lecture10

# lecture10 - Lecture 10 We begin today’s lecture with a...

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Unformatted text preview: Lecture 10 We begin today’s 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....
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lecture10 - Lecture 10 We begin today’s lecture with a...

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