332 FOR F&Auml;&deg;NAL3

# 332 FOR F&Auml;&deg;NAL3 - MATH 467/MAST 669/837...

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MATH 467/MAST 669/837 – Measure Theory Solutions to Assignment #1 1. ( Ex. 6, page 34) Let A be the set of irrational numbers in the interval [0 , 1]. Prove that m ( A ) = 1. Proof. Let B = Q [0 , 1], the set of rational numbers in [0 , 1], and note that B is countable as a countable subset of a countable set, namely Q . Recall that it was proved in class that the outer measure of a countable set of real numbers is zero. Hence m ( B ) = 0. Now, A B = [0 , 1] and A B = . Thus, by the subadditivity property of the outer measure among other things, 1 = length ([0 , 1]) = m ([0 , 1]) m ( A ) + m ( B ) = m ( A ) + 0, or m ( A ) 1. On the other hand, A [0 , 1], and by the monotonicity property of the outer measure, we also have m ( A ) m ([0 , 1]) = 1. In conclusion, m ( A ) = 1. 2. ( Ex. 9, page 34) Prove that if m ( A ) = 0, then m ( A B ) = m ( B ). Proof. From the subadditivity of m , m ( A B ) m ( A ) + m ( B ) = m ( B ). On the other hand, B A B and, by the monotonicity of m , we also have m ( B ) m ( A B ). Hence m ( A B ) = m ( B ) is proved. 3. Suppose that A E B , where A and B are measurable sets of finite measure. Prove that if m ( A ) = m ( B ), then E is measurable. Proof. Since m ( B ) < , and A B , we have m ( B \ A ) = m ( B ) m ( A ) = 0 . Recall that m is m restricted to M , the σ -algebra of measurable sets. Thus any subset of B \ A will also have outer measure zero and, consequently, it will be measurable. In particular, B \ E B \ A , so B \ E ∈ M . On the other hand, as B = E ( B \ E ), we have E = B \ ( B \ E ) = [ B ( B \ E ) C ] ∈ M due to the fact that M is an algebra and that both B and B \ E are measurable. 4. (Ex. 24, page 47) Show that if E 1 and E 2 are measurable, then m ( E 1 E 2 ) + m ( E 1 E 2 ) = m ( E 1 ) + m ( E 2 ) Proof. Suppose that we have two measurable sets E 1 and E 2 that they are disjoint. Since E 1 E 2 = , then, by Proposition 13 (page 43) foe example, we have m ( E 1 E 2 ) + m ( E 1 E 2 ) = m ( E 1 ) + m ( E 2 ) + 0 = m ( E 1 ) + m ( E 2 ) Now suppose that E 1 E 2 ̸ = . Then we have the following (by splitting E 1 E 2 into the disjoint sets E 1 E 2 C , E 1 E 2 and E 1 C E 2

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