1.4. Abstract measure spaces 81 function that takes on finitely many values a1,...,ak. We then define the simple integral Simp X f dμ by the formula Simp X f dμ := k j=1 ajμ(f −1({aj})). In order to set out the basic properties of the simple integral, the fol- lowing preliminary result is handy: Exercise 1.4.32. Let f : X → [0, +∞] be a simple function on a measurable space (X, B), and suppose that there are disjoint measurable sets E1,...,Em such that f is supported on E1 ∪ . . . ∪ Em and equals ci on each Ei for some ci ∈ [0, +∞]. Show that Simp X f dμ = m j=1 cjμ(Ej). Now we can deduce the following properties of the simple integral. As with the Lebesgue theory, we say that a property P (x) of an element x ∈ X of a measure space (X, B,μ) holds μ-almost everywhere if it holds outside of a sub-null set. Exercise 1.4.33 (Basic properties of the simple integral). Let (X, B,μ) be a measure space, and let f, g : X → [0, +∞] be simple functions. (i) (Monotonicity) If f ≤g pointwise, then Simp X f dμ≤Simp X g dμ. (ii) (Compatibility with measure) For every B-measurable set E, we have Simp X 1E dμ = μ(E). (iii) (Homogeneity) For every c ∈ [0, +∞], one has Simp X cf dμ = c × Simp X f dμ. (iv) (Finite additivity) Simp X (f+g) dμ=Simp X f dμ+Simp X g dμ. (cf. Exercise 1.1.2.) (v) (Insensitivity to refinement) If (X, B , μ ) is a refinement of (X, B,μ) (which means that B is a σ-algebra containing B, and μ : B → [0, +∞] is the restriction of μ : B → [0, +∞] to B), then we have Simp X f dμ = Simp X f dμ . (vi) (Almost everywhere equivalence) If f(x) = g(x) for μ-almost every x ∈ X, then Simp X f dμ = Simp X g dμ. (vii) (Finiteness) Simp X f dμ ∞ if and only if f is finite almost everywhere, and is supported on a set of finite measure. (viii) (Vanishing) Simp X f dμ = 0 if and only if f is zero almost every- where.

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