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Unformatted text preview: # # # Lecture 13 Let A be an open set in R n , and let f : A R be a continuous function. For the → moment, we assume that f ≥ 0. Let D ⊆ A be a compact and rectifiable set. Then f  D is bounded, so f is welldefined. Consider the set of all such integrals: D # = { f : D ⊆ A, D compact and rectifiable } . (3.122) D Definition 3.22. The improper integral of f over A exists if ∗ is bounded, and we define the improper integral of f over A to be its l.u.b. f ≡ l.u.b. f = improper integral of f over A . (3.123) A D Claim. If A is rectifiable and f : A R is bounded, then → f = f. (3.124) A A Proof. Let D ⊆ A be a compact and rectifiable set. So, f (3.125) f ≤ D A = ⇒ sup f (3.126) f ≤ D D A = ⇒ A f ≤ f. (3.127) A The proof of the inequality in the other direction is a bit more complicated. ¯ Choose a rectangle Q such that A ⊆ Int Q . Define f A : Q → R by f A ( x ) = f ( 0 x ) if x ∈ A , if x / ∈ A . (3.128) By definition, f = f A . (3.129) A Q Now, let P be a partition of Q , and let R 1 , . . . , R k be rectangles belonging to a partition of A . If R is a rectangle belonging to P not contained in A , then R − A = φ ....
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 Fall '04
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 Calculus, Topology, Integrals, Continuous function

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