1/2 12 1. Let VCR be a subset of Rn. We say that V is measure-0...

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1/2 12 1. Let VCR be a subset of Rn. We say that V is measure-0 if there exists a sequence of sets U1, U2, . such that each U; is a union of cubes [a1, b] .. [an, bn] which only overlap on their boundaries, VC U for all i, and vol(U) < 1/i for all i. The idea here is that because V is contained inside sets with arbitrarily small volume, it is "negligible." (a) Show that a set consisting of finitely many points is measure-0. (b) If n 2, show that the set [1,0] {0} xx {0} is measure-0. (c) If n 2, show that the entire axis R {0} {0} has measure-0. (Hint: try this first for n = 2. Remember that there are functions whose integral over the entire real line is finite.) (d) (Optional) Let a : [0, 1] Rn be a smooth curve. Show that the image of a is measure-0. (e) The rational numbers are countable, in the sense that we can write a list a1, a2,... which will contain all rational numbers. (This is not true for the irrationals, or all real numbers.) Prove that the rationals are measure-0. (f) I mentioned in class that the correct version of the theorem on integrability is the following: Theorem 1 A bounded function f : Rn R which is continuous everywhere but a measure-0 set is integrable. This implies that the function 1 f(x) n n = if x = m in lowest terms, otherwise is integrable. Find its integral. (Hint: show that it is nonnegative, then use a cleverly-constructed Riemann sum to show that it is less than any positive number. Hint to the hint: if f(x) g(x) for all x then f(x) dx fog(x) dx, so it suffices to bound f by some function g with a small integral.) 1/2 12 1. Let VCR be a subset of Rn. We say that V is measure-0 if there exists a sequence of sets U1, U2, . such that each U; is a union of cubes [a1, b] .. [an, bn] which only overlap on their boundaries, VC U for all i, and vol(U) < 1/i for all i. The idea here is that because V is contained inside sets with arbitrarily small volume, it is "negligible." (a) Show that a set consisting of finitely many points is measure-0. (b) If n 2, show that the set [1,0] {0} xx {0} is measure-0. (c) If n 2, show that the entire axis R {0} {0} has measure-0. (Hint: try this first for n = 2. Remember that there are functions whose integral over the entire real line is finite.) (d) (Optional) Let a : [0, 1] Rn be a smooth curve. Show that the image of a is measure-0. (e) The rational numbers are countable, in the sense that we can write a list a1, a2,... which will contain all rational numbers. (This is not true for the irrationals, or all real numbers.) Prove that the rationals are measure-0. (f) I mentioned in class that the correct version of the theorem on integrability is the following: Theorem 1 A bounded function f : Rn R which is continuous everywhere but a measure-0 set is integrable. This implies that the function 1 f(x) n n = if x = m in lowest terms, otherwise is integrable. Find its integral. (Hint: show that it is nonnegative, then use a cleverly-constructed Riemann sum to show that it is less than any positive number. Hint to the hint: if f(x) g(x) for all x then f(x) dx fog(x) dx, so it suffices to bound f by some function g with a small integral.)