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Unformatted text preview: Real Analysis - Math 630 Homework Set #4 - Chapter 4 by Bobby Rohde 9-28-00 Problem 1 & a) Show that if f & x Â¡ & Â¢ 0, x is irrational 1, x is rational Then R Â£ Â¡Â¡Â¡ a b f & x Â¡ Â¢ x & b Â£ a and R Â£ Â¡Â¡Â¡ a b f & x Â¡ Â¢ x & & Proof R & &&& a b f Â¡ x Â¢ Â¡ x Â¢ inf & a b Â£ Â¡ x Â¢ Â¡ x , Â¤ step functions Â£ ( x ) Â¥ f ( x ), Â¤ x . But each step function is composed of a finite collection of open intervals, and in the domain each open interal Â¦ an rational number since the rationals are dense. Thus Â£ ( x ) Â¥ f ( x ), Â¤ x implies that the value over any open interval of the step function Â£ ( x ) is Â¥ 1. Thus the infimum of the integral of all such step functions must be when Â£ ( x ) = 1, Â¤ x . Â§ R & &&& a b f Â¡ x Â¢ Â¡ x Â¢ inf & a b Â£ Â¡ x Â¢ Â¡ x Â¢ Â¡ b Â¨ a Â¢ Â© 1 Â¢ b Â¨ a . R & &&& a b f Â¡ x Â¢ Â¡ x Â¢ sup & a b Âª Â¡ x Â¢ Â¡ x , Â¤ step functions Âª ( x ) Â« f ( x ), Â¤ x . Since irrationals are dense, each interval in the composition of Âª has an irrational in its domain. Thus has an irrational in its domain....
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- Spring '08
- Math, Mathematical analysis, Riemann, Step function