Real Analysis Notes. Thomas Goller


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1 Real Analysis Notes Thomas Goller September 4, 2011
2 Contents 1 Abstract Measure Spaces Basic Definitions Measurable Functions Integration Counterexamples Banach Spaces Basic Definitions L p Spaces l p Spaces Uniform Boundedness Principle 10 1
3 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 nonempty collection M of subsets of X closed under complements and countable unions and intersections (a σalgebra), which are the measurable sets. 2. A measure M µ [0, ] with the property that if E 1, E 2,... is a countable family of disjoint sets in M, then ( ) µ E n = µ(e n ) n=1 Now we state an assumption on a measure space that allows us to develop integration as in the case of R d. Definition 2. The measure space (X, M, µ) is σfinite if X can be written as the union of countably many measurable sets with finite measure. (To see that R d is σfinite, consider balls centered at the origin with radii going to infinity.) n=1 From now on, we assume (X, M, µ) is a σfinite measure space. 1.2 Measurable Functions Definition 3. A function X f [, ] is measurable if f 1 ([, a)) is measurable for all a R. Proposition 1. Some properties of measurable functions. 2
4 If f, g are measurable and finitevalued a.e., then f + g and fg are measurable. If {f n } n=1 is a sequence of measurable functions, then lim f n(x), sup f n (x), n are measurable (lim only if it exists). inf f n(x), lim sup f n (x), n lim inf f n(x) 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 a k χ Ek, k=1 where E k are measurable sets of finite measure and a k are real numbers. Proposition 2. If f is a nonnegative measurable function, then there exists a sequence of simple functions {ϕ k } k=1 that satisfies ϕ k (x) ϕ k+1 (x) and lim k ϕ k(x) = f(x) for all x. In general, if f is only measurable, then we conclude that ϕ k (x) ϕ k+1 (x) and lim k ϕ k(x) = f(x) for all x. Proposition 3 (Egorov). Suppose {f k } k=1 is a sequence of measurable functions defined on a measurable set E X with µ(e) <, and f k f a.e. Then for each ɛ > 0, there is a set A ɛ E, such that µ(e A ɛ ) ɛ and f k f uniformly on A ɛ. 1.3 Integration 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. (i) If f(x) = N k=1 a kχ Ek (x) is a simple function, then f := N a k µ(e k ). k=1 (ii) If f is bounded and supported on a set of finite measure, then choose a sequence of simple functions {ϕ k } k=1 such that ϕ k(x) ϕ k+1 (x) and lim k ϕ k (x) = f(x) for all x. Then f := lim ϕ n, 3
5 (iii) If f is nonnegative, then 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 nonnegative measurable functions. Then f := f + f. Definition 6. Let f be measurable. Then f is Lebesgue integrable if f < in the sense of (iii) above. Proposition 4. Key properties of the Lebesgue integral are: Linearity. If f, g integrable and a, b R, then (af + bg) = a f + b g. Additivity. If E, F are disjoint measurable subsets of X, then f = f + f. E F E F Monotonicity. If f g, then f g. Triangle inequality. f f. 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 f and f f. Proposition 5. Key limit theorems. Fatou s lemma. If {f n } is a sequence of nonnegative measurable functions on X, then lim inf f n lim inf f n. 4
6 Monotone convergence. If {f n } is a sequence of nonnegative measurable functions with f n f, then f n = f. lim Dominated convergence. If {f n } is a sequence of measurable functions with f n f a.e., and such that f n g for some integrable g, then f n f 0 as n, and consequently f n f as n. Proof. Fatou s lemma: Set f := lim inf f n. Suppose 0 g f, where g is bounded and supported on a set E of finite measure. Then g n (x) := min{g(x), f n (x)} is measurable, supported on E, and g n (x) g(x), so that gn g by the bounded convergence theorem (combine boundedness with Egorov to show g n g 0 as n ). Since g n f n, monotonicity gives gn f n, whence g lim inf f n. Since f = sup g g for such g, we get the result. Monotone convergence: f n f by monotonicity, whence lim sup fn f. Conversely, Fatou s lemma implies f lim inf fn. Combining these inequalities gives the result. Dominated convergence: we will use the reverse Fatou s lemma, which states that if {f n } is a sequence of measurable functions, such that f n g for some integrable function g, then lim sup f n lim sup This is proved by applying Fatou s lemma to the nonnegative sequence {g f n }. Since f n f < 2g, the reverse Fatou s lemma implies lim sup f n f lim sup f n f = 0, whence we get the desired result using the linearity of the integral and the triangle inequality. 1.4 Counterexamples A nonmeasurable subset of R (Stein 24). For x, y [0, 1], consider the equivalence relation x y whenever x y is rational. Using the axiom f n. 5
7 of choice, let N be the set obtained by choosing one element from each equivalence class. Then if {r k } is an enumeration of the rationals in [ 1, 1], we have [0, 1] (N + r k ) [ 1, 2], k=1 but the measure of each N + r k is the same since they are translates of N. But 1 k=1 m(n + r k) 3, a contradiction. A nonmeasurable function R f R such that f is measurable. Let N R be nonmeasurable, e.g. as above. Then { 1 x N f(x) = 1 x / N is not measurable since f 1 ((, 0)) = N c is not measurable, but f 1 is measurable. A sequence {ϕ n } of nonnegative bounded, measurable functions on R that converge pointwise to an integrable function, where the ϕ n do not converge to f. Let ϕ n (x) = { 1 2n n x n 0 otherwise. Then ϕ n 0 pointwise, but ϕ n = 1 for each n. See also Stein 61. A Cauchy sequence {f n } in the L 1 norm that does not converge pointwise a.e. Let f 1 = χ [0,1], f 2 = χ [0, 1 2 ], f 3 = χ [ 1 2,1], f 4 = χ [0, 1 4 ], etc. (Stein 92, 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 {r k } is an enumeration of the rationals, F (x) = 2 n f(x r n ) 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 9293, Ex. 15). 6
8 Chapter 2 Banach Spaces 2.1 Basic Definitions Definition 7. A metric (distance measure) on a set S is a function S S 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). d R Definition 8. A norm on a vector space X is a function X R such that 1. x 0; 2. x = 0 x = 0; 3. αx = α x, α F; 4. x + y x + y. One can use a norm to define a metric by setting d(x, y) := x y. The metric triangle inequality holds since d(x, z) = x z = (x y) + (y z) x y + y z = 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. A Hilbert space is an innerproduct space that is complete in the associated norm, hence also a Banach space. A separable Hilbert space has a countable spanning subset, hence has a (countable) orthonormal basis by GramSchmidt. 7
9 2.2 L p Spaces Proposition 6. Let 1 p < and (X, µ) a measure space. The set of equivalence classes of measurable functions X f C (that agree a.e.) such that ( ) 1/p f p := f p < is a normed vector space L p (X, µ). X Proof. First we check that L p is a vector space. It is clearly closed under (e.g. complex) scalar multiplication. For vector addition, note that f(x) + g(x) 2 max{ f(x), g(x) }, whence f(x) + g(x) p 2 p ( f(x) p + g(x) p ). Now we check the properties of the norm. Clearly f p 0 for all f, and f p = 0 if and only if f = 0. Likewise, λf p = λ f p is clear by the linearity of the integral. The triangle inequality is true but difficult to prove look up Minkowski s inequality. For p =, consider the equivalence classes of measurable functions that are bounded except on a set of measure 0, with the norm Then f := inf{c 0: f(x) C a.e. x}. f = lim p f p holds if f is in L p (X, µ) for some p <. (If f L p and f L, then f L q for each p q. For the set E on which f 1 must have finite measure, so we can bound f on this set by some M, and obtain f q f p + µ(e) M.) The fact that L p (X, µ) for 1 p is complete (and hence a Banach space) is the RieszFischer theorem. When p <, L p is separable (for X = R d, the countable collection is rχ R (x), where r is a complex number with rational real and imaginary parts and R is a rectangle with rational coordinates). In the special case when p = 2, L 2 is in fact a Hilbert space under the inner product (f, g) = f(x)g(x)dx. A useful tool in the study of L p spaces is Hölder s inequality, which states that if f and g are measurable and 1 p q such that 1 p + 1 q = 1, then fg 1 f p g q, namely ( fg f p ) 1/p ( g q ) 1/q. In particular, if f L p and g L q, then fg L 1. 8
10 2.3 l p Spaces Consider the vector space of sequences over R or C. For 0 < p <, l p is the subspace of all sequences x = {x k } such that x k p <. k 1 If p 1, then x p := x k p k 1 1/p defines a norm on l p, and l p is complete, hence a Banach space (it s the L p space on the integers with the counting measure). If 0 < p < 1, then l p does not have a norm, but still has the metric d(x, y) = k 1 x k y k p. (To see that the natural attempt to define a norm fails to satisfy the triangle inequality, let p = 1 2 and set x = ( 1 2, 0, 0,... ), y = (0, 1, 0,... ). 2 Then while x + y = ( ) 2 = 2, x + y = = 1.) If p =, then l is the space of all bounded sequences, which is a Banach space under the norm x := sup x k. k Note that l p l q l for 1 p < q < (since only finitely many elements of the sequence can have absolute value 1). Also, l 2 is the only l p space that is a Hilbert space. 9
11 Chapter 3 Uniform Boundedness Principle The discussion in this chapter, based on Chapter 8 of Munkres, is geared toward proving the Uniform Boundedness Principle. Definition 10. A topological space X is a Baire space if: Given any countable collection {A n } of closed sets of X each of which has empty interior in X, their union A n also has empty interior in X. Example 1. Q in the standard topology is not a Baire space. Every point is closed, and the union of countably many points makes up the whole space. Z + is a Baire space. Every subset is open, so the only subset with empty interior is the empty set. Every closed subspace of R is a complete metric space, hence a Baire space. In fact, the irrationals in R also form a Baire space. (See Exercise 6.) Lemma 1. X is a Baire space if and only if given any countable collection {U n } of open sets in X, each of which is dense in X, their intersection U n is also dense in X. Proof. The result follows easily from the facts that complements of open sets are closed sets (and vice versa) and A X is dense if and only if X A has empty interior. Before proving that complete metric spaces are Baire spaces, we will need a simple result. Lemma 2. If X is a complete metric space, then X is regular. Namely for each pair of a point x and a closed set B disjoint from x, there exist disjoint open sets containing x and B, respectively. 10
12 Proof. Since x B c and B c is open, there is an open ball B δ (x) of radius δ > 0, centered at x, whose closure is contained in B c. Then the open sets B δ (x) and X B δ (x) give the result. Now we come to the first big result. Theorem 1 (Baire Category Theorem). If X is a complete metric space, then X is a Baire space. Proof. Let {A n } be a countable collection of closed subsets of X, each with empty interior. We need to show that any nonempty open set U 0 contains a point not in any of the A n. Since A 1 does not contain U 0, choose y U 0 not in A 1. Since X is regular and A 1 is closed, we can in fact choose a neighborhood U 1 of y of diameter < 1 whose closure is contained in U 0 and disjoint from A 1. Now repeat by induction, where at the nth step we choose a point of U n 1 not in A n, and a neighborhood U n of that point with diameter < 1/n and with closure contained in U n 1 and disjoint from A n. Then we claim that the intersection U n is nonempty, which will give the result. This is a result of the following lemma. Lemma 3. Let C 1 C 2 be a nested sequence of nonempty closed sets in the complete metric space X. If diam C n 0, then C n. Proof. Choose x n C n for each n. Then {x n } is Cauchy, and eventually lies inside each C k, whence the limit x is in each C k and thus in the intersection. Now we come to a crucial theorem. Theorem 2 (Uniform boundedness principle). Let X be a Banach space and Y be a normed vector space. Suppose F is a collection of continuous (bounded) linear operators from X to Y, such that for each x X, there is an M x 0 satisfying sup T x M x. T F Then sup T <. T F The gist of the theorem is that we want a sufficient condition to ensure the norms of a collection of operators are bounded. An obvious necessary condition is that the norms are pointwise bounded in the assumed sense, which turns out to be sufficient as well. Proof. For n Z 1, define A n := { } x X : sup T x n. T F 11
13 Then A n is closed (since X is complete and each T F is continuous), contains 0, and is symmetric about the origin. Thus if some A n contains a nonempty open set, then it contains a ball of radius δ > 0 centered at the origin, whence for any x X, T x = 2 x ( ) x δ T δ 2 x n, x 2 δ whence T 2n/δ, giving the result. So we may assume that each A n has empty interior. But X = A n by the pointwise boundedness assumption, so by the Baire category theorem, X = A n has empty interior, contradiction. Corollary 1. If a sequence of bounded linear operators on a Banach space converges pointwise, then these pointwise limits define a bounded linear operator. 12
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