{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture13

lecture13 - Lecture 13 Let A be an open set in R n and let...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 well-defined. 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 ....
View Full Document

{[ snackBarMessage ]}

Page1 / 5

lecture13 - Lecture 13 Let A be an open set in R n and let...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online