86 1. Measure theory where f+ := max(f, 0), f− := max(−f, 0) are the magnitudes of the posi- tive and negative components of f. If f is complex-valued and absolutely integrable, we define the integral X f dμ by the formula X f dμ := X Re f dμ + i X Im f dμ where the two integrals on the right are interpreted as real-valued integrals. It is easy to see that the unsigned, real-valued, and complex-valued inte- grals defined in this manner are compatible on their common domains of definition. Clearly, this definition generalises the Definition 1.3.17. We record some of the key facts about the absolutely convergent integral: Exercise 1.4.40. Let (X, B,μ) be a measure space. (i) Show that L1(X, B,μ) is a complex vector space. (ii) Show that the integration map f → X f dμ is a complex-linear map from L1(X, B,μ) to C. (iii) Establish the triangle inequality f + g L1(μ) ≤ f L1(μ) + g L1(μ) and the homogeneity property cf L1(μ) = |c|f L1(μ) for all f, g ∈ L1(X, B,μ) and c ∈ C. (iv) Show that if f, g ∈ L1(X, B,μ) are such that f(x) = g(x) for μ- almost every x ∈ X, then X f dμ = X g dμ. (v) If f ∈ L1(X, B,μ), and (X, B , μ ) is a refinement of (X, B,μ), then f ∈ L1(X, B , μ ), and X f dμ = X f dμ. (Hint: It is easy to get one inequality. To get the other inequality, first work in the case when f is both bounded and has finite measure support (i.e. is both vertically and horizontally truncated).) (vi) Show that if f ∈ L1(X, B,μ), then f L1(μ) = 0 if and only if f is zero μ-almost everywhere. (vii) If Y ⊂ X is B-measurable and f ∈ L1(X, B,μ), then f Y ∈ L1(Y, B Y , μ Y ) and Y f Y dμ Y = X f1Y dμ. As before, by abuse of notation we write Y f dμ for Y f Y dμ Y . 1.4.5. The convergence theorems. Let (X, B,μ) be a measure space, and let f1,f2,... : X → [0, +∞] be a sequence of measurable functions. Suppose that as n → ∞, fn(x) converges pointwise either everywhere, or μ- almost everywhere, to a measurable limit f. A basic question in the subject is to determine the conditions under which such pointwise convergence would imply convergence of the integral: X fn dμ ? → X f dμ.

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