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Lectures on the theory of functions of real variables

James, Pierpont

History

BiblioLife

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1912 Excerpt: ...For let S, J£ be a pair of complementary e/4 normal enclosures belonging to 21; let g, F be similar enclosures of SB. Let e = Z»(g, 1?), f-Dv% F). Then e e/2, f e/2, by 355, 2. Now ® = Dv((£, 5) is a normal metric enclosure of 35. Moreover its cells g which contain points of 35 and C(35) lie among the cells of e, f. Hence ae+T+ Thus by 357, 35 is measurable. 2. Let?(, S 6e measurable. Let Then For Hence Forlet o,-at-aI, o.-a.-a,.. For uniformity let us set at = 8. Then a = 2o». As each Q„ is measurable 3 = 2 = lim(ai+ +o.)--00 = lira in. 361. Let 2Ij, S be measurable and their union 21 limited. If 3) = Dv 2l„S 0, ft 18 measurable. For let 8 lie in the metric set 2ft; let ® + D=m, an+4.=a» as usual. Now 3) denoting the points common to all the a„, no point of D can lie in all of the a„, hence it lies in some one or more of the An. Thus DAn. (1 On the other hand, a point of An lies in some Am, hence it does not lie in a„. Hence it does not lie in 3). Thus it lies in D. Hence AnD. (2 From 1), 2) we have D= A As each An is measurable, so is D. Hence 3) is. 362. If aa is an enumerable set of measurable aggregates, their divisor 3) is measurable, and © = liman. B=ae For as usual let 2, An be the complements of 3), 8, with respect to some metric set 9ft. Then 2=M»j, 4.4.+1. Hence by 360, 3-Km A.. Let era=(emil, em2, e,„,8...) + em, (3 then em is measurable. By this process the metric or measurable cell em falls into an enumerable set of non-overlapping measurable cells, as indicated in 3). If we suppose this decomposition to take place for each cell of ©, we shall say we have superimposed 8 on S. 364. (TF. H. Young. Let S be any complete set in limited SJ. Then = Max S. (1 For le...

Lectures on the theory of functions of real variables Volume 2

RareBooksClub.com

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1912 Excerpt: ... enclosure of 21. Thus I 2, 2n + £) = 2?„ + e, which establishes 1). 357. Let @ be a distinct measurable enclosure of 21. Let f denote those cells containing points of the complement A. If for each e 0 there exists an g such that J e, then 21 is measurable. For let (g = e + f. Then c 21. Hence c 21 by 348. But ii=t+fa+«. Hence %-%«, and thus IF--w 358. 1. The divisor D of two measurable sets 21, S3 t also measurable. For let (g, E be a pair of complementary c/4 normal enclosures belonging to 21; let g, F be similar enclosures of 83. Let e = Zi(£, J?), f = ZK& &) Then e e/2, f e/2, by 355, 2. Now ® = Dv(@, 5) is a normal metric enclosure of 35. Moreover its cells 9 which contain points of 35 and C(35) lie among the cells of e, f. Hence Thus bv 357, 35 is measurable. 2. Let 21, 58 be measurable. Let Then For Hence S3 is measurable by 353, 1, and the relation 1) holds by the same theorem. 360. Let fl1fl%---bea set of measurable aggregates whose union 21 is limited. Then 21 is measurable, and a=uma,. then em is measurable. By this process the metric or measurable cell em falls into an enumerable set of non-overlapping measurable cells, as indicated in 3). If we suppose this decomposition to take place for each cell of 6, we shall say we have superimposed & on (g. 364. (W. H. Young. Let 6 be any complete set in limited 91. Then a = Max 6. (1 For let 81 lie within a cube 9K, and let A =. 9W-a, C= 2K-6 be as usual the complementary sets. Let S= b„ be a border set of S 328. It is also a nonoverlapping enclosure of 0; we may suppose it is a standard enclosure of C. Let E be a standard e-enclosure of A. Let us superimpose E on SB, getting a measurable enclosure A of both (7 »nd A. Then C=7AAA. Hpncs 6 = aw _ c= s&...

Lectures on the Theory of Functions of Real Variables Volume 2

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1905 Excerpt: ...in two sets. In one, call it V0, we put the points for which Ax = 0. Then Vo = ± ± 1 ni m + 1 In the other set, call it Vu we put all the other points of V. We can now show for the function y = f(x) in the above theorem, that 1) is true for each one of these sets of points, and therefore true for both together. 379. We give now the proof of 378. Let t be any point in T let x he the corresponding point in X. Let As, Ay he the increments of x, y, corresponding to the increment At of t. Case 1. There exists a V(t), in which AxO. The identity Ay = Ay At,j At Ax ' At does not involve a division by 0, as Ax 0. which proves the theorem for this case. Case 2. Ax = 0 for some point in every V(t. Let FJ, be the points of V(t), for which Ax = 0. Let Fj be the remaining points of V(t). If we show linif, and lim, (3 At dx At have one and the same value for every sequence of points whose limit is t, we have proved 2) for this case. Let A be any sequence in V0. Then 11=0, (4 At For, Ax being 0 for every point of A, y=f(x) receives no increment, and hence Ay = 0 in A. Thus, for every sequence A, the two limits in 3) have the same value, viz. 0. Let now B be any sequence in Vl which = t. Let the image of the points B be the points C, on the z-axis. Then, by 292, lim--f = lim lim-r-r j At o Ax « At ax B At Thus the two limits of 3) are the same for each sequence B. It remains to show that one is 0. Now, by 4), for any sequence whose limit is the point f. Hence the right side of 5) is 0. Thus the two limits 3) have the value 0 for every sequence A or B. These limits therefore have the value 0 for any sequence, whether its points all lie in V0, or in Vv or partly in V0 and partly in Vy 380. The demonstration, as ordinarily given, rests on the identity... Ay _ Ay...

Lectures on the Theory of Functions of Real Variables Volume 1

<p>The ever-imaginative Gary Schocker has provided flutists with a fresh approach to a shepherds dozen of favorite Christmas melodies, sacred and secular a collection for all occasions: church services, holiday parties, and just fun duet playing! The book includes: Jingle Bells, It Came upon the Midnight Clear, Lo, How a Rose Eer Blooming, O Little Town of Bethlehem, Angels We Have Heard on High, We Wish You a Merry Christmas, What Child Is This? &amp; Adeste Fidelis, The First Noël, Joy to the World, Silent Night, Sing We Now of Christmas, Deck the Halls, and God Rest Ye Merry, Gentlemen.</p>

Two Pipers Piping (for 2 Flutes)

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