Real Analysis Notes Thomas Goller September 4, 2011 Contents 1 Abstract Measure Spaces 2 1.1 Basic Definitions . .2 1.2 Measurable Functions . .2 1.3 Integration . .3 1.4 Counterexamples . .5 2 Banach Spaces 7 2.1 Basic Definitions . .7 2.2 Lp Spaces . .8 2.3 `p Spaces . .9 3 Uniform Boundedness Principle 10 1 Chapter 1 Abstract Measure Spaces 1.1 Basic Definitions The discussion will be based on Stein's Real Analysis. Definition 1. A measure space consists of a set X equipped with: 1. A non-empty collection M of subsets of X closed under complements and countable unions and intersections (a σ-algebra), which are the \measur- able" sets. µ 2. A measure M / [0; 1] with the property that if E1;E2;::: is a countable family of disjoint sets in M, then 1 ! 1 [ X µ En = µ(En) n=1 n=1 Now we state an assumption on a measure space that allows us to develop integration as in the case of Rd. Definition 2. The measure space (X; M; µ) is σ-finite if X can be written as the union of countably many measurable sets with finite measure. (To see that Rd is σ-finite, consider balls centered at the origin with radii going to infinity.) From now on, we assume (X; M; µ) is a σ-finite measure space. 1.2 Measurable Functions f Definition 3. A function X / [−∞; 1] is measurable if f −1([−∞; a)) is measurable for all a 2 R. Proposition 1. Some properties of measurable functions. 2 • If f; g are measurable and finite-valued a.e., then f + g and fg are mea- surable. 1 • If ffngn=1 is a sequence of measurable functions, then lim fn(x); sup fn(x); inf fn(x); lim sup fn(x); lim inf fn(x) n!1 n n n!1 n!1 are measurable (lim only if it exists). • If f is measurable and f(x) = g(x) for a.e. x, then g is measurable. Definition 4. A simple function on X is of the form N X akχEk ; k=1 where Ek are measurable sets of finite measure and ak are real numbers. Proposition 2. If f is a non-negative measurable function, then there exists 1 a sequence of simple functions f'kgk=1 that satisfies 'k(x) ≤ 'k+1(x) and lim 'k(x) = f(x) for all x: k!1 In general, if f is only measurable, then we conclude that j'k(x)j ≤ j'k+1(x)j and lim 'k(x) = f(x) for all x: k!1 1 Proposition 3 (Egorov). Suppose ffkgk=1 is a sequence of measurable func- tions defined on a measurable set E ⊆ X with µ(E) < 1, and fk ! f a.e. Then for each > 0, there is a set A ⊆ E, such that µ(E −A) ≤ and fk ! f uniformly on A. 1.3 Integration R R We define the Lebesgue integral X f(x) dµ(x), often written f, of a general measurable function f by a four part process: Definition 5. Let f be a measurable function. PN (i) If f(x) = k=1 akχEk (x) is a simple function, then N Z X f := akµ(Ek): k=1 (ii) If f is bounded and supported on a set of finite measure, then choose a 1 sequence of simple functions f'kgk=1 such that j'k(x)j ≤ j'k+1(x)j and limk!1 'k(x) = f(x) for all x. Then Z Z f := lim 'n; n!1 3 (iii) If f is non-negative, then Z Z f := sup g; g where the supremum is taken over all measurable functions g such that 0 ≤ g ≤ f and g is bounded and supported on a set of finite measure. (iv) For general f, write f = f + − f −, a decomposition into non-negative measurable functions. Then Z Z Z f := f + − f −: Definition 6. Let f be measurable. Then f is Lebesgue integrable if R jfj < 1 in the sense of (iii) above. Proposition 4. Key properties of the Lebesgue integral are: • Linearity. If f; g integrable and a; b 2 R, then Z Z Z (af + bg) = a f + b g: • Additivity. If E; F are disjoint measurable subsets of X, then Z Z Z f = f + f: EtF E F • Monotonicity. If f ≤ g, then Z Z f ≤ g: • Triangle inequality. Z Z f ≤ jfj: Proof. Prove these step by step, as in the definition of the integral, starting with the simple functions. The triangle inequality follows from monotonicity and linearity by considering f ≤ jfj and −f ≤ jfj. Proposition 5. Key limit theorems. • Fatou's lemma. If ffng is a sequence of non-negative measurable functions on X, then Z Z lim inf fn ≤ lim inf fn: n!1 n!1 4 • Monotone convergence. If ffng is a sequence of non-negative measurable functions with fn % f, then Z Z lim fn = f: n!1 • Dominated convergence. If ffng is a sequence of measurable functions with fn ! f a.e., and such that jfnj ≤ g for some integrable g, then Z jfn − fj ! 0 as n ! 1; and consequently Z Z fn ! f as n ! 1: Proof. Fatou's lemma: Set f := lim infn!1 fn. Suppose 0 ≤ g ≤ f, where g is bounded and supported on a set E of finite measure. Then gn(x) := minfg(x); fn(x)g is measurable, supported on E, and gn(x) ! g(x), so that R R gn ! g by the bounded convergence theorem (combine boundedness with R Egorov to show jgn − gj ! 0 as n ! 1). Since gn ≤ fn, monotonicity gives R R gn ≤ fn, whence Z Z g ≤ lim inf fn: n!1 Since R f = sup R g for such g, we get the result. g R R R Monotone convergence: fn ≤ f by monotonicity, whence lim sup fn ≤ R R R n!1 f. Conversely, Fatou's lemma implies f ≤ lim infn!1 fn. Combining these inequalities gives the result. Dominated convergence: we will use the reverse Fatou's lemma, which states that if ffng is a sequence of measurable functions, such that fn ≤ g for some integrable function g, then Z Z lim sup fn ≥ lim sup fn: n!1 n!1 This is proved by applying Fatou's lemma to the non-negative sequence fg−fng. Since jfn − fj < 2g, the reverse Fatou's lemma implies Z Z lim sup jfn − fj ≤ lim sup jfn − fj = 0; n!1 n!1 whence we get the desired result using the linearity of the integral and the triangle inequality. 1.4 Counterexamples • A non-measurable subset of R (Stein 24). For x; y 2 [0; 1], consider the equivalence relation x ∼ y whenever x − y is rational. Using the axiom 5 of choice, let N be the set obtained by choosing one element from each equivalence class. Then if frkg is an enumeration of the rationals in [−1; 1], we have 1 [ [0; 1] ⊆ (N + rk) ⊆ [−1; 2]; k=1 but the measure of each N + rk is the same since they are translates of P1 N . But 1 ≤ k=1 m(N + rk) ≤ 3, a contradiction. f • A non-measurable function R / R such that jfj is measurable. Let N ⊆ R be non-measurable, e.g. as above. Then ( 1 x 2 N f(x) = −1 x2 = N is not measurable since f −1((−∞; 0)) = N c is not measurable, but jfj ≡ 1 is measurable. • A sequence f'ng of non-negative bounded, measurable functions on R R that converge pointwise to an integrable function, where the 'n do not converge to R f. Let ( 1 2n −n ≤ x ≤ n 'n(x) = 0 otherwise: R Then 'n ! 0 pointwise, but 'n = 1 for each n. See also Stein 61. 1 • A Cauchy sequence ffng in the L norm that does not converge pointwise a.e. Let f1 = χ[0;1], f2 = χ 1 , f3 = χ 1 , f4 = χ 1 , etc. (Stein 92, [0; 2 ] [ 2 ;1] [0; 4 ] Ex. 12). • A function f integrable but discontinuous at every x, even when corrected on a set of measure 0. Set ( x−1=2 if 0 < x < 1 f(x) = 0 otherwise: Then if frkg is an enumeration of the rationals, 1 X −n F (x) = 2 f(x − rn) n=1 is integrable, so the series converges a.e., but the series is unbounded on every interval, and any function that agrees with F a.e. is unbounded in any interval (Stein 92-93, Ex. 15). 6 Chapter 2 Banach Spaces 2.1 Basic Definitions d Definition 7. A metric (distance measure) on a set S is a function S × S / R such that 1. d(x; y) ≥ 0; 2. d(x; y) = 0 () x = y; 3. d(x; y) = d(y; x); 4. d(x; z) ≤ d(x; y) + d(y; z). k·k Definition 8. A norm on a vector space X is a function X / R such that 1. kxk ≥ 0; 2. kxk = 0 () x = 0; 3. kαxk = jαjkxk, α 2 F; 4. kx + yk ≤ kxk + kyk. One can use a norm to define a metric by setting d(x; y) := kx − yk: The metric triangle inequality holds since d(x; z) = kx − zk = k(x − y) + (y − z)k ≤ kx − yk + ky − zk = d(x; y) + d(y; z): In general, a metric is not sufficient to define a norm. Definition 9. A Banach space is a complete normed vector space.