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1.4. Abstract measure spaces 83 (vii) (Finiteness) If X f dμ ∞, then f(x) is finite for μ-almost every x. (viii) (Vanishing) If X f dμ = 0, then f(x) is zero for μ-almost every x. (ix) (Vertical truncation) We have limn→∞ X min(f, n) dμ = X f dμ. (x) (Horizontal truncation) If E1 ⊂ E2 ⊂ . . . is an increasing sequence of B-measurable sets, then lim n→∞ X f1En dμ = X f1 ∞ n=1 En dμ. (xi) (Restriction) If Y is a measurable subset of X, then X f1Y dμ = Y f Y dμ Y , where f Y : Y → [0, +∞] is the restriction of f : X → [0, +∞] to Y , and the restriction μ Y was defined in Example 1.4.25. We will often abbreviate Y f Y dμ Y (by slight abuse of notation) as Y f dμ. As before, one of the key properties of this integral is its additivity: Theorem 1.4.37. Let (X, B,μ) be a measure space, and let f, g : X → [0, +∞] be measurable. Then X (f + g) dμ = X f dμ + X g dμ. Proof. In view of superadditivity, it suffices to establish the subadditivity property X (f + g) dμ ≤ X f dμ + X g dμ We establish this in stages. We first deal with the case when μ is a finite measure (which means that μ(X) ∞) and f, g are bounded. Pick an ε 0, and let fε be f rounded down to the nearest integer multiple of ε, and let f ε be f rounded up to the nearest integer multiple. Clearly, we have the pointwise bounds fε(x) ≤ f(x) ≤ f ε(x) and f ε(x) − fε(x) ≤ ε. Since f is bounded, fε and f ε are simple. Similarly, define gε,gε. We then have the pointwise bound f + g ≤ f ε + gε ≤ fε + gε + 2ε,