A set of real numbers D is called dense if and only if every open interval contains at least one point of D. Assume a set E and its complement E = {x € R : x ¢ E} are both dense. Let f : [0, 1] → R be given by f(x) = 1 if x E E and f(x) = 0 if x ¢ E. (Thus f(x) = XE (x) for x E [0, 1].) Prove that f is not integrable on [0, 1].

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.2: Mappings
Problem 23E: Let a and b be constant integers with a0, and let the mapping f:ZZ be defined by f(x)=ax+b. Prove...
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A set of real numbers D is called dense if and only if every open interval contains at least
one point of D. Assume a set E and its complement E = {x € R : x ¢ E} are both dense.
Let f : [0, 1] → R be given by f(x) = 1 if x E E and f(x) = 0 if x ¢ E.
(Thus f(x) = XE (x) for x E [0, 1].)
Prove that f is not integrable on [0, 1].
Transcribed Image Text:A set of real numbers D is called dense if and only if every open interval contains at least one point of D. Assume a set E and its complement E = {x € R : x ¢ E} are both dense. Let f : [0, 1] → R be given by f(x) = 1 if x E E and f(x) = 0 if x ¢ E. (Thus f(x) = XE (x) for x E [0, 1].) Prove that f is not integrable on [0, 1].
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