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84 1. Measure theory hence by Exercise 1.4.35 and the properties of the simple integral, X f + g dμ ≤ X fε + gε + 2ε dμ = Simp X fε + gε + 2ε dμ = Simp X fε dμ + Simp X gε dμ + 2εμ(X). From (1.14) we conclude that X f + g dμ ≤ X f dμ + X g dμ + 2εμ(X). Letting ε → 0 and using the assumption that μ(X) is finite, we obtain the claim. Now we continue to assume that μ is a finite measure, but now we do not assume that f, g are bounded. Then for any natural number n, we can use the previous case to deduce that X min(f, n) + min(g, n) dμ ≤ X min(f, n) dμ + X min(g, n) dμ. Since min(f + g, n) ≤ min(f, n) + min(g, n), we conclude that X min(f + g, n) ≤ X min(f, n) dμ + X min(g, n) dμ. Taking limits as n → ∞ using vertical truncation, we obtain the claim. Finally, we no longer assume that μ is of finite measure, and also do not require f, g to be bounded. If either X f dμ or X g dμ is infinite, then by monotonicity, X f +g dμ is infinite as well, and the claim follows so we may assume that X f dμ and X g dμ are both finite. By Markov’s inequality (Exercise 1.4.35(vi)), we conclude that for each natural number n, the set En := {x ∈ X : f(x) 1 n } ∪ {x ∈ X : g(x) 1 n } has finite measure. These sets are increasing in n, and f, g, f + g are supported on ∞ n=1 En, and so by horizontal truncation X (f + g) dμ = lim n→∞ X (f + g)1En dμ. From the previous case, we have X (f + g)1En dμ ≤ X f1En dμ + X g1En dμ. Letting n → ∞ and using horizontal truncation we obtain the claim. Exercise 1.4.36 (Linearity in μ). Let (X, B,μ) be a measure space, and let f : X → [0, +∞] be measurable. (i) Show that X f d(cμ) = c × X f dμ for every c ∈ [0, +∞].