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Course: MATH 8601, Fall 2010
School: Minnesota
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8601: Math REAL ANALYSIS. Fall 2010 Homework #5. Problems and Solutions. #1. Let E be a Lebesgue measurable set in R1 with Lebesgue measure m(E) &gt; 0. Show that for any &lt; 1, there is an open interval I = (a, b) such that m(E I) &gt; m(I). Proof. Suppose that this property is not true, i.e. m(E I) m(I) for any interval I = (a, b), where = const &lt; 1. Fix a small &gt; 0, such that...

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8601: Math REAL ANALYSIS. Fall 2010 Homework #5. Problems and Solutions. #1. Let E be a Lebesgue measurable set in R1 with Lebesgue measure m(E) > 0. Show that for any < 1, there is an open interval I = (a, b) such that m(E I) > m(I). Proof. Suppose that this property is not true, i.e. m(E I) m(I) for any interval I = (a, b), where = const < 1. Fix a small > 0, such that (1 + ) < 1. By Theorem 1.18, there is an open set U E such that m(U ) < (1 + ) m(E). By Proposition 0.21, the open set U is represented as a finite or countable union of disjoint open intervals Ij . Therefore, m(U ) = m(E U ) = m E j Ij =m j E Ij = j m(E Ij ) j m(Ij ) = m(U ) < (1 + ) m(E) < m(E). This contradiction proves the desired property. #2. Let (X, M) be a measurable space, and let fn : X R1 be a sequence of M-measurable functions on X. Show that the set E := {x X : lim fn (x)} is M-measurable. n Proof. By Proposition 2.7, the functions g(x) := lim sup fj (x) and h(x) := lim inf fj (x) j j are M-measurable. Obviously, the function -h(x) is also M-measurable, hence by Proposition 2.6, the function F (x) := g(x) - h(x) = g(x) + (-h(x)) is M-measurable as well. Finally, E = {x X : g(x) = h(x)} = F -1 ({0}) M. #3. Let (X, M, ) be a measure space, and let f L+ - the space of all measurable functions from X to [0, ], with f d < . Show that for every > 0, there exists a set E M such that (E) < and E f d > f d - . 1 Proof. By definition on p.50, for any positive > 0, there is a simple function n 0 := j=1 aj Ej f, such that f d d > f d - . Here aj > 0 for all j, and Ej are disjoint sets in M. Denote n := min > aj 0, j and E := j=1 Ej M. Then E , hence (E) = E d d f d < , so that (E) < . Moreover, f d E E d = X d > X f d - . Another Proof. We have En := {f 1/n} {f > 0}, fn := f En f as n . By the Monotone Convergence Theorem 2.14, fn d = En f d f d as n . Therefore, > 0, n such that f d > En f d - . Moreover, by Chebyshev's inequality 6.17, p.193, (En ) n f d < . #4. Let (X, M, ) be a measure space with (X) < , and let f be as in the previous problem. Give a direct proof, based on the definition of f d on p.50, of the following fact: > 0, > 0 such that from E M and (E) < it follows E f d < . 2 Proof. Fix a constant > 0. Similarly to the previous proof, choose a simple function n 0 := j=1 aj Ej f, such that d > f d - . 2 Denote h := f - 0. Then from f = + h it follows 0 Finally, set h d = f d - d < . 2 , where A := max aj < . j 2A Then from E M and (E) < it follows := f d = E E n h d + E d < + 2 n aj Ej d j=1 E n + aj (E Ej ) + A = 2 j=1 2 = + A (E) < + A = . 2 2 (E Ej ) j=1 #5. Let (X, M, ) be a measure space with (X) < . Show that d(f + g) d(f ) + d(g) for all measurable functions f, g where d(f ) := inf + {x X : |f (x)| } . >0 on X, Proof. For arbitrary 1 > 0 and 2 > 0, the set {|f + g| 1 + 2 } is contained in {|f | 1 } {|g| 2 }. Therefore, {|f + g| 1 + 2 } {|f | 1 } + {|g| 2 } . From the definition of d(f + g) with = 1 + 2 it follows d(f + g) 1 + 2 + {|f + g| 1 + 2 } 1 + {|f | 1 } + 2 + {|g| 2 } . Taking the inf over 1 > 0 and 2 > 0, we obtain the desired inequality d(f + g) d(f ) + d(g). 3
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