This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 202A Homework 13 Roman Vaisberg November 28, 2007 Problem 17. Suppose f is defined on R 2 as follows: f ( x, y ) = a n if n x < n + 1 and n y < n + 1, ( n 0); f ( x, y ) = a n if n x < n + 1 and n + 1 y < n + 2, ( n 0); while f ( x, y ) = 0 elsewhere. Here a n = k b k , with { b k } a positive sequence such that k =0 b k = s < . (a) Verify that each slice f y and f x is integrable. Also for all x , R f x ( y ) dy = 0, and hence R ( R f ( x, y ) dy ) dx = 0. Proof. If y < 0 then f y ( x ) = 0 If 0 y < 1 then f y ( x ) = a x < 1 elsewhere If n y < n + 1 then f y ( x ) =  a n 1 n 1 x < n a n n x < n + 1 elsewhere Clearly f y is integrable for every y . If x < 0 then f x ( y ) = 0 If n x < n + 1 then f x ( y ) = a n n y < n + 1 a n n + 1 y < n + 2 elsewhere It is also clear that f x is integrable for every x and R f x ( y ) dy = 0. Thus R ( R f ( x, y ) dy ) dx = 0. (b) However, R f y ( x ) dx = a if 0 y < 1, and R f y ( x ) dx = a n a n 1 if n y < n + 1 with n 1. Hence y 7 R f y ( x ) dx is integrable on (0 , ) and Z Z f ( x, y ) dx dy = s. Proof. R (R f ( x, y ) dx ) dy = a + ( a 1 a ) + ( a 2 a 1 ) + = b k = s. (c) Note that R R R  f ( x, y )  dxdy = . Proof. In the extended sense we have R R R  f ( x, y )  dxdy = 2 a k and since { a n } is a positive increasing sequence we have the desired conclusion....
View Full
Document
 Spring '10
 tao/analysis
 Math

Click to edit the document details