1.4. Abstract measure spaces 93 integrable limit f. Show that X fn X f f fn L1(μ) 0 as n ∞. (Hint: Apply the dominated convergence theorem (Theorem 1.4.48) to min(fn,f).) Informally, this result (first established in [BrLi1983]) tells us that the gap between the left and right-hand sides of Fatou’s lemma can be measured by the quantity f fn L1(μ). Exercise 1.4.48. Let (X, B,μ) be a measure space, and let g : X [0, +∞] be measurable. Show that the function μg : B [0, +∞] defined by the formula μg(E) := X 1Eg = E g is a measure. (Such measures are studied in greater detail in §1.2 of An epsilon of room, Vol. I.) The monotone convergence theorem is, in some sense, a defining property of the unsigned integral, as the following exercise illustrates. Exercise 1.4.49 (Characterisation of the unsigned integral). Let (X, B) be a measurable space, and let I : f I(f) be a map from the space U(X, B) of unsigned measurable functions f : X [0, +∞] to [0, +∞] that obeys the following axioms: (i) (Homogeneity) For every f U(X, B) and c [0, +∞], one has I(cf) = cI(f). (ii) (Finite additivity) For every f, g U(X, B), one has I(f + g) = I(f) + I(g). (iii) (Monotone convergence) If 0 f1 f2 . . . are a non-decreasing sequence of unsigned measurable functions, then I(limn→∞ fn) = limn→∞ I(fn). Then there exists a unique measure μ on (X, B) such that I(f) = X f for all f U(X, B). Furthermore, μ is given by the formula μ(E) := I(1E) for all B-measurable sets E. Exercise 1.4.50. Let (X, B,μ) be a finite measure space (i.e. μ(X) ∞), and let f : X R be a bounded function. Suppose that μ is complete (see Definition 1.4.31). Suppose that the upper integral X f := inf g≥f g simple X g and lower integral X f := sup h≤f h simple X h
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