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By Morgan F.

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2. Corollary. The set T D fT 2 Im : spt T ² Bn 0,c1 , M T Ä c2 , and M ∂T Ä c3 g is totally bounded under F. Proof. 1(4). Since there are only finitely many such P, T is totally bounded. 3. 10]. If T 2 Im Rn with ∂T D 0, then there exists S 2 ImC1 Rn with ∂S D T and MS Here m/ mC1 Ä MT. D 2n2mC2 as in the Deformation Theorem. Remarks. That T bounds some rectifiable current S is shown by taking the cone over T. The value of the isoperimetric inequality lies in the numerical estimate on M(S). It was long conjectured that the worst case (exhibiting the best constant) was the sphere, in all dimensions and codimensions.

13. 3(1): Nm D fT 2 Rm : M ∂T < 1g, Rm D fT 2 Fm : M T < 1g. 14. Prove that, in analogy with the definitions of Im and Fm , Nm D fT 2 Rm : ∂T 2 Rm 1 g, Fm D fT C ∂S: T 2 Rm , S 2 RmC1 g. 15. Prove that Im ² Nm , Rm ² Rm , and Fm ² Fm . 16. 3 to which T belongs: (a) (b) (c) (d) (e) TD TDH TDH TDH TDH 1 kD1 2 1 0 2 H 1 f k 1 , y : 0 Ä y Ä 2 k g ^ j. f x, y : 0 Ä x Ä 1, 0 Ä y Ä 1g ^ i. f x, 0 : 0 Ä x Ä 1g ^ j. fag ^ i. f x, y : x 2 C y 2 Ä 1g ^ i. 17. Let E be the modification of the Cantor set obtained by starting with the unit interval and removing 2n 1 middle intervals, each of length 4 n , at the nth step (n D 1, 2, 3 .

Similarly replace the second Lipschitz function by a nice one with disjoint image. Continuing through all the original Lipschitz functions yields 1% of the set E. Countably many repetitions of the whole process prove the proposition. The following proposition shows that in a certain sense a rectifiable set has a tangent plane at almost every point. 5, p. 12. 19]. If W is an m-dimensional rectifiable subset of Rn , then for almost all points a in W, the density m W, a D 1 and Tanm W, a is an m-dimensional plane.

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Geometric Measure Theory, A Beginner's Guideimportante by Morgan F.

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