1.3. The Lebesgue integral 49 This implies that, for any λ, the set {x Rd : f(x) λ} is equal to M0 N0 {x Rd : sup n≥N fn(x) λ + 1 M } outside of a set of measure zero this set in turn is equal to M0 N0 n≥N {x Rd : fn(x) λ + 1 M } outside of a set of measure zero. But as each fn is an unsigned simple function, the sets {x Rd : fn(x) λ + 1 M } are Lebesgue measurable. Since countable unions or countable intersections of Lebesgue measurable sets are Lebesgue measurable, and modifying a Lebesgue measurable set on a null set produces another Lebesgue measurable set, we obtain (v). To obtain the equivalence of (v) and (vi), observe that {x Rd : f(x) λ} = λ ∈Q+:λ λ {x Rd : f(x) λ } for λ (0, +∞] and {x Rd : f(x) λ} = λ ∈Q+:λ λ {x Rd : f(x) λ } for λ [0, +∞), where Q+ := Q [0, +∞] are the non-negative rationals. The claim then easily follows from the countable nature of Q+ (treating the extreme cases λ = 0, +∞ separately if necessary). A similar argument lets one deduce (v) or (vi) from (ix). The equivalence of (v), (vi) with (vii), (viii) comes from the observation that {x Rd : f(x) λ} is the complement of {x Rd : f(x) λ}, and {x Rd : f(x) λ} is the complement of {x Rd : f(x) λ}. A similar argument shows that (x) and (xi) are equivalent. By expressing an interval as the intersection of two half-intervals, we see that (ix) follows from (v)–(viii), and so all of (v)–(ix) are now shown to be equivalent. Clearly (x) implies (vii), and hence (v)–(ix). Conversely, because every open set in [0, +∞) is the union of countably many open intervals in [0, +∞), (ix) implies (x). The only remaining task is to show that (v)–(xi) implies (iv). Let f obey (v)–(xi). For each positive integer n, we let fn(x) be defined to be the largest integer multiple of 2−n that is less than or equal to min(f(x),n) when |x| n, with fn(x) := 0 for |x| n. From construction it is easy to see that the fn : Rd [0, +∞] are increasing and have f as their supre- mum. Furthermore, each fn takes on only finitely many values, and for each non-zero value c it attains, the set fn −1(c) takes the form f −1(Ic)∩{x Rd :
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