals where every point has w(x) < e0. lemma B cuts each of them into pieces with osc ≤ e0. their total part of the Gap: at most e0 times (b − a), which is e/2
Gap ≤ e. by the criterion, f is integrable.