Lebesgue Integration on R n

Transcription

1 Lebesgue Integration on R n The treatment here is based loosely on that of Jones, Lebesgue Integration on Euclidean Space We give an overview from the perspective of a user of the theory Riemann integration is based on subdividing the domain of f This leads to the requirement of some smoothness of f for the Riemann integal to be defined: for x, y close, f(x) and f(y) need to have something to do with each other Lebesgue integration is based on subdividing the range space of f: it is built on inverse images Typical Example For a set E R n, define the characteristic function of the set E to be { if x E χ E (x) = if x / E Consider χ Q(x)dx, where Q R is the set of rational numbers: at rationals at irrationals Riemann: The upper Riemann integral is the inf of the upper sums : χ Q(x)dx = The lower Riemann integral is the sup of the lower sums : χ Q(x)dx = Since χ Q(x)dx χ Q(x)dx, χ Q is not Riemann integrable 65

2 66 Lebesgue Integration on R n Lebesgue: Let λ(e) denote the Lebesgue measure ( size ) of E (to be defined) Then χ Q (x)dx = λ(q [, ]) + λ(q c [, ]) = + = First, we must develop the theory of Lebesgue measure to measure the size of sets Advantages of Lebesgue theory over Riemann theory: Can integrate more functions (on finite intervals) 2 Good convergence theorems: lim n fn (x)dx = lim n f n (x)dx under mild assumptions Completeness of L p spaces Our first task is to construct Lebesgue measure on R n For A R n, we want to define λ(a), the Lebesgue measure of A, with λ(a) This should be a version of n- dimensional volume for general sets However, it turns out that one can t define λ(a) for all subsets A R n and maintain all the desired properties We will define λ(a) for [Lebesgue] measurable subsets of R n (very many subsets) We define λ(a) for increasingly complicated sets A R n See Jones for proofs of the unproved assertions made below Step Define λ( ) = Step We call a set I R n a special rectangle if I = [a, b ) [a 2, b 2 ) [a n, b n ), where < a j < b j < (Note: Jones leaves the right ends closed) Define λ(i) = (b a )(b 2 a 2 ) (b n a n ) Step 2 We call a set P R n a special polygon if P is a finite union of special rectangles Fact: Every special polygon is a disjoint union of finitely many special rectangles For P = N k= I k, where the I k s are disjoint (ie, for j k, I j I k = ), define λ(p) = N λ(i k ) Note that a special polygon may be written as a disjoint union k= of special rectangles in different ways Fact: λ(p) is independent of the way that P is written as a disjoint union of special rectangles

3 Lebesgue Integration on R n 67 Step Let G R n be a nonempty open set Define λ(g) = sup{λ(p) : P is a special polygon, P G} (Approximation by special polygons from the inside) Remark: Every nonempty open set in R n can be written as a countable disjoint union of special rectangles Step 4 Let K R n be compact Define λ(k) = inf{λ(g) : G open, K G} (Approximation by open sets from the outside) Fact: If K = P for a special polygon P, then λ(k) = λ(p) Now for A R n, A arbitrary, define λ (A) = inf{λ(g) : G open, A G} (outer measure of A) λ (A) = sup{λ(k) : K compact, K A} (inner measure of A) Facts: If A is open or compact, then λ (A) = λ(a) = λ (A) Hence for any A, λ (A) λ (A) Step 5 A bounded set A R n is said to be [Lebesgue] measurable if λ (A) = λ (A) In this case we define λ(a) = λ (A) = λ (A) Step 6 An arbitrary set A R n is said to be [Lebesgue] measurable if for each R >, A B(, R) is measurable, where B(, R) is the open ball of radius R with center at the origin If A is measurable, define λ(a) = sup R> λ(a B(, R)) Let L denote the collection of all Lebesgue measurable subsets of R n Fact L is a σ-algebra of subsets of R n That is, L has the properties: (i), R n L (ii) A L A c L (iii) If A, A 2, L is a countable collection of subsets of R n in L, then A k L k=

4 68 Lebesgue Integration on R n Fact If S is any collection of subsets of a set X, then there is a smallest σ-algebra A of subsets of X containing S (ie, with S A), namely, the intersection of all σ-algebras of subsets of X containing S This smallest σ-algebra A is called the σ-algebra generated by S Definition The smallest σ-algebra of subsets of R n containing the open sets is called the collection B of Borel sets Closed sets are Borel sets Fact Every open set is [Lebesgue] measurable Thus B L Fact If A L, then λ (A) = λ(a) = λ (A) Caution: However, λ (A) = λ (A) = does not imply A L Properties of Lebesgue measure λ is a measure This means: λ( ) = 2 ( A L) λ(a) ( ) If A, A 2, L are disjoint then λ A k = λ(a k ) (countable additivity) k= k= Consequences: (i) If A, B L and A B, then λ(a) λ(b) ( ) (ii) If A, A 2, L, then λ A k λ(a k ) (countable subadditivity) k= k= Remark: Both (i) and (ii) are true of outer measure λ on all subsets of R n Sets of Measure Zero Fact If λ (A) =, then λ (A) λ (A) =, so = λ (A) = λ (A), so A L Thus every subset of a set of measure zero is also measurable (we say λ is a complete measure)

5 Lebesgue Integration on R n 69 Characterization of Lebesgue measurable sets Definition A set is called a G δ if it is the intersection of a countable collection of open sets A set is called an F σ if it is the union of a countable collection of closed sets G δ sets and F σ sets are Borel sets Fact A set A R n is Lebesgue measurable iff a G δ set G and an F σ set F for which F A G and λ(g\f) = (Note: G\F = G F c is a Borel set) Examples () If A = {a} is a single point, then A L and λ(a) = () If A = {a, a 2, } is countable, then A is measurable, and λ(a) j= λ({a j}) =, so λ(a) = For example, λ(q) = (2) λ(r n ) = () Open sets in R Every nonempty open set G R is a (finite or) countable disjoint union of open intervals (a j, b j ) ( j J or j < ), and λ(g) = j λ(a j, b j ) = j (b j a j ) (4) The Cantor Set is a closed subset of [, ] Let ( G =, 2 ), λ(g ) = ( G 2 = 9, 2 ) ( 7 9 9, 8 ), λ(g 2 ) = ( G = 27, 2 ) ( , 26 ), λ(g ) = ) etc (note λ(g k ) = 2k Let G = k= [ ] 9 [ 2 9 ] (middle thirds of remaining subintervals) G k, so G is an open subset of (, ) Then λ(g) = = k [ 2 ] 7 9 [ 8 9 ( + 2 ( ) 2 2 ) + + = ( 2) = ] Define the Cantor set C = [, ]\G Since λ(c) + λ(g) = λ([, ]) =, we have λ(c) =

6 7 Lebesgue Integration on R n Fact For x [, ], x C iff x has a base expansion with only s and 2 s, ie, x = j= d j j with each d j {, 2} For example: = ( ) = ( ) = (222 ) 2 = (2 ) = (222 ) = (222 ) 4 is in C, but it is not an endpoint of any interval in any G k Despite the fact that λ(c) =, C is not countable In fact, C can be put in correspondence with [, ] (and thus also with R) Invariance of Lebesgue measure () Translation For a fixed x R n and A R n, define x + A = {x + y : y A} Fact If x R n and A L, then x + A L, and λ(x + A) = λ(a) (2) If T : R n R n is linear and A L, then T(A) L, and λ(t(a)) = det T λ(a) Measurable Functions We consider functions f on R n with values in the extended real numbers [, ] We extend the usual arithmetic operations from R to [, ] by defining x ± = ± for x R; a (± ) = ± for a > ; a (± ) = for a < ; and (± ) = The expressions +( ) and ( )+ are usually undefined, although we will need to make some convention concerning these shortly A function f : R n [, ] is called Lebesgue measurable if for every t R, f ([, t]) L (in R n ) Recall: Inverse images commute with unions, intersections, and complements: [ ] f [B c ] = f [B] c, f A α = [ ] f [A α ], f A α = f[a α ] α α α α Fact For any function f : R n [, ], the collection of sets B [, ] for which f [B] L is itself a σ-algebra of subsets of [, ] Note The smallest σ-algebra of subsets of [, ] containing all sets of the form [, t] for t R contains also { }, { }, and all sets of the form [, t), [t, ], (t, ], (a, b), etc It is the collection of all sets of the form B, B { }, B { }, or B {, } for Borel subsets B of R

7 Lebesgue Integration on R n 7 Comments If f and g : R n [, ] are measurable, then f + g, f g, and f are measurable (Here we need to make a convention concerning +( ) and ( )+ This statement concerning measurability is true so long as we define both of these expressions to be the same, arbitrary but fixed, number in [, ] For example, we may define + ( ) = ( ) + = ) Moreover, if {f k } is a sequence of measurable functions f k : R n [, ], then so are sup k f k (x), inf k f k (x), lim sup f k (x), lim inf k f k (x) Thus k }{{} =inf k sup j k f j (x) if lim k f k (x) exists x, it is also measurable Definition If A R n, A L, and f : A [, ], we say that f is measurable (on A) if, when we extend f to be on A c, f is measurable on R n Equivalently, we require that fχ A is measurable for any extension of f Definition If f : R n C (not including ), we say f is Lebesgue measurable if Ref and Imf are both measurable Fact f : R n C is measurable iff for every open set G C, f [G] L Integration First consider integration of a non-negative function f : R n [, ], with f measurable Let N be a positive integer, and define S N = k2 N λ ( {x : k2 N f(x) < (k + )2 N } ) + λ({x : f(x) = + }) k= In the last term on the right-hand-side, we use the convention = The quantity S N can be regarded as a lower Lebesgue sum approximating the volume under the graph of f by subdividing the range space [, ] rather than by subdividing the domain R n as in the case of Riemann integration R (k + )2 N k2 N f R n {x : k2 N f(x) < (k + )2 N }

8 72 Lebesgue Integration on R n Claim S N S N+ Proof We have {x : k2 N f(x) < (k + )2 N } = { x : k2 N f(x) < ( k + 2) } 2 N { x : ( ) k N f(x) < (k + )2 N} and the union is disjoint Thus k2 N λ({x : k2 N f(x) < (k + )2 N }) k2 N λ ({ x : k2 N f(x) < ( ) k + }) 2 2 N + ( k + 2) 2 N λ ({ x : ( ) k N f(x) < (k + )2 N}) and the claim follows after summing and redefining indices Definition The Lebesgue integral of f is defined by: f = lim S N R n N This limit exists (in [, ]) by the monotonicity S N S N+ Other notation for the integral is f = R n f(x)dx = R n fdλ R n General Measurable Functions Let f : R n [, ] be measurable Define f + (x) = { { f(x) if f(x) if f(x) <, f (x) = if f(x) > f(x) if f(x) Then f + and f are non-negative and measurable, and ( x) f(x) = f + (x) f (x) The integral of f is only defined if at least one of f + < or f < holds, in which case we define f = f + f R n R n R n Definition A measurable function is called integrable if both f + < and f < Since f = f + + f, this is equivalent to f <

9 Lebesgue Integration on R n 7 Properties of the Lebesgue Integral (We will write f L to mean f is measurable and f < ) () If f, g L and a, b R, then af + bg L, and (af + bg) = a f + b g We will write f = g ae (almost everywhere) to mean λ{x : f(x) g(x)} = (2) If f, g L and f = g ae, then f = g () If f and f <, then f < ae Thus if f L, then f < ae In integration theory, one often identifies two functions if they agree ae, eg, χ Q = ae (4) If f and f =, then f = ae (This is not true if f can be both positive and negative, eg, x dx = ) +x 4 (5) If A is measurable, χ A = λ(a) Definition If A is a measurable set and f : A [, ] is measurable, then A f = R n fχ A (6) If A and B are disjoint and fχ A B L, then f = f + A B A B f Definition If f : R n C is measurable, and both Ref and Imf L, define R n f = R n Ref + i R n Imf (7) If f : R n C is measurable, then Ref and Imf L iff f L Moreover, f f Comparison of Riemann and Lebesgue integrals If f is bounded and defined on a bounded set and f is Riemann integrable, then f is Lebesgue integrable and the two integrals are equal Theorem If f is bounded and defined on a bounded set, then f is Riemann integrable iff f is continuous ae Note: The two theories vary in their treatment of infinities (in both domain and range) For R sinx sinx example, the improper Riemann integral lim R dx exists and is finite, but is x x not Lebesgue integrable over [, ) since sinx x dx =

10 74 Lebesgue Integration on R n Convergence Theorems Convergence theorems give conditions under which one can interchange a limit with an integral That is, if lim k f k (x) = f(x) (maybe only ae), where f k and f are measurable, give conditions which guarantee that lim k fk = f This is not true in general: Examples () Let f k = χ [k, ) Then f k, lim f k =, and f k =, so lim f k lim f k (2) Let f k = χ [k,k+] Then again lim f k =, and f k =, so lim f k lim f k Monotone Convergence Theorem (Jones calls this the Increasing Convergence Theorem ) If f f 2 ae, f = lim f k ae, and f k and f are measurable, then lim k fk = f Here all the limits are non-negative extended real numbers Note that lim f k exists ae by monotonicity Fatou s Lemma If f k are nonnegative ae and measurable, then lim inf f k lim inf f k k k Lebesgue Dominated Convergence Theorem Suppose {f k } is a sequence of complexvalued (or extended-real-valued) measurable functions Assume lim k f k = f ae, and assume that there exists a dominating function, ie, an integrable function g such that f k (x) g(x) ae Then f = lim f k k A corollary is the Bounded Convergence Theorem Let A be a measurable set of finite measure, and suppose f k M in A Assume lim k f k exists ae Then lim k A f k = f (Apply Dominated A Convergence Theorem with g = Mχ A ) The following result illustrates how Fatou s Lemma can be used together with a dominating sequence to obtain convergence Extension of Lebesgue Dominated Convergence Theorem Suppose g k, g are all integrable, and g k g, and g k g ae Suppose f k, f are all measurable, f k g k ae (which implies that f k is integrable), and f k f ae (which implies f g ae) Then fk f (which implies f k f) Proof f k f f k + f g k + g ae Apply Fatou to g k + g f k f (which is ae) Then lim inf(g k + g f k f ) lim inf (g k + g f k f ) So 2g lim g k + g lim sup fk f = 2 g lim sup f k f Since g <, lim sup f k f Thus f k f

11 Lebesgue Integration on R n 75 Example the Cantor Ternary Function The Cantor ternary function is a good example in differentiation and integration theory It is a nondecreasing continuous function f : [, ] [, ] defined as follows Let C be the Cantor set If x C, say x = k= d k k with d k {, 2}, set f(x) = ( k= d 2 k) 2 k Recall that [, ]\C is the disjoint union of open intervals, the middle thirds which were removed in the construction of C Define f to be a constant on each of these open intervals, namely f = on (, ) 2 2 2, f = on (, ) and f = on ( 7, ), etc The general definition is: for x [, ], write x = k= d k k where d k {,, 2}, let K be the smallest k for which d k =, and define f(x) = 2 K + K ( k= d ) 2 k 2 k The graph of f looks like: Let us calculate f(x)dx using our convergence theorems Define a sequence of functions f k, k, inductively by: f = 2 χ (, 2 ) f 2 = f + 4 χ ( 9, 2 9) + 4 χ ( 7 9, 8 9), etc Then each f k is a simple function, ie, a finite linear combination of characteristic functions of measurable sets Note that if ϕ = N j= a jχ Aj is a simple function, where A j L and λ(a j ) <, then ϕ = N j= a jλ(a j ) Also, f k (x) f(x) for x [, ]\C, and f k (x) = for x C( k) Since λ(c) =, f k f ae on [, ] So by the MCT or LDCT or BCT, f(x)dx = lim k f k(x)dx Now f k = ( + ) + 2( ) + + k 2 ( k (2k ))

12 76 Lebesgue Integration on R n Recall that So f = lim k f k = m= (2j ) = j 2 ( = + 2 ( ) 2 2 m 2 m22m 2 6 ) + + = ( ) 6 2 = 2 (An easier way to see this is to note that f( x) = f(x), so f( x)dx = f(x)dx But changing variables gives f( x)dx = f(x)dx, so f(x)dx = ) 2 Multiple Integration via Iterated Integrals Suppose n = m + l, so R n = R m R l For x R n, write x = (y, z), y R m, z R l Then fdλ R n n = f(x)dλ R n n (x) = f(y, z)dλ R n n (y, z) Write dx for dλ n (x), dy for dλ m (y), dz for dλ l (z) (λ n denotes Lebesgue measure on R n ) Consider the iterated integrals [ ] [ ] f(y, z)dy dz and f(y, z)dz dy R l R m R m R l Questions: () When do these iterated integrals agree? (2) When are they equal to R n f(x)dx? There are two key theorems, usually used in tandem The first is Tonelli s Theorem, for non-negative functions Tonelli s Theorem Suppose f is measurable on R n Then for ae z R l, the function f z (y) f(y, z) is measurable on R m (as a function of y), and [ ] f(x)dx = f(y, z)dy dz R n R l R m It can happen in Tonelli s Theorem that for some z, the slice function f z is not measurable: Example Let A R m be non-measurable Pick z R l and define f : R n R by: { (z z ) f(y, z) = χ A (y) (z = z ) Then f is measurable on R n (since λ n ({x : f(x) }) = ) But f z (y) = f(y, z ) = χ A (y) is not measurable on R m However, since the set of z s for which f z (y)dy is undefined has measure zero, the iterated integral still makes sense and is

13 Lebesgue Integration on R n 77 2 Fubini s Theorem Suppose f is integrable on R n (ie, f is measurable and f < ) Then for ae z R l, the slice functions f z (y) are integrable on R m, and [ ] f(x)dx = f(y, z)dy dz R n R l R m Typically one wants to calculate R n f(x)dx by doing an iterated integral One uses Tonelli to verify the hypothesis of Fubini as follows: (i) Since f is non-negative, Tonelli implies that one can calculate R n f by doing either iterated integral If either one is <, then the hypotheses of Fubini have been verified: R n f(x) dx = [ f(y, z) dz ] dy < (ii) Having verified now that R n f <, Fubini implies that R n f can be calculated by doing either iterated integral Example Fubini s Theorem can fail without the hypothesis that f < Define f on (, ) (, ) by { x 2 < y x < f(x, y) = y 2 < x < y < Then f(x, y)dydx = ( x x 2 dy ) x y 2 dy dx = (x + x )dx = Similarly, (,) (,) f(x, y)dxdy = Note that by Tonelli, f(x, y) dλ 2 (x, y) = f(x, y) dydx = ( x + x ) y 2 dy dx = L p spaces p < Fix a measurable subset A R n Consider measurable functions f : A C for which A f p < Define f p = ( A f p) p On this set of functions, f p is only a seminorm: f p (but f p = does not imply f =, only f(x) = ae) αf p = α f p f + g p f p + g p (Minkowski s Inequality) (Note: f p = A f p = f = ae on A) Define an equivalence relation on this set of functions: f g means f = g ae on A Set f = {g measurable on A : f = g ae} to be the equivalence class of f Define f p = f p ; this is independent of the choice of representative in f Define L p (A) = { f : f p < } A

14 78 Lebesgue Integration on R n Then p is a norm on L p (A) We usually abuse notation and write f L p (A) to mean f L p (A) Example We say f L p (R n ) is continuous if g L p (R n ) for which g : R n C is continuous and f = g ae Equivalently, there exists g f such that g is continuous In this case, one typically works with the representative g of f which is continous This continuous representative is unique since two continuous functions which agree ae must be equal everywhere p = Let A R n be measurable Consider essentially bounded measurable functions f : A C, ie, for which M < so that f(x) M ae on A Define f = inf{m : f(x) M ae on A}, the essential sup of f If < f <, then for each ǫ >, λ{x A : f(x) > f ǫ} > As above, is a seminorm on the set of essentially bounded measurable functions, and is a norm on L (A) = { f : f < } Fact For f L (A), f(x) f ae This is true since {x : f(x) > f } = {x : f(x) > f + m }, m= and each of these latter sets has measure So the infimum is attained in the definition of f Fact L (R n ) is not separable (ie, it does not have a countable dense subset) Example For each α R, let f α (x) = χ [α,α+] (x) For α β, f α f β = So {B(f α ) : α R} is an uncountable collection of disjoint nonempty open subsets in L (R) Conjugate Exponents If p, q, and + = (where p q p 2 that p and q are conjugate exponents Examples: q 2 2 ), we say Hölder s Inequality If p, q, and + p q fg f p g q =, then (Note: if fg <, also fg fg f p g q ) { { p = p = Remark The cases q = and are obvious When p = 2, q = 2, this is the q = Cauchy-Schwarz inequality fg f 2 g 2

15 Lebesgue Integration on R n 79 Completeness Theorem (Riesz-Fischer) Let A R n be measurable and p Then L p (A) is complete in the L p norm p The completeness of L p is a crucially important feature of the Lebesgue theory Locally L p Functions Definition Let G R n be open Define L p loc (G) to be the set of all equivalence classes of measurable functions f on G such that for each compact set K G, f K L p (K) There is a metric on L p loc which makes it a complete metric space (but not a Banach space; the metric is not given by a norm) The metric is constructed as follows Let K, K 2, be a compact exhaustion of G, ie, a sequence of nonempty compact subsets of G with K m Km+ (where Km+ denotes the interior of K m+ ), and K m = G (eg, K m = {x G : dist(x, G c ) and x m}) Then for any compact set K G, K m Km+, so m for which K K m The distance in L p loc (G) is m= d(f, g) = m= m= 2 m f g p,k m + f g p,km m= K m It is easy to see that d(f j, f) iff ( K compact G) f j f p,k To see that d is a metric, one uses the fact that if (X, p) is a metric space, and we define σ(x, y) = p(x,y), then σ is a metric on X, and (X, p) is uniformly equivalent to (X, σ) To +p(x,y) show that σ satisfies the triangle inequality, one uses that t t is increasing on [, ) +t Note that σ(x, y) < for all x, y X Continuous Functions not closed in L p Let G R n be open and bounded Consider C b (G), the set of bounded continuous functions on G Clearly C b (G) L p (G) But C b (G) is not closed in L p (G) if p < Example Take G = (, ) and let f j have graph: 2 Then {f j } is Cauchy in p for p < But there is no continuous function f for which f j f p as j Facts Suppose p < and G open R n () The set of simple functions (finite linear combinations of characteristic functions of measurable sets) with support in a bounded subset of G is dense in L p (G) 2 j

16 8 Lebesgue Integration on R n (2) The set of step functions (finite linear combinations of characteristic functions of rectangles) with support in a bounded subset of G is dense in L p (G) () C c (G) is dense in L p (G), where C c (G) is the set of continuous functions f whose support {x : f(x) } is a compact subset of G (4) C c (G) is dense in L p (G), where C c (G) is the set of C functions whose support is a compact subset of G (Idea: mollify a given f L p (G) We will discuss this when we talk about convolutions) Consequence: For p <, L p (G) is separable (eg, use (2), taking rectangles with rational endpoints and linear combinations with rational coefficients) Another consequence of the density of C c (R n ) in L p (R n ) for p < is the continuity of translation For f L p (R n ) and y R n, define f y (x) = f(x y) (translate f by y) Claim If p <, the map y f y from R n into L p (R n ) is uniformly continuous Proof Given ǫ >, choose g C c (R n ) for which g f p < ǫ Let By uniform continuity of g, δ > for which M = λ({x : g(x) }) < ǫ z y < δ ( x) g z (x) g y (x) < (2M) p Then for z y < δ, ( ) p g z g y p p = g z g y p ǫ ǫ p λ({x : g z (x) or g y (x) }) (2M) (2M) p p (2M), ie, g z g y p ǫ Thus f z f y p f z g z p + g z g y p + g y f y p < ǫ + ǫ + ǫ = ǫ L p convergence and pointwise ae convergence p = f k f in L on the complement of a set of measure, f k f uniformly (Let A k = {x : f k (x) f(x) > f k f }, and A = A k Since each λ(a k ) =, also λ(a) = On A c, ( k) f k (x) f(x) f k f, so f k f uniformly on A c ) p < Let A R n be measurable Here f k f in L p (A) (ie, f k f p ) does not imply that f k f ae Example: A = [, ], f = χ [,], f 2 = χ [, 2], f = χ [,], 2 f 4 = χ [, 4], etc k= f f 2 f f 4 f 5 f 6 f 7

17 Lebesgue Integration on R n 8 Clearly f k p, so f k in L p, but for no x [, ] does f k (x) So L p convergence for p < does not imply ae convergence However: Fact If p < and f k f in L p (A), then a subsequence f kj for which f kj f ae as j Example Suppose A R n is measurable, p <, f k, f L p (A), and f k f ae Question: when does f k f in L p (A) (ie f k f p )? Answer: In this situation, f k f in L p (A) iff f k p f p Proof ( ) If f k f p, then f k p f p f k f p, so f k p f p ( ) First, observe: If x, y, then (x+y) p 2 p (x p +y p ) (Proof: let z = max{x, y}; then (x+y) p (2z) p = 2 p z p 2 p (x p +y p )) We will use Fatou s lemma with a dominating sequence We have f k f p ( f k + f ) p 2 p ( f k p + f p ) Apply Fatou to 2 p ( f k p + f p ) f k f p By assumption, f k p f p, so fk p f p We thus get 2 p ( f p + f p ) lim inf 2 p ( f k p + f p ) f k f p = 2 p ( f p + f p ) lim sup f k f p Thus lim sup f k f p So f k f p p = f k f p So f k f p Intuition for growth of functions in L p (R n ) Fix n, fix p with p <, and fix a R Define f (x) = x aχ {x: x <} f 2 (x) = x aχ {x: x >} So f blows up near x = for a >, but vanishes near And f 2 vanishes near but grows/decays near at a rate depending on a To calculate the integrals of powers of f and f 2, use polar coordinates on R n Polar Coordinates in R n rs n = {x : x = r} n dim volume element dv = r n dr dσ

18 82 Lebesgue Integration on R n Here dσ is surface area measure on S n Evaluating the integral in polar coordinates, [ ( ) p f (x) p dx = r n dr]dσ = ω R r n a n r n ap dr, S n where ω n = σ(s n ) This is < iff n ap >, ie, a < n p So f L p (R n ) iff a < n p Similarly, f 2 L p (R n ) iff a > n p Conclusion For any p q with p, q, L p (R n ) L q (R n ) However, for sets A of finite measure, we have: Claim If λ(a) < and p < q, then L q (A) L p (A), and f p λ(a) p q f q Proof This is obvious when q = So suppose p < q < Let r = q p Then ) ( < r < Let s be the conjugate exponent to r, so + = Then = p = p r s s q p q By Hölder, f p p = A f p = χ A f p χ A s f p r = λ(a) s = λ(a) p( p q) f p q ( f q ) p q Now take p th roots Remark This is in sharp contrast to what happens in l p For sequences {x k } k=, the l norm is x = sup k x k, and for p <, the l p norm is x p = ( k x k p ) p Claim For p < q, l p l q In fact x q x p Proof This is obvious when q = So suppose p < q < Then x q q = k x k q = k x k q p x k p x q p xk p x p q p x p p = x q p Take q th roots to get x q x p