82 1. Measure theory Exercise 1.4.34 (Inclusion-exclusion principle). Let (X, B,μ) be a measure space, and let A1,...,An be B-measurable sets of finite measure. Show that μ n i=1 Ai = J⊂{1,...,n}:J=∅ (−1)|J|−1μ i∈J Ai . (Hint: Compute Simp X (1 − n i=1 (1 − 1Ai )) dμ in two different ways.) Remark 1.4.35. The simple integral could also be defined on finitely addi- tive measure spaces, rather than countably additive ones, and all the above properties would still apply. However, on a finitely additive measure space one would have diﬃculty extending the integral beyond simple functions, as we will now do. From the simple integral, we can now define the unsigned integral, in analogy to the way the unsigned Lebesgue integral was constructed in Sec- tion 1.3.3. Definition 1.4.36. Let (X, B,μ) be a measure space, and let f : X → [0, +∞] be measurable. Then we define the unsigned integral X f dμ of f by the formula (1.14) X f dμ := sup 0≤g≤f g simple Simp X g dμ. Clearly, this definition generalises Definition 1.3.13. Indeed, if f : Rd → [0, +∞] is Lebesgue measurable, then Rd f(x) dx = Rd f dm. We record some easy properties of this integral: Exercise 1.4.35 (Easy properties of the unsigned integral). Let (X, B,μ) be a measure space, and let f, g : X → [0, +∞] be measurable. (i) (Almost everywhere equivalence) If f = g μ-almost everywhere, then X f dμ = X g dμ (ii) (Monotonicity) If f ≤ g μ-almost everywhere, then X f dμ ≤ X g dμ. (iii) (Homogeneity) We have X cf dμ = c X f dμ for every c ∈ [0, +∞]. (iv) (Superadditivity) We have X (f + g) dμ ≥ X f dμ + X g dμ. (v) (Compatibility with the simple integral) If f is simple, then we have X f dμ = Simp X f dμ. (vi) (Markov’s inequality) For any 0 λ ∞, one has μ({x ∈ X : f(x) ≥ λ}) ≤ 1 λ X f dμ. In particular, if X f dμ ∞, then the sets {x ∈ X : f(x) ≥ λ} have finite measure for each λ 0.

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