Integration on Measure Spaces

Transcription

1 Chapter 3 Integration on Measure Spaces In this chapter we introduce the general notion of a measure on a space X, define the class of measurable functions, and define the integral, first on a class of positive simple functions, then on general positive, measurable functions, and finally on measurable, integrable functions. We establish several key results about integrals of limits of sequences of functions. It is useful to select the crucial features of Lebesgue measure, as developed in Chapter 2, and apply them to more general situations; this leads to the notion of a measure space. To define this, we start with a set X and a σ-algebra F of subsets. As in Chapter 2, this means F is a nonempty family of subsets of X, closed under countable unions and under complements, and hence also under countable intersections. We call (X, F) a measurable space. Note that, if S F, then X \ S F, so X = S (X \ S) F, and hence X c = F. A measure µ on (X, F) is a function µ : F [0, ] satisfying the following two conditions: (3.1) µ( ) = 0, ( ) (3.2) S j F disjoint, countable = µ S j = j j µ(s j ). We call (X, F, µ) a measure space. It is easy to verify that µ automatically has the following four additional properties: (3.3) S j F, S 1 S 2 = µ(s 1 ) µ(s 2 ), 25

2 26 3. Integration on Measure Spaces ( ) (3.4) S j F countable = µ S j j j µ(s j ), (3.5) S j F, S 1 S 2 S 3 = µ(s j ) µ( S j ), and, provided µ(s j ) < for some j, (3.6) S j F, S 1 S 2 S 3 = µ(s j ) µ( S j ). Property (3.3) is called monotonicity, and property (3.4) is called countable subadditivity; the contrast with countable additivity in (3.2) is that in (3.4) the sets S j need not be disjoint. To prove (3.3), note that, if S 1 S 2, then Y = S 2 \S 1 = S 2 (X\S 1 ) F, and S 2 = S 1 Y is a disjoint union; hence, by (3.2), µ(s 2 ) = µ(s 1 ) + µ(y ), a sum of terms that are 0. We leave the proofs of (3.4) (3.6) as exercises. Compare Exercises 2 and 3 in Chapter 2. So far we have as examples of measure spaces X = [a, b] and X = R, with F either the class of Lebesgue measurable subsets of X or the class of Borel subsets of X, and µ = Lebesgue measure. More generally, we could let X be any Lebesgue measurable subset of R. We will study further constructions of measures in Chapters 5 7 and Chapters Now we look at the issue of integrating a function f : X R, when (X, F, µ) is a measure space. We first define what it means for f to be measurable. The most liberal-sounding characterization is that f 1 (J) X belongs to F for every open interval J R. If this property holds, it is an automatic consequence that f 1 (S) F for every Borel set S R. (Proof: The collection of subsets S R such that f 1 (S) F is a σ-algebra. If it contains all open intervals, then it contains all Borel subsets of R.) We regard this latter characterization as the definition of measurability. To emphasize the role of F, we sometimes say f is F-measurable. Note that, if X is a topological space and B is the σ-algebra of Borel sets in X, i.e., the smallest σ-algebra containing the closed subsets of X, then any continuous f : X R is B-measurable. By the definition, f : R R is Lebesgue measurable provided f 1 (S) L for every S B, where B is the class of Borel sets in R and L is the class of Lebesgue measurable subsets of R. Recall that B L. We note, however, that it does not follow that f 1 (S) L for all S L. Indeed, there exist continuous f : R R and S L such that f 1 (S) / L. (See Exercise 21.) We can also define f : X [, ] to be F-measurable provided f 1 (S) F for all Borel sets S in [, ]. It is valuable to know that certain basic operations preserve measurability. For example, we have the following.

3 3. Integration on Measure Spaces 27 Proposition 3.1. If f j : X R are F-measurable, so are f 1 +f 2 and f 1 f 2. Proof. Let J be an open interval in R. If g = f 1 + f 2, then (3.7) g 1 (J) = {x X : ( f 1 (x), f 2 (x) ) O}, where O = {(x, y) R 2 : x + y J}. Now we can write O as a countable disjoint union of rectangles, of the form O k = I k I k (where the intervals I k, I k are open). Then { x X : f1 (x) I k, f 2 (x) I k} (3.8) g 1 (J) = k = k [ f 1 1 (I k) f 1 2 (I k )] is clearly an element of F, provided f 1 and f 2 are F-measurable. The measurability of f 1 f 2 is proved similarly. Remark 3.1. The argument used above to prove Proposition 3.1 readily extends to the following more general setting. Assume f j : X R are F- measurable, for j = 1,..., n, and that Φ : R n R is continuous, and define g : X R by g(x) = Φ(f 1 (x),..., f n (x)). Then g is F-measurable. The following result on measurability of sups, infs, etc., will be very valuable. Proposition 3.2. If f j are F-measurable, for j Z +, then (3.9) g 1 (x) = sup j g 3 (x) = lim sup j f j (x), f j (x), g 2 (x) = inf j f j (x), g 4 (x) = lim inf j f j (x) are all F-measurable functions, from X to [, + ]. If g(x) = lim f j (x) exists for all x X, then g is F-measurable. Proof. Here the σ-algebra of Borel sets in [, + ] is generated by intervals, either of the form (a, + ] or of the form [, a). Note that (3.10) g1 1 ( ) (a, ] = f 1 ( ) j (a, ], j 1

4 28 3. Integration on Measure Spaces so the measurability of g 1 follows. That of g 2 is similar. Now (3.11) g 3 (x) = inf k 1 h k(x), where h k (x) = sup f j (x), j k so measurability of g 3 follows, and measurability of g 4 is similar. Finally, if g(x) = lim f j (x) exists everywhere, then g 3 = g 4 = g, so we have measurability of g. If (X, F, µ) is a measure space and S F, then the characteristic function χ S, defined as in (1.22) by (3.12) χ S (x) = 1 if x S, 0 if x X \ S, is F-measurable, as is any finite linear combination (3.13) ϕ(x) = N a j χ Sj (x), a j R, S j F. j=1 Such a function is called a simple function. The sets S j in (3.13) can overlap. It is convenient to rewrite (3.13) using characteristic functions of disjoint sets. We can do this as follows. For each subset σ {1,..., N}, let ( (3.14) Sσ = j σ ) ( S j j / σ S c j ), S c j = X \ S j. Then the sets S σ are mutually disjoint, and for each j {1,..., N}, we have (3.15) S j = σ j S σ. Furthermore, if ϕ is given by (3.13), then (3.16) ϕ(x) = N a j χ esσ (x) = σ j=1 σ j ã σ χ esσ (x), where ã σ = j σ a j. Let us denote by S + (X) the class of simple functions that are 0 on X. It follows from (3.16) that such functions can be put in the form (3.13) with a j 0. It also follows from (3.16) that S + (X) is the class of F-measurable functions with range in a finite subset of [0, ).

5 3. Integration on Measure Spaces 29 Given ϕ S + (X), we define the integral (3.17) N ϕ dµ = a j µ(s j ). j=1 This quantity is possibly +, if µ(s j ) = for some S j. However, if a j = 0, then by convention we take a j µ(s j ) = 0 whether or not µ(s j ) is finite. (3.18) Note that by (3.16) a j µ(s j ) = j j a j µ( S σ ) = σ σ j ã σ µ( S σ ). Note also that if the set of nonzero values taken on by ϕ is {b 1,..., b M }, then (3.19) ϕ(x) = M b k χ Tk (x), T k = {x X : ϕ(x) = b k }. k=1 Each T k is a union of sets S σ in (3.16), and each corresponding coefficient ã σ is equal to b k. It follows that (3.20) ϕ dµ = ã σ µ( S σ ) = b k µ(t k ), σ k the first identity by (3.18) and the second by additivity of µ. This identity shows that ϕ dµ is well defined, independently of the particular representation of ϕ in (3.13). If ϕ, ψ S + (X), one also verifies immediately that (3.21) cϕ dµ = c ϕ dµ, (ϕ + ψ) dµ = ϕ dµ + ψ dµ, and (3.22) ϕ ψ = ϕ dµ ψ dµ. The following is also easy, though worthy of the status of a lemma. Lemma 3.3. If ϕ S + (X), then (3.23) λ(a) = ϕ dµ = A χ A ϕ dµ

6 30 3. Integration on Measure Spaces is a measure on F. Here, the last identity is a definition of A ϕ dµ. Note that if ϕ is given by (3.13), then (3.24) λ(a) = N a j µ(a S j ). j=1 The proof of Lemma 3.3 is an exercise. Now let M + (X) denote the set of all measurable functions f : X [0, ]. For f M + (X), we define the integral { } (3.25) f dµ = sup ϕ dµ : 0 ϕ f, ϕ S + (X). This is somewhat analogous to the definition of the lower Riemann integral I(f). One difference is that the quantity defined by (3.25) might be. Our hypothesis that f is measurable makes it unnecessary to introduce also an analogue of the upper integral. That (3.25) agrees with (3.17) for f = ϕ S + (X) follows from (3.22). Note that, for any f M + (X), we can find a sequence (3.26) ϕ j S + (X), ϕ j f. For example, set ϕ j (x) = k/2 j on {x X : k2 j f(x) < (k + 1)2 j } if k < j 2 j, and set ϕ j (x) = j on {x X : j f(x)}. If K < and 0 f(x) K on S X, then ϕ j f uniformly on S, as j. When (3.26) holds, one should expect ϕ j dµ to converge to f dµ. That this is true is a special case of the following central result, known as the Monotone Convergence Theorem. Theorem 3.4. If f j M + (X) and f j, then f j f M + (X), and (3.27) f dµ = lim f n dµ. n Proof. Given that f j (x), the limit certainly exists for each x, though it might be +. The measurability of the limit f follows from Proposition 3.2. Clearly f n dµ f dµ for all n, and f n dµ is increasing, so Φ = lim f n dµ exists, and Φ f dµ. We need the reverse inequality. To get it, take α (0, 1). Let ϕ S + (X), ϕ f. Let E n = {x X : f n (x) αϕ(x)}. Then {E n } is an increasing sequence of measurable sets whose union is X, and (3.28) f n dµ f n dµ α ϕ dµ = αλ(e n ), E n E n

7 3. Integration on Measure Spaces 31 where λ(e n ) is as in (3.23). By Lemma 3.3, together with the property (3.5), we have (3.29) lim ϕ dµ = ϕ dµ. E n Hence (3.28) implies Φ α ϕ dµ, for all α (0, 1). Letting α 1, we obtain Φ ϕ dµ, for all ϕ S + (X) such that ϕ f. In view of the definition (3.25), this implies Φ f dµ, and the proof is done. We next establish an additivity result: Proposition 3.5. If f n M + (X) is a countable sequence and f = n 1 f n, then (3.30) f dµ = f n dµ. n 1 Proof. First consider the case f = f 1 + f 2. As in (3.26), take ϕ nj S + (X) such that ϕ nj f n as j. Then ϕ 1j + ϕ 2j f 1 + f 2, so, by the Monotone Convergence Theorem together with (3.21), we have (f 1 + f 2 ) dµ = lim (ϕ 1j + ϕ 2j ) dµ (3.31) = lim ϕ 1j dµ + lim ϕ 2j dµ = f 1 dµ + f 2 dµ. By induction, (3.32) ( N ) N f n dµ = n=1 n=1 f n dµ, given f n M + (X), for any N <, and applying the Monotone Convergence Theorem again yields (3.30). The next result is known as Fatou s Lemma. Proposition 3.6. If f n M + (X), then (lim ) (3.33) inf fn dµ lim inf f n dµ.

8 32 3. Integration on Measure Spaces Proof. For each k Z +, j k inf n k f n f j, so (3.34) inf f n dµ inf n k j k f j dµ. Of course, inf n k f n lim inf f n as k. Thus, letting k in (3.25) and applying the Monotone Convergence Theorem, we have (3.35) which gives (3.33). ( (lim inf f n ) dµ = lim inf f ) n dµ lim inf k n k f j dµ, We now tackle the definition of f dµ when f(x) is not necessarily 0 everywhere. For a general measurable function f : X R, we can write f = f + f, where f + (x) = f(x) when f(x) 0, f + (x) = 0 otherwise, and f (x) = f(x) where f(x) 0, f (x) = 0 otherwise. Thus f +, f M + (X). We define the integral (3.36) f dµ = f + dµ f dµ, provided at least one of the terms on the right is finite. If they are both finite or, equivalently, if (3.37) f dµ <, we say f is integrable and write (3.38) f L 1 (X, µ). For short, we might write f L 1 (X) or simply f L 1. The following basic result is parallel in statement, though not in proof, to Proposition 1.1. Proposition 3.7. The set L 1 (X, µ) is a linear space, and the map (3.39) I : L 1 (X, µ) R, I(f) = f dµ is a linear map.

9 3. Integration on Measure Spaces 33 Proof. From f 1 + f 2 f 1 + f 2, it easily follows that f j L 1 (X, µ) f 1 + f 2 L 1 (X, µ). Also, clearly a j R a j f j L 1 (X, µ) in this case. Let us establish that, if f j L 1 (X, µ), then (3.40) (f 1 + f 2 ) dµ = f 1 dµ + f 2 dµ. Indeed, if g = f 1 +f 2, then g + g = f 1 + f 1 +f+ 2 f 2, so g+ +f1 +f 2 = g + f f 2 +. Hence, by Proposition 3.5, (g + + f1 + f 2 ) dµ = g + dµ + f1 dµ + f2 dµ (3.41) = (g + f f 2 + ) dµ = g dµ + f 1 + dµ + f 2 + dµ. Equating the right sides, we get (3.40). The other ingredient in linearity, (3.42) af dµ = a f dµ, for f L 1 (X, µ), a R, is easy. We can also integrate a complex-valued function. If f : X C, write f = f 1 + if 2 where f j are real valued. We say f is measurable if f 1 and f 2 are both measurable. If f 1, f 2 L 1 (X, µ), we set (3.43) f dµ = f 1 dµ + i f 2 dµ. Often, we use the same notation, L 1 (X, µ), to denote the set of such complexvalued integrable functions. The rest of the results of this section will deal specifically with real-valued functions, but the reader should be able to formulate extensions to complex-valued functions. The next result is known as the Dominated Convergence Theorem. Theorem 3.8. Let f n L 1 (X, µ), for n Z +. Suppose f : X R and (3.44) f n (x) f(x) for all x X, and suppose that there is a function g satisfying (3.45) g L 1 (X, µ), f n g, for all n. Then (3.46) f L 1 (X, µ) and f dµ = lim n f n dµ.

10 34 3. Integration on Measure Spaces Proof. We know f is measurable; of course then f g f L 1. Now if g satisfies (3.45), then g + f n 0. Thus, by Fatou s Lemma and (3.40), g dµ + f dµ = (g + f) dµ (3.47) lim inf (g + f n ) dµ = g dµ + lim inf f n dµ, and hence (3.48) f dµ lim inf f n dµ, whenever the hypotheses of Theorem 3.8 hold. Then the same argument applies with f n replaced by f n, so ( f) dµ lim inf ( f n ) dµ or, equivalently, (3.49) f dµ lim sup f n dµ. The inequalities (3.48) (3.49) yield (3.46). The next result is known as Egoroff s Theorem. It has many uses, including an alternative approach to the proof of the Dominated Convergence Theorem (explored in some of the exercises below). Proposition 3.9. Let (X, F, µ) be a measure space such that µ(x) <. Assume f j are F-measurable and f j (x) f(x) for all x X. Then, given any δ > 0, there exists B F such that (3.50) µ(b) δ and f j f uniformly on X \ B. Proof. For ε > 0, n Z +, set (3.51) F nε = {x X : for some j n, f j (x) f(x) ε}. Note that each set F nε belongs to F. The hypothesis implies n 1 F nε =, for each ε > 0, so µ(f nε ) 0 as n. In particular, given any δ > 0, if we set ε = 2 k, there exists n = n(k) such that (3.52) µ ( F n(k),2 k) 2 k δ.

11 3. Integration on Measure Spaces 35 Let B = k 1 F n(k),2 k, B F. We see that µ(b) δ and (3.53) x / B, j n(k) = f j (x) f(x) < 2 k. This establishes (3.50). We mention a point which is quite simple but important for applications. Namely, in all the convergence theorems established above, the hypothesis that a sequence of measurable functions f j converges for all x X can be weakened to the hypothesis that f j converges almost everywhere, i.e., f j converges for all x X 1, where X 1 X is a measurable subset such that µ(x \ X 1 ) = 0. The proof is easy; just restrict attention to the measure space (X 1, F 1, µ 1 ), where F 1 = {S F : S X 1 } = {S X 1 : S F} and µ 1 is µ restricted to F 1. To close this chapter, we relate the Riemann integral, considered in Chapter 1, to the Lebesgue integral, which is the integral defined in this chapter when µ is Lebesgue measure. The theory of the Lebesgue integral allows us to specify precisely which bounded functions on an interval I = [a, b] are Riemann integrable. Indeed, let ϕ : I R be a bounded function, and recall the definition (1.3) (1.6) of I(ϕ) and I(ϕ). Let P j be the sequence of partitions of I such that P 0 = {I} and P j+1 is obtained by dividing each interval of P j in half. Define f j to be sup J ϕ(x) on each interval J P j and define g j to be inf J ϕ(x) on each such interval J. (Here, omit the right endpoint of J, unless it is the point b.) Then f j while g j and, by (1.42), (3.54) f j dx I(ϕ), g j dx I(ϕ). On the other hand, we see that I (3.55) f j f ϕ, g j g ϕ, where f and g are bounded and measurable, and, by the Monotone Convergence Theorem, together with (3.54), we have (3.56) I(ϕ) = f dx, I(ϕ) = g dx. I Consequently ϕ is Riemann integrable if and only if (f g) dx = 0, hence if and only if f(x) = g(x) for almost all x I. (See Exercise 4 below.) On the other hand, if E is the collection of endpoints of all the intervals in the partitions P j (a countable set, hence of Lebesgue measure zero), we have (3.57) x / E, f(x) = g(x) ϕ is continuous at x. Thus we have the following proposition. I I

12 36 3. Integration on Measure Spaces Proposition If ϕ : I R is bounded, then ϕ is Riemann integrable if and only if the set of points x I at which ϕ is discontinuous has Lebesgue measure zero. If this condition holds, then the Riemann integral and the Lebesgue integral of ϕ coincide. Exercises 1. Suppose F is an algebra of subsets of X. Show that F is a σ-algebra, provided it is closed under countable disjoint unions. 2. Establish the properties (3.3) (3.6) of a measure. 3. Prove Lemma 3.3. Hint. Show that, if µ 1,..., µ N are measures on X and a j [0, ), then λ = a 1 µ a N µ N is a measure. Also show that, if µ is a measure on X and E j X are measurable, then (3.58) µ j (S) = µ(s E j ) = χ Ej dµ are measures. 4. Let (X, F, µ) be a measure space and f M + (X). Show that (3.59) f = 0 µ-a.e. f dµ = 0. To say that f = 0 µ-a.e. 0}, µ(n ) = 0. is to say that, for N = {x X : f(x) 5. A measure µ on (X, M) is said to be complete provided A M, µ(a) = 0, S A = S M, and µ(s) = 0. Suppose µ is a measure on (X, F) that is not complete. Show that (3.60) F = {E S : E F, S F F, µ(f ) = 0} is a σ-algebra, that there is a unique measure µ on (X, F) such that µ = µ on F, and that µ is complete. We call µ the completion of µ.

13 3. Integration on Measure Spaces Let µ be a measure on (X, F), as above, and µ its completion, on (X, F). If f : X R is F-measurable, show that there exists f 0 that is F- measurable, such that f 0 = f outside a set of µ-measure zero. Hint. First treat the case of simple functions, then positive functions, via the approximation (3.26). 7. Let f M + (X) and set, for A F, (3.61) λ(a) = f dµ = A χ A f dµ. Show that λ is a measure. Hint. Use the Monotone Convergence Theorem. Note that this result is much stronger than Lemma Let f L 1 (X, µ) be given. Show that, for every ε > 0, there is a δ > 0 such that (3.62) S F, µ(s) < δ = f dµ < ε. Hint. Pick ϕ S + (X) such that 0 ϕ f and ϕ dµ f dµ ε/2. Then A = sup ϕ <, so pick δ < ε/2a. 9. Give an alternative proof of the Dominated Convergence Theorem, deducing it from Egoroff s Theorem and the result of Exercise 8, rather than from Fatou s Lemma. Hint. Given f n g L 1 (X, µ), f n f, ε > 0, pick a measurable set Y X such that (3.63) µ(y ) <, g dµ < ε, X\Y and pick a measurable S Y, such that µ(s) < ε and f n f uniformly on Y \ S. 10. Let f j be measurable functions on (X, F, µ). We say f j f in measure provided µ(e jε ) 0 as j, for each ε > 0, where E jε = {x X : f j (x) f(x) > ε}. Show that, if f j f in measure, then some subsequence f jν f almost everywhere. S

14 38 3. Integration on Measure Spaces Hint. Arange that µ(e jν,2 ν) 2 ν. 11. Show that the Dominated Convergence Theorem continues to hold if the hypothesis (3.44) that f n f pointwise is changed to f n f in measure. 12. Let (X, F, µ) be a measure space and assume µ(x) <. Let M(X, F) denote the space of measurable functions and let M(X, F, µ) denote the set of equivalence classes of measurable functions, where f g f(x) = g(x), µ-a.e. For measurable f and g, set (3.64) d(f, g) = X f g 1 + f g dµ. Show that this is a distance function, making M(X, F, µ) a metric space, and that f n f in this metric if and only if f n f in measure, as defined in Exercise Show that, if µ(x) <, then f n f, µ-a.e. implies f n f in measure. Give a counterexample for X = Z, with counting measure. In Exercises 14 18, we deal with Lebesgue measure on the line R or on an interval I = [a, b]. A set S R is called an F σ set if S = j 1 F j, F j closed. S is called a G δ set if S = j 1 G j, G j open. 14. If O I is open, show that there exist f j C(I) such that 0 f j (x) χ O (x) for all x I. If K I is closed, show that there exist f j C(I) such that 1 f j (x) χ K (x) for all x I. 15. If S is an F σ set or a G δ set, show that there exist f j C(I) such that 0 f j 1 and f j (x) χ S (x) for almost all x I. Hint. Tackle the question of convergence in measure. 16. Let S I be Lebesgue measurable. Show that there exists an F σ set S 0 and a G δ set S 1 such that S 0 S S 1 and m(s 1 \ S 0 ) = 0. Deduce that there exist f j C(I) such that 0 f j 1 and f j χ S almost everywhere.

15 3. Integration on Measure Spaces Let f L 1 (I, dx). Show that there exist f j C(I) such that f j f almost everywhere. Hint. First approximate f by simple functions, as in (3.26). 18. Given f L 1 (I, dx) and δ > 0, show that there is a Lebesgue measurable S I such that m(s) < δ and f I\S is continuous. Hint. Apply Egoroff s Theorem to the results of Exercise 17. This result is known as Lusin s Theorem. 19. If (X, F, µ) is a measure space and f : X [0, ] is measurable, show that there exist A j F such that (3.65) f = j 1 1 j χ A j. Hint. Set A 1 = {x X : f(x) 1} and, inductively, for k 2, A k = {x X : f(x) 1 k 1 k + 1 } j χ A j (x). 20. Let (X, F, µ) be a measure space. Given f M + (X), t (0, ), set j=1 (3.66) S f (t) = {x X : f(x) > t}, Ψ f (t) = µ ( S f (t) ). (3.67) Show that X f dµ = 0 Ψ f (t) dt. Hint. First verify this for simple functions. Then use (3.26) and the Monotone Convergence Theorem. Show that, if ϕ j f, then, for each t (0, ), S ϕ1 (t) S ϕ2 (t) S f (t) to deduce that Ψ ϕj (t) dt Ψ f (t) dt. 21. Let F : [0, 1] [0, 1] be the homeomorphism described in Exercise 10 of Chapter 2, satisfying F (L) = K, m(l) > 0, m(k) = 0. Show that there is a subset S K such that F 1 (S) is not Lebesgue measurable.

16 40 3. Integration on Measure Spaces Hint. Apply Exercise 7 of Chapter 2 to X = L. 22. Let u n be measurable functions on (X, F, µ). Assume u n u pointwise, u n 0, u n dµ u dµ <. Prove that u u n dµ 0. Hint. Write u u n = 2(u u n ) + (u u n ) and apply the Dominated Convergence Theorem to the first term on the right side. 23. Below we display the guts (though not the skeleton or the flesh) of the derivation of Fatou s Lemma and the Dominated Convergence Theorem (DCT) from the Monotone Convergence Theorem (MCT), as done in the proofs of Proposition 3.6 and Theorem 3.8. Familiarize yourself with the details of these arguments, to the point where you can look at the material displayed below and see the entire argument. MCT = Fatou = DCT. (Fatou) ( lim inf f n dµ = lim inf f n) dµ lim inf k n k MCT f j dµ, (DCT) (g + f) dµ lim inf (g + f n ) dµ = g dµ + lim inf f n dµ Fatou (analogue for f n replaced by f n ).