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 “measurable” sets.

µ 2. A measure M / [0, ∞] with the property that if E1,E2,... is a countable family of disjoint sets in M, then

∞ ! ∞ [ 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 / [−∞, ∞] is measurable if f −1([−∞, a)) is measurable for all a ∈ 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 measurable.

∞ • If {fn}n=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→∞ n n n→∞ n→∞ 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 ∞ a sequence of simple functions {ϕk}k=1 that satisﬁes

ϕk(x) ≤ ϕk+1(x) and lim ϕk(x) = f(x) for all x. k→∞ In general, if f is only measurable, then we conclude that

|ϕk(x)| ≤ |ϕk+1(x)| and lim ϕk(x) = f(x) for all x. k→∞ ∞ Proposition 3 (Egorov). Suppose {fk}k=1 is a sequence of measurable functions deﬁned on a measurable set E ⊆ X with µ(E) < ∞, 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 ∞ sequence of simple functions {ϕk}k=1 such that |ϕk(x)| ≤ |ϕk+1(x)| and limk→∞ ϕk(x) = f(x) for all x. Then Z Z f := lim ϕn, n→∞

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 |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 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 ≤ |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 {fn} is a sequence of non-negative measurable functions on X, then Z Z lim inf fn ≤ lim inf fn. n→∞ n→∞

4 • Monotone convergence. If {fn} is a sequence of non-negative measurable functions with fn % f, then Z Z lim fn = f. n→∞

• Dominated convergence. If {fn} is a sequence of measurable functions with fn → f a.e., and such that |fn| ≤ g for some integrable g, then Z |fn − f| → 0 as n → ∞,

and consequently Z Z fn → f as n → ∞.

Proof. Fatou’s lemma: Set f := lim infn→∞ fn. Suppose 0 ≤ g ≤ f, where g is bounded and supported on a set E of ﬁnite measure. Then gn(x) := min{g(x), fn(x)} 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 |gn − g| → 0 as n → ∞). Since gn ≤ fn, monotonicity gives R R gn ≤ fn, whence Z Z g ≤ lim inf fn. n→∞ 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→∞ f. Conversely, Fatou’s lemma implies f ≤ lim infn→∞ fn. Combining these inequalities gives the result. Dominated convergence: we will use the reverse Fatou’s lemma, which states that if {fn} 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→∞ n→∞

This is proved by applying Fatou’s lemma to the non-negative sequence {g−fn}. Since |fn − f| < 2g, the reverse Fatou’s lemma implies Z Z lim sup |fn − f| ≤ lim sup |fn − f| = 0, n→∞ n→∞ 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 ∈ [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 {rk} is an enumeration of the rationals in [−1, 1], we have ∞ [ [0, 1] ⊆ (N + rk) ⊆ [−1, 2], k=1

but the measure of each N + rk is the same since they are translates of P∞ N . But 1 ≤ k=1 m(N + rk) ≤ 3, a contradiction.

f • A non-measurable function R / R such that |f| is measurable. Let N ⊆ R be non-measurable, 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 {ϕn} of non-negative bounded, measurable functions on R R that converge pointwise to an integrable function, where the ϕn do not converge to R f. Let

( 1 2n −n ≤ x ≤ n ϕn(x) = 0 otherwise. R Then ϕn → 0 pointwise, but ϕn = 1 for each n. See also Stein 61.

1 • A Cauchy sequence {fn} 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 {rk} is an enumeration of the rationals,

∞ 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 = |α|kxk, α ∈ 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. A Hilbert space is an inner-product 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 Gram-Schmidt.

7 2.2 Lp Spaces

Proposition 6. Let 1 ≤ p < ∞ and (X, µ) a measure space. The set of f equivalence classes of measurable functions X / C (that agree a.e.) such that Z 1/p p kfkp := |f| < ∞ X is a normed vector space Lp(X, µ). Proof. First we check that Lp 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 ≤ 2p(|f(x)|p + |g(x)|p). Now we check the properties of the norm. Clearly kfkp ≥ 0 for all f, and kfkp = 0 if and only if f = 0. Likewise, kλfkp = |λ|kfkp 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

kfk∞ := inf{C ≥ 0: |f(x)| ≤ C a.e. x}.

Then kfk∞ = lim kfkp p→∞ holds if f is in Lp(X, µ) for some p < ∞. (If f ∈ Lp and f ∈ L∞, then f ∈ Lq 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 R |f|q ≤ R |f|p + µ(E) · M.) The fact that Lp(X, µ) for 1 ≤ p ≤ ∞ is complete (and hence a Banach space) is the Riesz-Fischer theorem. When p < ∞, Lp is separable (for d X = R , the countable collection is rχR(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, L2 is in fact a Hilbert space under the inner product Z (f, g) = f(x)g(x)dx.

A useful tool in the study of Lp spaces is H¨older’sinequality, which states 1 1 that if f and g are measurable and 1 ≤ p ≤ q ≤ ∞ such that p + q = 1, then

kfgk1 ≤ kfkpkgkq, namely Z Z 1/p Z 1/q |fg| ≤ |f|p |g|q .

In particular, if f ∈ Lp and g ∈ Lq, then fg ∈ L1.

8 2.3 `p Spaces

Consider the vector space of sequences over R or C. For 0 < p < ∞, `p is the subspace of all sequences x = {xk} such that

X p |xk| < ∞. k≥1

If p ≥ 1, then  1/p X p kxkp :=  |xk|  k≥1 deﬁnes a norm on `p, and `p is complete, hence a Banach space (it’s the Lp space on the integers with the counting measure). If 0 < p < 1, then `p does not have a norm, but still has the metric

X p d(x, y) = |xk − yk| . k≥1

(To see that the natural attempt to deﬁne a norm fails to satisfy the triangle 1 inequality, let p = 2 and set 1 1 x = ( , 0, 0,... ), y = (0, , 0,... ). 2 2 Then 1 1 2 kx + yk = √ + √ = 2, 2 2 while 1 1 kxk + kyk = + = 1.) 2 2 If p = ∞, then `∞ is the space of all bounded sequences, which is a Banach space under the norm kxk∞ := sup |xk|. k Note that `p ⊂ `q ⊂ `∞ for 1 ≤ p < q < ∞ (since only ﬁnitely many elements of the sequence can have absolute value ≥ 1). Also, `2 is the only `p space that is a Hilbert space.

9 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 {An} of closed sets of X each of which has empty interior in X, their S union An 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 T {Un} of open sets in X, each of which is dense in X, their intersection Un 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 c c Proof. Since x ∈ B and B is open, there is an open ball Bδ(x) of radius δ > 0, c centered at x, whose closure is contained in B . Then the open sets Bδ(x) and X − Bδ(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 {An} be a countable collection of closed subsets of X, each with empty interior. We need to show that any non-empty open set U0 contains a point not in any of the An. Since A1 does not contain U0, choose y ∈ U0 not in A1. Since X is regular and A1 is closed, we can in fact choose a neighborhood U1 of y of diameter < 1 whose closure is contained in U0 and disjoint from A1. Now repeat by induction, where at the nth step we choose a point of Un−1 not in An, and a neighborhood Un of that point with diameter < 1/n and with closure contained in Un−1 and disjoint from An. T Then we claim that the intersection Un is non-empty, which will give the result. This is a result of the following lemma.

Lemma 3. Let C1 ⊇ C2 ⊇ · · · be a nested sequence of nonempty closed sets in T the complete metric space X. If diam Cn → 0, then Cn 6= ∅.

Proof. Choose xn ∈ Cn for each n. Then {xn} is Cauchy, and eventually lies inside each Ck, whence the limit x is in each Ck 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 Mx ≥ 0 satisfying sup kT xk ≤ Mx. T ∈F Then sup kT k < ∞. 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 An := x ∈ X : sup kT xk ≤ n . T ∈F

11 Then An is closed (since X is complete and each T ∈ F is continuous), contains 0, and is symmetric about the origin. Thus if some An contains a non-empty open set, then it contains a ball of radius δ > 0 centered at the origin, whence for any x ∈ X, 2kxk x δ 2kxk kT xk = T ≤ n, δ kxk 2 δ whence kT k ≤ 2n/δ, giving the result. So we may assume that each An has empty interior. S But X = An by the pointwise boundedness assumption, so by the Baire S category theorem, X = An 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.