lecture10

lecture10 - Lecture 10 We begin todays lecture with a...

Info iconThis preview shows pages 1–3. 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 DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lecture 10 We begin todays lecture with a simple claim. Claim. Let Q R n be a rectangle and f, g : Q R be bounded functions such that f g . Then Q f Q g. (3.49) Proof. Let P be a partition of Q , and let R be a rectangle belonging to P . Clearly, m R ( f ) m R ( g ), so L ( f, P ) = m R ( f ) v ( R ) (3.50) R L ( g, P ) = m R ( g ) v ( R ) (3.51) R = L ( f, P ) L ( g, P ) g, (3.52) Q for all partitions P . The lower integral f (3.53) Q is the l.u.b. of L ( f, P ), so g. (3.54) f Q Q Similarly, g. (3.55) f Q Q It follows that if f g , then g. (3.56) f Q Q This is the monotonicity property of the R. integral. 3.4 Fubini Theorem In one-dimensional calculus, when we have a continuous function f : [ a, b ] R , then we can calculate the R. integral b f ( x ) dx = F ( b ) F ( a ) , (3.57) a 1 where F is the anti-derivative of f . When we integrate a continuous function f : Q R over a two-dimensional region, say Q = [ a 1 , b 1 ] [ a 2 , b 2 ], we can calculate the R. integral b 1 b 2 b 1 b 2 f = f ( x, y ) dxdy = f ( x, y ) dxdy (3.58) Q a 1 a 2 a 1 a 2 That is, we can break up Q into components and integrate separately over those components. We make this more precise in the following Fubini Theorem. components....
View Full Document

Page1 / 5

lecture10 - Lecture 10 We begin todays lecture with a...

This preview shows document pages 1 - 3. Sign up to view the full document.

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