Let (X, µ) be a measure space. As in Chapter 3, we say a measurable function f belongs to L 1 (X, µ) provided

Transcription

1 Chapter 4 L p Spaces Let (, µ) be a measure space. As in Chapter 3, we say a measurable function f belongs to L 1 (, µ) provided (4.1) f L 1 = f(x) dµ(x) <. Elements of L 1 (, µ) consist of equivalence classes of elements of L 1 (, µ), where we say (4.2) f f f(x) = f(x) for µ-almost every x. With a slight abuse of notation, we denote by f both a measurable function in L 1 (, µ) and its equivalence class in L 1 (, µ). Also we say f, defined only almost everywhere on, belongs to L 1 (, µ), if there exists f L 1 (, µ), equal µ-almost everywhere to f. The quantity f L 1 defined by (4.1) is called the L 1 norm of f. In general, a normed linear space is a vector space equipped with a positive function v having the properties (4.3) av = a v, for v V, a C (or R), v + w v + w, v > 0, unless v = 0. The second of these conditions is called the triangle inequality. Given a norm on V, setting d(u, v) = u v defines a distance function on V, making it a metric space. It is easy to see that L 1 (, µ) is a vector space and that f L 1 satisfies the first two conditions in (4.3). However, f L 1 = 0 if and only if f = 0 41

2 42 4. L p Spaces almost everywhere. (Recall Exercise 4 of Chapter 3.) That is the reason we define L 1 (, µ) to consist of equivalence classes defined by (4.2), so L 1 (, µ) becomes a normed linear space. Generally speaking, a sequence (v j ) in a normed linear space is said to be a Cauchy sequence if v j v k 0 as j, k. If every Cauchy sequence has a limit in V, then V is said to be complete; a complete normed linear space is called a Banach space. Theorem 4.1. L 1 (, µ) is a Banach space. The proof of completeness of L 1 (, µ) makes use of the following two lemmas, which are essentially restatements of the Monotone Convergence Theorem and the Dominated Convergence Theorem, respectively. Lemma 4.2. If f j L 1 (, µ), 0 f 1 (x) f 2 (x), and f j L 1 C <, then lim f j (x) = f(x), with f L 1 (, µ) and f j f L 1 0 as j j. Proof. We know that f M + (). The Monotone Convergence Theorem implies f j dµ f dµ. Thus f dµ C. Since f j f L 1 = f dµ fj dµ in this case, the lemma follows. Lemma 4.3. If f j L 1 (, µ), lim f j (x) = f(x) µ-a.e., and if there is an F L 1 (, µ) such that f j (x) F (x) µ-a.e., for all j, then f L 1 (, µ), and f j f L 1 0. Proof. Apply the Dominated Convergence Theorem to g j = f j f 0 a.e. Note that g j 2F. To show L 1 (, µ) is complete, suppose (f n ) is Cauchy in L 1. Passing to a subsequence, we can assume f n+1 f n L 1 2 n. Consider the infinite series (4.4) f 1 (x) + [ fn+1 (x) f n (x) ]. n=1 Now the partial sums are dominated by (4.5) f 1 (x) + m f n+1 (x) f n (x) = f 1 (x) + G m (x), n=1 and since 0 G 1 G 2 and G m L 1 2 n 1, we deduce from Lemma 4.2 that G m G µ-a.e. and in L 1 -norm. Hence the infinite series

3 4. L p Spaces 43 (4.4) is convergent a.e., to a limit f(x), and via Lemma 4.3 we deduce that f n f in L 1 -norm. This proves completeness. Continuing with a description of L p spaces, we define L (, µ) to consist of bounded measurable functions, L (, µ) to consist of equivalence classes of such functions, via (4.2), and we define f L to be the smallest sup of f f. It is easy to show that L (, µ) is a Banach space. For p (1, ), we define L p (, µ) to consist of measurable functions f such that [ ] 1/p (4.6) f(x) p dµ(x) is finite. L p (, µ) consists of equivalence classes, via (4.2), and the L p -norm f L p is given by (4.6). This time it takes a little work to verify the triangle inequality. That this holds is the content of Minkowski s inequality: (4.7) f + g L p f L p + g L p. One neat way to establish this is by the following characterization of the L p -norm. Suppose p and q are related by (4.8) 1 p + 1 q = 1. We claim that, if f L p (, µ), (4.9) f L p = sup{ fh L 1 : h L q (, µ), h L q = 1}. We can apply (4.9) to f + g, which belongs to L p (, µ) if f and g do, since f + g p 2 p( f p + g p). Given this, (4.7) follows easily from the inequality (f + g)h L 1 fh L 1 + gh L 1. The identity (4.9) can be regarded as two inequalities. The part can be proved by choosing h(x) to be an appropriate multiple C f(x) p 1. We leave this as an exercise. The converse inequality,, is a consequence of Hölder s inequality: (4.10) f(x)g(x) dµ(x) f L p g L q, 1 p + 1 q = 1. Hölder s inequality can be proved via the following inequality for positive numbers: (4.11) ab ap p + bq q for a, b > 0,

4 44 4. L p Spaces assuming that p (1, ) and (4.8) holds. a, b > 0, 1/p + 1/q = 1, In fact, we claim that, given (4.12) ϕ(t) = ap p tp + bq q t q = inf ϕ(t) = ab, t>0 which implies (4.11) since the right side of (4.11) is ϕ(1). As for (4.12), note that ϕ(t) as t 0 and as t, and the unique critical point occurs for a p t p = b q t q, i.e., for t = b 1/p /a 1/q, giving the desired conclusion. (4.13) Applying (4.11) to the integrand in (4.10) gives f(x)g(x) dµ(x) 1 p f p L p + 1 q g q L q. This looks weaker than (4.10), but now replace f by tf and g by t 1 g, so that the left side of (4.13) is dominated by (4.14) t p p f p L p + 1 qt q g q L q, for all t > 0. Another application of (4.12) then gives Hölder s inequality. Consequently (4.6) defines a norm on L p (, µ). Completeness follows as in the p = 1 case discussed above. In detail, given (f n ) Cauchy in L p (, µ), we can pass to the case f n+1 f n L p 2 n and define G m as in (4.5). We have G m L p 1, and hence (via the Monotone Convergence Theorem) deduce that G m G, µ-a.e., and in L p -norm. Hence the series (4.4) converges, µ-a.e., to a limit f(x). Since f f m+1 G G m, we have by the Dominated Convergence Theorem that f f m+1 p dµ (G G m ) p dµ 0, as m. Hence L p (, µ) is complete. To summarize, we have Theorem 4.4. For p [1, ), L p (, µ), with norm given by (4.6), is a Banach space. It is frequently useful to show that a certain linear subspace L of a Banach space V is dense. We give an important case of this here; C() denotes the space of continuous functions on. Proposition 4.5. If µ is a finite Borel measure on a compact metric space, then C() is dense in L p (, µ) for each p [1, ).

5 4. L p Spaces 45 Proof. First, let K be any compact subset of. The functions (4.15) f K,n (x) = [ 1 + n dist(x, K) ] 1 C() are all 1 and decrease monotonically to the characteristic function χ K equal to 1 on K, 0 on \ K. The Monotone Convergence Theorem gives f K,n χ K in L p (, µ) for 1 p <. Now let A be any measurable set. Any Borel measure on a compact metric space is regular, i.e., (4.16) µ(a) = sup{µ(k) : K A, K compact}. In case = I = [a, b] and µ = m is Lebesgue measure, this follows from (2.20) together with the consequence of Theorem 2.11, that all Borel sets in I are Lebesgue measurable. The general case follows from results that will be established in the next chapter; see (5.60). Thus there exists an increasing sequence K j of compact subsets of A such that µ(a \ j K j) = 0. Again, the Monotone Convergence Theorem implies χ Kj χ A in L p (, µ) for 1 p <. Thus all simple functions on are in the closure of C() in L p (, µ) for p [1, ). Construction of L p (, µ) directly shows that each f L p (, µ) is a norm limit of simple functions, so the result is proved. Using a cut-off, we can easily deduce the following. Let C 00 (R) denote the space of continuous functions on R with compact support. Corollary 4.6. For 1 p <, the space C 00 (R) is dense in L p (R). The case L 2 (, µ) is special. In addition to the L 2 -norm, there is an inner product, defined by (4.17) (f, g) L 2 = f(x)g(x) dµ(x). This makes L 2 (, µ) into a Hilbert space. It is worthwhile to consider the general notion of Hilbert space in some detail. We devote the next few pages to this and then return to the specific consideration of L 2 (, µ). Generally, a Hilbert space H is a complete inner product space. That is to say, first the space H is a linear space provided with an inner product, denoted (u, v), for u and v in H, satisfying the following defining conditions: (4.18) (au 1 + u 2, v) = a(u 1, v) + (u 2, v), (u, v) = (v, u), (u, u) > 0 unless u = 0.

6 46 4. L p Spaces To such an inner product there is assigned a norm, denoted by (4.19) u = (u, u). To establish that the triangle inequality holds for u + v, we can expand u + v 2 = (u + v, u + v) and deduce that this is [ u + v ] 2, as a consequence of Cauchy s inequality: (4.20) (u, v) u v, a result that can be proved as follows. The fact that (u v, u v) 0 implies 2 Re (u, v) u 2 + v 2 ; replacing u by e iθ u with e iθ chosen so that e iθ (u, v) is real and positive, we get (4.21) (u, v) 1 2 u v 2. Now in (4.21) we can replace u by tu and v by t 1 v to get (4.22) (u, v) t 2 u t v 2. Minimizing over t gives (4.20). This establishes Cauchy s inequality, so we can deduce the triangle inequality. Thus (4.19) defines a norm on H. Note the parallel between this argument and the proof of (4.7), via (4.10). The completeness hypothesis on H is that, with this norm, H is a Banach space. The nice properties of Hilbert spaces arise from their similarity with familiar Euclidean space, so a great deal of geometrical intuition is available. For example, we say u and v are orthogonal and write u v, provided (u, v) = 0. Note that the Pythagorean Theorem holds on a general Hilbert space: (4.23) u v = u + v 2 = u 2 + v 2. This follows directly from expanding (u + v, u + v). Another useful identity is the following, called the parallelogram law, valid for all u, v H : (4.24) u + v 2 + u v 2 = 2 u v 2. This also follows directly by expanding (u+v, u+v)+(u v, u v), observing some cancellations. One important application of this simple identity is to the following existence result. Let K be any closed, convex subset of H. Convexity implies x, y K (x + y)/2 K. Given x H, we define the distance from x to K to be (4.25) d(x, K) = inf { x y : y K}.

7 4. L p Spaces 47 Proposition 4.7. If K H is a nonempty, closed, convex set in a Hilbert space H and if x H, then there is a unique z K such that d(x, K) = x z. Proof. We can pick y n K such that x y n d = d(x, K). It will suffice to show that (y n ) must be a Cauchy sequence. Use (4.24) with u = y m x, v = x y n, to get y m y n 2 = 2 y n x y m x 2 4 x 1 2 (y n + y m ) 2. Since K is convex, (y n + y m )/2 K, so x (y n + y m )/2 d. Therefore which implies convergence. lim sup y n y m 2 2d 2 + 2d 2 4d 2 0, m,n In particular, this result applies when K is a closed linear subspace of H. In this case, for x H, denote by P K x the point in K closest to x. We have (4.26) x = P K x + (x P K x). We claim that x P K x belongs to the closed linear space K, called the orthogonal complement of K, defined by (4.27) K = {u H : (u, v) = 0 for all v K}. Indeed, take any v K. Then (t) = x P K x + tv 2 = x P K x 2 + 2t Re (x P K x, v) + t 2 v 2 is minimal at t = 0, so (0) = 0, i.e., Re (x P K x, v) = 0, for all v K. Replacing v by iv shows that (x P K x, v) also has vanishing imaginary part for any v K, so our claim is established. The decomposition (4.26) gives (4.28) x = x 1 + x 2, x 1 K, x 2 K, with x 1 = P K x, x 2 = x P K x. Clearly such a decomposition is unique. This implies that H is an orthogonal direct sum of K and K ; we write (4.29) H = K K.

8 48 4. L p Spaces From this it is clear that (4.30) ( K ) = K, that (4.31) x P K x = P K x, and that P K and P K are linear maps on H. We call P K the orthogonal projection of H on K. Note that P K x is uniquely characterized by the condition (4.32) P K x K, (P K x, v) = (x, v) for all v K. We remark that if K is a linear subspace of H that is not closed, then K coincides with K, and (4.30) becomes ( K ) = K. Using the orthogonal projection discussed above, we can establish the following result. Proposition 4.8. If H is a Hilbert space and ϕ : H C is a continuous linear map, there exists a unique f H such that (4.33) ϕ(u) = (u, f) for all u H. Proof. Consider K = Ker ϕ = {u H : ϕ(u) = 0}, a closed linear subspace of H. If K = H, ϕ = 0 and we can take f = 0. Otherwise, K 0; select a nonzero x 0 K, such that ϕ(x 0 ) = 1. We claim K is one dimensional in this case. Indeed, given any y K, y ϕ(y)x 0 is annihilated by ϕ, so it belongs to K as well as to K, so it is zero. The result is now easily proved by setting f = ax 0 with a C chosen so that (4.33) works for u = x 0, namely a(x 0, x 0 ) = 1, i.e., a = x 0 2. We note that the correspondence ϕ f gives a conjugate linear isomorphism (4.34) H H, where H denotes the space of all continuous linear maps ϕ : H C. Recall that our interest in Hilbert spaces arises from our interest in L 2 (, µ). Let us record the content of Proposition 4.8 in that case.

9 4. L p Spaces 49 Corollary 4.9. If ϕ : L 2 (, µ) C is a continuous linear map, there exists a unique f L 2 (, µ) such that (4.35) ϕ(u) = u(x)f(x) dµ(x), u L 2 (, µ). We can use orthogonal projection operators to construct an orthonormal basis of a Hilbert space H. Let us assume H is separable, i.e., H has a countable dense subset S = {v j : j 1}. See the exercises for results on separability. In such a case, pick a countable subset {w j : j 1} of S such that each w k is linearly independent of {w j : j < k} and such that the linear span of {w j } is dense in H. Let L k = span {w j : 1 j k}. We define an orthonormal set {u j : j 1} inductively, as follows. Set u 1 = w 1 / w 1. Suppose you have {u j : 1 j k}, an orthonormal basis of L k. Then set (4.36) u k+1 = w k+1 P k w k+1 w k+1 P k w k+1, where P k is the orthogonal projection onto L k. One does not need the construction involving Proposition 4.7 to get P k here. We can simply set (4.37) P k f = k (f, u j )u j. j=1 The set {u j : j 1} so constructed is orthonormal, i.e., (u k, u l ) = δ kl. Also its linear span is dense in H, since the linear span coincides with that of {w j : j 1}. We claim that, for each f H, P k f f as k. To see this, note that (4.38) f 2 = P k f 2 + f P k f 2 P k f 2 = k (f, u j ) 2. We also see that, if n k, then P n f P k f 2 = k<j n (f, u j) 2, and hence that (P k f) is a Cauchy sequence in H, for each f. We claim that the limit is equal to f, hence that j=1 (4.39) f = j (f, u j )u j. Indeed, we now know that the right side of (4.39) defines an element of H; call it g. Then, f g has inner product 0 with each u j, hence with all

10 50 4. L p Spaces elements of the linear span of {u j }, hence with all elements of the closure, i.e., with all elements of H, so f g = 0. In the case H = L 2 (I, m), with I = [ π, π] and m = dx/2π, an orthonormal basis is given by (4.40) e k (x) = e ikx, k Z. See the exercises for a proof of this. In such a case, (4.39) is the expansion of a function in a Fourier series. We use Corollary 4.9 to prove an important result known as the Radon- Nikodym Theorem. Let µ and ν be two finite measures on (, F). Let (4.41) α = µ + 2ν, ω = 2µ + ν. On the Hilbert space H = L 2 (, α), consider the linear functional ϕ : H C given by (4.42) ϕ(f) = f(x) dω(x). Note that ϕ(f) 2 f dα 2 α() f L 2 (,α). By Corollary 4.9, there exists g L 2 (, α) such that, for any f L 2 (, α) = L 2 (, µ) L 2 (, ν), (4.43) f(x) dω(x) = f(x)g(x) dα(x). In particular, this holds for any bounded measurable f. Note that this identity is equivalent to (4.44) f(2g 1) dν = f(2 g) dµ, for all f L 2 (, α). If we let f be the characteristic function of S 1l = {x : g(x) < 1/2 1/l} or of S 2l = {x : g(x) > 2 + 1/l}, we see that µ(s jl ) = ν(s jl ) = 0. As a consequence, we can arrange that 1/2 g(x) 2, for all x. We also see that Z = {x : g(x) = 1/2} must have µ-measure zero. (Similarly, {x : g(x) = 2} has ν-measure zero.) Also, (4.44) holds for all f M + (), by the Monotone Convergence Theorem. We say that ν is absolutely continuous with respect to µ and write ν << µ, provided (4.45) µ(s) = 0 = ν(s) = 0.

11 4. L p Spaces 51 In such a case, we see that Z = {x : g(x) = 1/2} has ν-measure zero. Given F M + (), we can set (4.46) f(x) = F (x) 2g(x) 1, and apply (4.44) to get (4.47) F (x) dν(x) = 2 g(x) h(x) = 2g(x) 1 F (x)h(x) dµ(x) for all positive measurable F. Note that taking F = 1 gives h L 1 (, µ). The result we have just obtained is known as the Radon-Nikodym Theorem. We record a formal statement. Theorem Let µ and ν be two finite measures on (, F). If ν is absolutely continuous with respect to µ, then (4.47) holds for some nonnegative h L 1 (, µ) and every positive measurable F. We mention that (4.47) also holds for every bounded measurable F. If we do not assume that ν << µ, we can still consider (4.48) h(x) = 2 g(x) 2g(x) 1 if g(x) 1 2, 0 if g(x) = 1 2, and we have (4.49) F dν = F h dµ Y Y for any positive measurable F, where (4.50) Y = \ Z = { x : g(x) 1 }. 2 Recall that µ( \ Y ) = 0. We can define the measure λ on (, F) by (4.51) λ(e) = ν(y E). Then we have (4.52) F dλ = F h dµ

12 52 4. L p Spaces for all positive measurable F. Write (4.53) ρ(e) = ν(e \ Y ) = ν(e Z), so (4.54) ν = λ + ρ. Now the measure λ is supported on Y, i.e., λ( \ Y ) = 0. Similarly, ρ is supported on Z. Thus λ and ρ have disjoint supports. Generally, two measures with disjoint supports are said to be mutually singular. When two measures λ and ρ are mutually singular, we write λ ρ. We have the following result, known as the Lebesgue decomposition of ν with respect to µ. Theorem If µ and ν are finite measures on (, F), then we can write (4.55) ν = λ + ρ, λ << µ, ρ µ. This decomposition is unique. Proof. The measures λ and ρ are given by (4.51) and (4.52). The fact that λ << µ is contained in (4.52). As we have noted, µ( \ Y ) = 0, so µ is supported on Y, which is disjoint from Z, on which ρ is supported; hence ρ µ. If also λ and ρ are measures such that ν = λ + ρ, λ << µ, ρ µ, we have Z F such that ρ is supported on Z and µ( Z) = 0. Now µ(z Z) = 0, so λ(z Z) = 0 and λ(z Z) = 0, and, for E M, This gives uniqueness. λ(e) = λ(e \ Z) = ν ( E \ (Z Z) ), λ(e) = λ(e \ Z) = ν ( E \ (Z Z) ). We say a measure µ on (, F) is σ-finite if we can write as a countable union j 1 j where j F and µ( j ) <. A paradigm case is Lebesgue measure on = R. There are routine extensions of Theorems to the case where µ and ν are σ-finite measures, which we leave to the reader. Exercises 1. Let V and W be normed linear spaces. Suppose we have linear transformations (4.56) T j : V W, T j v W C v V,

13 4. L p Spaces 53 with C independent of j. (We say {T j } is uniformly bounded.) Suppose also T : V W satisfies this bound. Let L be a dense subspace of V. Then show that (4.57) T j v T v, v L = T j v T v, v V. 2. Define (4.58) τ s : L p (R) L p (R), τ s f(x) = f(x s). Show that, for p [1, ), (4.59) f L p (R) = τ s f f in L p -norm, as s 0. Hint. Apply Exercise 1, with V = W = L p (R), L = C 00 (R), as in Corollary 4.6. Note that τ s f L p = f L p. One says a metric space is separable if it has a countable dense subset. 3. If I = [a, b] R and a α < β b, define ϕ αβ (x) = dist ( x, I \ [α, β] ). Show that the linear span over Q of {ϕ αβ : α, β Q I} is dense in C(I), and deduce that C(I) is separable. From the denseness and continuity of the inclusion ι : C(I) L p (I), prove that L p (I) is separable, for 1 p <. Then prove that L p (R) is separable, for 1 p <. 4. Let be a compact metric space; has a countable dense subset {z j : j 1}. Given 0 < ρ < (diam )/2, set ψ jρ (x) = dist ( x, \ B ρ (z j ) ). Show that the algebra generated by {ψ j,ρ : j Z +, ρ Q + } and 1 is dense in C() and deduce that C() is separable. Conclude, from Proposition 4.5, that L p (, µ) is separable, for p [1, ), if µ is a finite measure on the σ-algebra of Borel sets in. 5. Let {u j : j 1} be a countable orthonormal set in a Hilbert space H. Show that (4.60) (f, u j )u j = P V f, j=1

14 54 4. L p Spaces where V is the closure of the linear span of {u j }. The Stone-Weierstrass Theorem states that, if is a compact Hausdorff space and A an algebra of functions in C R () (the space of real-valued continuous functions on ), such that 1 A, and if A has the property of separating points, i.e., for any two distinct p, q, there exists f A such that f(p) f(q), then A is dense in C R (). If A is an algebra in C C () with these properties, plus the property that f A f A, then A is dense in C C (). A proof is given in Appendix A. 6. Use the Stone-Weierstrass Theorem to show that, if e k (x) = e ikx, as in (4.40), then the linear span E of {e k : k Z} is dense in C(S 1 ), where S 1 = R/2πZ; hence E is dense in L p (S 1, dx/2π), for p [1, ). Hence {e k : k Z} is an orthonormal basis of L 2 (S 1, dx/2π). 7. For f, u C0 (R), the space of smooth functions with compact support in R, set (4.61) K f u(x) = f u(x) = f(y)u(x y) dy. Show that, for 1 p <, K f has a unique bounded extension: (4.62) K f : L p (R) L p (R), K f u L p f L 1 u L p. The operation in (4.61) is called convolution. Hint. For f, u C0 (R), if f is supported in [a, b], show that n f u(x) = lim n b a n j=0 ( j ( f n b + 1 j ) ) a τ j u(x), n where τ j u(x) = u(x jb/n (1 j/n)a). Use the triangle inequality (4.7) to estimate norms. 8. Show that there is a unique extension from f C 0 (R) to f L1 (R) of K f u, with (4.62) continuing to hold, giving a continuous linear map K : L 1 (R) L(L p (R)). 9. Let f j be a sequence of nonnegative functions in L 1 (R) such that (4.63) f j dx = 1, supp f j {x R : x < 1/j}. Show that, for 1 p <, (4.64) u L p (R) = f j u u in L p norm, as j.

15 4. L p Spaces 55 Derive the same conclusion, upon weakening the second hypothesis in (4.63) to f j dx = β j (ε) 1 as j, ε > 0. x <ε Hint. First verify the conclusion for u C 0 (R). Then use Exercise Given f L 1 (S 1 ), 0 < r < 1, define (4.65) P r f(θ) = Show that n= (4.66) P r f(θ) = p r f(θ) = 1 2π where (4.67) p r (θ) = Show that ˆf(n)r n e inθ 1 2π, ˆf(n) = f(θ)e inθ dθ. 2π 0 n= r n e inθ = 2π 0 p r (θ ϕ)f(ϕ) dϕ, 1 r 2 1 2r cos θ + r 2. (4.68) 1 2π p r (θ) dθ = 1. 2π If f L p (S 1 ), 1 p <, show that (4.69) P r f f, as r 1, in L p -norm. If f C(S 1 ), show that you have uniform convergence in (4.69). This is known as Abel convergence of Fourier series. Hint. Use a variant of the analysis needed for Exercise Show that Exercises provide an alternative proof of the conclusion of Exercise 6, that the linear span of e k (θ) = e ikθ, k Z, is dense in C(S 1 ) and in L p (S 1 ), for 1 p <, and hence that {e k : k Z} is an orthonormal basis of L 2 (S 1, dθ/2π). 13. Suppose (, F, µ) is the completion of (, F, µ). Show that L p (, µ) and L p (, µ) are identical.

16 56 4. L p Spaces Hint. Consult Exercise 6 of Chapter Show that if f L p (, µ) and p [1, ), then, for λ (0, ), (4.70) µ ({ x : f(x) > λ }) λ p f p L p. Deduce that if f k 0 in L p -norm, for some p [1, ), then f k 0 in measure, as defined in Exercise 10 of Chapter 3. Hint. Denote the set being measured in (4.70) by E λ and note that E λ f p dµ λ p µ(e λ ). The inequality (4.70) is called Tchebychev s inequality.