This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 447 Solutions to Assignment 4 Spring 2009 Exercise 17.6 on page 297 Suppose f : D → R , where D is measurable. Show that f is measurable if and only if { f > α } is measurable for each rational α . Solution If the function f is measurable, then by the definition at the top of page 296, the set { x ∈ D : f ( x ) > α } is measurable for every real number α and a fortiori for every rational number. Conversely, suppose that { x ∈ D : f ( x ) > α } is measurable for every rational number α . If β is an arbitrary real number, then there is a decreasing sequence ( α n ) of rational numbers having limit β . Each set { x ∈ D : f ( x ) > α n } is measurable by hypothesis. Since { x ∈ D : f ( x ) > β } = ∞ [ n =1 { x ∈ D : f ( x ) > α n } , and since by Corollary 16.20 on page 281 the union of a countable number of measurable sets is measurable, the set { x ∈ D : f ( x ) > β } is measurable too. Exercise 17.7 on page 298 If f : D → R is measurable and g : R → R is continuous, show that g ◦ f is measurable. Solution According to the definition on page 296, what needs to be shown is that for every real number α , the set { x ∈ D : g ( f ( x )) > α } is measurable. This set can be interpreted as the set of points x in D such that f ( x ) lies in the set { u ∈ R : g ( u ) > α } . In other words, the set at issue is the inverse image under f of the inverse image under g of the interval ( α, ∞ ). Since g is continuous, the inverse image set g 1 (( α, ∞ )) is open [by Theorem 5.1 on page 63], and the inverse image of this open set under the measurable[by Theorem 5....
View
Full Document
 Spring '08
 Staff
 Math, Rational number, Zero, measure, Measurable function, Dr. Boas

Click to edit the document details