Real Analysis Oral Exam study notes, Exams of Calculus

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Real Analysis Oral Exam study notes

Notes transcribed by Mihai Nica

Families of sets

These are notes from Chapter 2 of [ 1 ].

1.1. Algebras and σ − algebras Fix a universal set X to work in.

Definition. (2.1) An algebra is a collection A of subsets of X that: i) ∅, X ∈ A ii) A ∈ A =⇒ Ac^ ∈ A iii) closed under nite unions and intersections A σ−algebra has: iv) closed under countable unions and intersections.

Lemma. (2.7.) Any arbitrary intersection of σ−algebras is a sigma algebra Definition. We dene the σ−algebra generated by a subset C to be the smallest σ − algebra cotaining C: i.e. the intersection of all σ − algebras containing C. This is denoted by σ(C).

Definition. The Borel σ − algebra is the σ − algebra that is generated by any of the following: 1) open intervals 2) closed intervals 3) half open intervals 4) semi-innite open intervals (a, ∞]

Proof. Its easy to check by doing some intersections unions that these all generate the same thing. 

1.2. Monotone Class Theorem Definition. (2.9.) A monotone class is a collection of subsets of X that is closed under increasing, and decreasing sets. I.e. An ↑ A =⇒ A ∈ M and An ↓ A =⇒ A ∈ M.

Theorem. (2.10) Suppose A 0 is an algebra and, A = σ(A 0 ) is the smallest σ − algebra containing A 0 and M is the smallest monotone class containing A 0. Then M = A. (Mihai's version) In other words: for an algebra A 0 , the smallest monotone class containing A 0 is in fact, a σ − algebra.

Proof. We must show A ⊂ M as the other inclusion is clear. To do this we have to show that the monotone class is closed under intersections and unions, for then M will be an algebra containgn A 0 and hence A = σ(A 0 ) ⊂ M. The idea of the proof is the usual trick with sigma algebras and so on: Dene the set you are interested in and show it contains what you want. Claim 1: Closed under complements 4

1.3. PI-LAMBDA THEOREM 5

Let N = {A ∈ M : Ac^ ∈ M}. Since A 0 is an algebra and closed under com- plements, A 0 ⊂ N. We claim now that N is a monotone class, for if An ↑ A with An ∈ N then by def'n Acn ∈ M and we have Acn ↓ Ac^ shows Ac^ ∈ M since M is a monotone class. Similar for An ↓ A. Hence N is a monotone calss containg A 0 and consequently M ⊂ N i.e. M = N. Claim 2: Closed under intersections with A 0 Let N = {A ∈ M : A ∩ B ∈ M ∀B ∈ A 0 }. Again N contains A 0 since A 0 is an algebra, and N is a monotne class since if An ↑ A then An ∩ B ↑ A ∩ B for every B ∈ A 0 , so the monotone-class-ness of M gives us A ∩ B ∈ M and this holds for every B ∈ A 0. (same idea for An ↓ A). As before, we get N = M. Claim 3: Closed under intersections Let N = {A ∈ M : A ∩ B ∈ M ∀B ∈ M}. This contain A 0 by the previous claim! It is a monotone class by the same argument as in claim 2. Hence, we have M = N. This shows that M is in fact a σ − algebra (nite unions/intersections + monotone limits yields countable unions/intersections by splitting up any countable union/intersection up into a monotne limit of nite unions/intersections) 

Remark. The sneaky thing in the proof is that you can't jump to closed under intersections right away, because you don't know that A 0 is in there a priori. The little extra step is to rst prove that A 0 is closed under intersections from M and then move on to the whole space.

1.3. Pi-Lambda Theorem I'm going to include a little thinger on the Pi-Lambda theorem, as this comes up occasionally in Probability theory.

Definition. A π−system is a collection of subsets which is closed under nite intersections.

Definition. A λ−system C is a collection of subsets which has: i) X ∈ C ii) A, B ∈ C and A ⊂ B =⇒ B − A ∈ C iii) An ∈ C , and An ↑ A =⇒ A ∈ C

Proposition. (Alternative denition of a λ−system) a) X ∈ C b) A ∈ C =⇒ Ac^ ∈ C c) If An ∈ C are disjoint then ˙

n=1An^ ∈ C Proof. Its easy to see that i), ii) and iii) imply a,b,c. We prove the converse statements one at a time assuming a), b) and c): ii): For any A, B with A ∩ B = ∅, we have that: A ∪ B ∈ C =⇒ Ac^ ∩ Bc^ ∈ C But Ac^ ∩Bc^ = Ac^ −B so this shows that whenver A∩B = ∅, we have Ac^ −B ∈ C. Noticing that A ∩ B = ∅ ⇐⇒ A ⊂ Bc^ and relabeling Bc^ = C gives the result ii) iii) Take set dierences Bn = An − An− 1 which are disjoint. 

Theorem. If C is both a π−system and a λ−system, then C is a σ−algebra

Measures

These are notes from Chapter 3 of [ 1 ]. Definition. (3.1) A measure on a set X and an σ − algebraA is a a function μ : A → [0, ∞] such that: i) μ(∅) = 0 ii) Countably additive for disjoint sets:

μ(

i= Ai) =

∑^ ∞

i=

μ(Ai) for disjoint A′ is

Proposition. (3.5.) (Basic Properties of Measures) i) A ⊂ B =⇒ μ(A) ≤ μ(B) ii) A = ∪∞ i=1Ai =⇒ μ(A) ≤

i=1 μ(Ai) iii) Ai ↑ A =⇒ limn→∞ μ(Ai) = μ(A) iv) If μ(A 1 ) < ∞, then ,Ai ↓ A =⇒ limn→∞ μ(Ai) = μ(A) Proof. These all follow by doing constructions involving complements and set dierences and so on in such a way as to reduce the sets in question as unions of disjoint sets, for which we know that measures play nice. Most of them use the Treat Disjointly Trick (See remark below) i) Holds since μ(A − B) ≥ 0 ii) Make Bi ⊂ Ai by taking set dierences so that ∪ni=1Bi = ∪ni=1Ai for each n and the Bi's are disjoint. Then μ(A) = μ(∪Bi) =

μ(Bi) ≤

μ(Ai) iii) μ(An) is a montone increasing function and is bounded above by μ(A), so it is convergent (to possibly ∞if μ(A) = ∞). Hence, by the monotone convergence theorem for real numbers, this sequence has a limit. To see that the limit is actually μ(A), take Bi = Ai+1 − Ai so the Bi's are disjoint and have ∪ni=1Bn = ∪ni=1Ai and apply the countable additivity to get the result. Explicitly:

μ(A) = μ(∪∞ n=1An) = μ(∪∞ n=1Bn)

=

∑^ ∞

n=

μ(Bn)

= lim n→∞

∑^ n

i=

μ(Bi)

= lim n→∞ μ (∪ni=1Bi) = lim n→∞ μ (∪ni=1Ai)

iv) Take complements of the result in iii) (or, more generally if you arent in a ntis measures spaceBk = A 1 − Ak). The hypothesis that μ(A 1 ) < ∞ is needed to

7

MEASURES 8

avoid a ∞ − ∞” problem. Counterexample to keep in mind: An = (n, ∞) which is decreasing to the empty set but always has innte Lebesgue measure. 

Remark. A common theme that comes up for this type of thing: The proof is easy for disjoint sets. The arbitary case can be reduced to the disjoint case by making sets Bi which are disjoint and have ∪ni=1Bi = ∪ni=1Ai for every n. I will label this trick as [Treat Disjointly]

Definition. (3.7.) Denition of nite and σ-nite measure spaces. A null set is a set which is a subset of a 0 measure measurable set. (i.e. null sets need not be measurable) A complete measurable space is one where all null sets are measurable. The completion of A is the smallest sigma algebra containing all the null sets.

3.4. OUTER MEASURES 10

Example. Lebesgue outer measure is what you get when you put in C =half open intervals and `(a, b] = b − a. The Lebesgue-Stieltjes outer measure is what you get if you again choose C =half open intervals and you have a non-decreasing and right continuous function α.

Definition. (4.5) A set A ⊂ X is measurable with respect to an outer measure μ∗^ (or more simply μ∗−measurable....I like the rst phrasing better be- cause it reminds me that outer measures are dened on all subsets of a space) if: μ∗(E) = μ∗(E ∩ A) + μ∗(E ∩ Ac) ∀E ⊂ X Remark. By the denition of an outer measure, it is ALWAYS TRUE for any set A that: μ∗(E) ≤ μ∗(E ∩ A) + μ∗(E ∩ Ac) ∀E ⊂ X So the property of being measurable is really about the other inequality. This is most often how one proves certain sets are measurable. One way to rememember the direction of the sign is to think that E ∩ A and E ∩ Ac^ together form a cover of E. Hence, since μ∗(E) is the inf over the measur of all covers, μ∗(E) ≤the sum. Of coruse, this is a mnemonic device only since stricly speaking the measure μ∗^ is dened only as the inf over covers formed by the special sets C.

Theorem. (4.6.) Let A = {measurble w.r.t. μ∗^ sets} .Then A is a σ −algebra. If we dene μ := μ∗|A, then μ is a measure on A. Finally, A contains all the sets of outer measure 0.

Remark. One of the main themes to keep in mind for this proof is that we are trying hard to get the inequalities of the form μ∗(E) ≥... , because this is the non-trivial inequality for being measurable. Where will these inequalities come from? The only way is to use the trivial side of the inequality in a clever way: you must manipulate μ∗(E) =... = μ∗(U ∩ V ) + μ∗(U ∩ V c) and then remark μ∗(U ∩ V ) + μ∗(U ∩ V c) ≥ μ∗(U ) since this always holds. (i.e. use property iii) from the denition of outer measure.)

Proof. Claim 1: A is closed under complements. Pf: Follows by the symmetry of the denition of measurable w.r.t. μ∗. Claim 2: A, B ∈ A =⇒ A ∪ B ∈ A Pf: For any E, use the measurablity property of A at the set E and use the measurability property of the set B once at the set E ∩A and once at the set E ∩Ac:

μ∗(E) = μ∗(E ∩ A) + μ∗(E ∩ Ac) = (μ∗(E ∩ A ∩ B) + μ∗(E ∩ A ∩ Bc) + (μ∗(E ∩ Ac^ ∩ B) + μ∗(E ∩ Ac^ ∩ Bc))) Now notice that the rst three terms are E ∩ A ∩ B, E ∩ Ac^ ∩ B, E ∩ A ∩ Bc. Since A ∪ B = (A ∩ B) ∪ (A ∩ Bc) ∪ (B ∩ Ac), so consequently the rst three sets here cover E ∩ (A ∪ B). Hence the sum of their measures is ≥ μ∗(E ∩ (A ∪ B)) and we get (alspo use Ac^ ∩ Bc^ = (A ∪ B)cby DeMorgan):

μ∗(E) ≥ μ∗(E ∩ (A ∪ B)) + μ∗(E ∩ (A ∪ B)c) So indeed, A ∪ B satisfy the non-trivial direction of the measurable inequality, and are hence measurable.

3.5. LEBESGUE-STIELTJES MEASURE 11

Claim 3: An ∈ A =⇒ ∪∞ n=1An ∈ A Pf: We rstly remark that is suces to check this for disjoint sets An, because by using the property of Claim 2, we can rewrite any countable union as a countable union of disjoint sets modulo some nite union operations. For disjoint sets An let Bn = ∪ni=1Ai and B = limn→∞ Bn. For E ⊂ X have: μ∗(E ∩ Bn) = μ∗(E ∩ Bn ∩ An) + μ∗(E ∩ Bn ∩ Acn) = μ∗(E ∩ An) + μ∗(E ∩ Bn− 1 ) Moving the μ∗(E ∩ Bn− 1 ) to the other side, and then using this as a telescoping sum, we get that:

μ∗(E ∩ Bn) − μ∗(E ∩ B 1 ) =

∑^ n

i=

μ∗(E ∩ Ai)

=⇒ μ(E ∩ Bn) ≥

∑^ n

i=

μ∗(E ∩ Ai)

Hence:

μ∗(E) = μ∗(E ∩ Bn) + μ∗(E ∩ Bnc) ≥

∑^ n

i=

μ∗(E ∩ Ai) + μ∗(E ∩ Bc)

Taking n → ∞, (ok since the sum is monotone increasing) we get:

μ∗(E) ≥

∑^ ∞

i=

μ∗(E ∩ Ai) + μ∗(E ∩ Bc)

≥ μ∗^ (∪∞ i=1E ∩ Ai) + μ∗(E ∩ Bc) = μ∗(E ∩ B) + μ∗(E ∩ Bc) ≥ μ∗(E) So indeed B is μ∗-measurable! These claims together show that A is a sigma algebra. Claim 4: μ := μ∗|A is a measure. Pf: We have only to show that it is countable additive. Following the discussion above, we had that μ∗(E) =

i=1 μ

∗(E ∩ Ai) + μ∗(E ∩ Bc) for any set E. Taking

E = B gives exactly the result for countable additivity. Claim 5: If μ∗(A) = 0 then A is μ∗measurable. Pf: This follows from the monotone property of the outer measure, for if μ∗(A) = 0 then:

μ∗(E ∩ A) + μ∗(E ∩ Ac) ≤ μ∗(A) + μ∗(E ∩ Ac) ≤ μ∗(A) + μ∗(E) = μ∗(E) Shows the non-trivial direction of the measurability inequality. 

3.5. Lebesgue-Stieltjes Measure Let α(x) be an increasing right continuous function, let C = {(a, b] : a, < b} and dene `((a, b]) = α(b) − α(a). Then dene the outer measure:

m∗(E) = inf

`(Ai) : Ai cover E

3.6. EXAMPLES AND RELATED RESULTS 13

Proof. By looking at a subset of the intervals, we may suppose WOLOG that the intervals are in order so to speak so that:

a 1 ≤ C ≤ b 1 and , an ≤ D ≤ bn and ak < bk− 1 < bk Then just write out both sums and see by comparison to a telescoping sum that the inequality holds. 

Proposition. (4.9) If e and f are nite and I = (e, f ] then m∗(I) = `(I)

Proof. (This is the one that uses compactness of closed intervals. The idea is to convert any cover by semi-open intervals to a cover using open intervals (this gives an  of error by the continuity of α) and then use compactness to get down to nitely many intervals of interest. The apply the previous lemma.) Clearly m∗(I) ≤ `(I) by the inf denition of m∗^ since I is a cover for itself. For the other inequalty, suppose that I ⊂ ∪Ai where each Ai = (ci, di] is a an interval. By the right continuity of the function α, choose C ∈ (e, f ) so that α(C) < α(e) + / 2. Let D = f. Choose d

′ i > d^ so that^ α(d

′ i)^ < α(di) +^ /^2

i+1 (^) and

let Bi = (ci, d′ i) (Each (ci, d′ i) aproximates the semi open interval (ci, di] and the whole thing is done so that the total error from the point of view of the α function is no more than .) Now use the compactness of [C, D] to nd a nite set of the Bi's that cover all of [C, D] still. By the previous lemma we get that:

`(I) ≤ α(D) − α(C) + / 2 ≤

`(Ai) + / 2 i+1^ + / 2

and since  is arbitary the result follows. 

Now that all the major points have been hit, we will drop the ∗ and refer to m as the Lebesgue-Stieltjes measure corresponding to α.

3.6. Examples and related results 3.6.1. Approximating Sets. Example. (4.10) For the Lebesgue measure its not hard to show that singletons are measure 0 and so the open, half-open, closed, half-closed intervals all have the same measure. All countable sets are Lebesgue measure 0 too.

Example. (4.11) The middle thirds Cantor set is uncontable but still has Lebesgye measure 0. The Cantor ternary function is also breiy described in this example.

Example. (4.13) By changing the fraction of removed set as you go, you can end up with a positive measure generalized Cantor set. This set is closed and contains no intervals, and every point is a limit point.

Proposition. (4.14) Let A ⊂ [0, 1] be a Borel measurable set and let m be the Lebesgue measure. Then: i) Given any  > 0 there is an open set G so that A ⊂ G and m(G − A) <  ii) Given any  > 0 there is a closed set F so that F ⊂ A and m(A − F ) <  iii) There is a Gδ set H so that A ⊂ H and m(H − A) = 0 iv) There is an Fσ set F so that F ⊂ A and m(F − A) = 0

3.7. NONMEASURABLE SETS 14

Remark. We used half open sets in our construction of the Riemann-Stietijes measure, so we need to adapt from our half open intervals to the open intervals that make up open sets. If one had used open intervals to construct the measure (which is totally legit btw) then this proposition would be straight from the inf denition of m∗; but some early results (mostly the proof that m∗^ = ` for intervals would be slightly harder)

Proof. i) Approximate A be a union of semi-open intervals so that the error is no more than / 2 (this holds by the inf denition of m∗) Then approximate each semi-open interval by an open interval so that th error is no more than / 2 n+1. ii) Take complements and use the result in i) iii) Choose  = (^) n^1 and apply the result in i) to get a sequence of open set Gn. WOLOG they are decreasing. Take their intersection to get the desired Gδ set. iv) Same procedure as in iii) applied to closed sets Fn generated from  = (^1) n in part ii) 

3.7. Nonmeasurable Sets Theorem. (4.15) [Existence of a non-measurable set] Let m∗be the outer mea- sure dened in the usual way with the collection C = {(a, b]} and `((a, b]) = b − a. The m∗^ is not a measure on the collection of all subsets of R.

Remark. Since we showed that m∗IS a measure when restricted to the m∗measurable sets, this is saying that there are some sets that are not m∗−measurable, and that these sets are badly behaived enough that they prohibit m∗from being a measure here.

Proof. (Here we have the construction using shifting by rationals. There is also the x ∼ y if x − y = kα mod 1 proof where α is irrational that works in a someqhat similar way.) Suppose m∗^ is a measure. Dene x ∼ y if x − y is a rational. This is an equivalence relation. Use the axiom of choice to get a representative from each equivalence class from the points in [0, 1]. Let A be the set of these representatives. Clearly: [0, 1] ⊂ ∪q∈[− 1 ,1]∩QA + q because every point in [0, 1] belongs to some equivalence class, and is hence a rational shift away from some point in A by denition. Moreover, each set A + q 1 is disjoint from A + q 2 for q 1 6 = q 2 for if there was a point w in there intersection then w − q 1 ∈ A and w − q 2 ∈ A are both in the same equivalence class and this contradicts the choice of A. Hence:

q

m∗(A + q)

q

m∗(A)

(m∗(A) = m∗(A+q) comes from the fact that (I) =(I +q) and the denition of m∗). On the other hand, ∪q∈[− 1 ,1]∩QA + q ⊂ [− 1 , 2] since A ⊂ [0, 1]. So we get:

1 ≤

q

m∗(A) ≤ 3

3.8. THE CARATHEODORY EXTENSION THEOREM 16

(4) The fact that the measure exists follows from our previous work: We know that A = {μ∗^ − measurable sets}is a σ−algebra, it contains A 0 by (3), and we showed earlier that μ = μ∗|A is a measure. Uniqueness is most easily done with the π − λ theorem: If μ, ν are two measures so that μ(A) = ν(A) = `(A) for all A ∈ A 0 , let P = {A : μ(A) = ν(A)}. This contains the algebra A 0 by construction, and we verify that it is a λ−system. Hence, by the π − λ theorem, it is in fact a σ-algebra! Since it contains A 0 , it must in fact be all of σ(A 0 ).

Measurable Functions

These are notes from Chapter 5 of [ 1 ].

4.9. Measurability Fix a measurable space (X, A)

Definition. (5.1.) A function f : X → R is measurable or A-measurable if f −^1 ((a, ∞)) ∈ A for all a ∈ R

Proposition. (5.5.)The following are equivalent (i) f −^1 ((a, ∞)) ∈ A for all a ∈ R ii) f −^1 ([a, ∞)) ∈ A for all a ∈ R iii) f −^1 ((∞, a)) ∈ A for all a ∈ R iv) f −^1 ((∞, a]) ∈ A for all a ∈ R

Proof. Use compleemnts and intersections/unions with a fudge factor of (^) n^1 

Proposition. (5.6.) If X is a metric space and A contains all the open sets, then continuous functions are always measurable

Proof. The preimage through any continuous function of any open set is open, so f −^1 ((a, ∞)) ∈ Open Sets ⊂ A 

Proposition. (5.7.) If f, g are measurable then so are f +g, −f, cf, f g, max(f, g) and min(f + g)

Proof. i) {f + g < a} = ∪r∈Q {f < r} ∪ {g < a − r} ii) {−f > a} = {f < −a} iii) Assume WOLOG by ii that c > 0 and then:{cf > a} = {f > a/c} iv) f g = (^12)

(f + g)^2 − f 2 − g^2

v) {max(f, g) < a} = {f < a} ∩ {g < a} vi) {min(f, g) > a} = {f > a} ∩ {g > a} 

Proposition. (5.8.) If fn are all measurable then so are sup fn, inf fn, lim sup fn, lim inf fn

Proof. {sup fn > a} = ∩n {fn > a} and {lim sup fn > a} = ∪n∩k>n{fk > a} = lim sup {fn > a}. Similar for inf 

Proposition. (5.10) If f is monotone, then f is Borel measurable.

Proof. Suppose WOLOG that f is increasing. Look at x 0 = sup {y : f (y) ≤ a} since f is increasing, for x > x 0 we have f (x) ≥ a by uisng the deniton of inf and the fact that f is increasing. Hence (x 0 , ∞) ⊂ {f > a}. Similarly, (−∞, x 0 ) ⊂ {f ≤ a}. So then {f > a} = [x 0 , ∞) or (x 0 , ∞) and in either case it is measur- able. 

17

4.11. LUSIN'S THEOREM 19

and since each gk is continuous, it converges to a continous function g. This function is equal to

k=1 hk^ on all of^ F^. But by denition,^

k=1 hk^ = limn→∞^ fn^ =^ f^ so this is exactly what we want. Finally, for arbitarty functions, split up f into f +^ and f −^ and do the dance for both pieces. 

Example. (5.16) A good exmaple to see what this theorem is and is not saying is to look at f = χ[0,1]−Q. Since the irrationals are dense in [0, 1], the function f takes values of both 1 and 0 in any interval, so you might be tempted to thing that its impossible for a continuous function to be equal to this on a set of positive measure. The resolution is that, since m(Q) = 0, we can nd an closed set F that contains no rationals with m(F ) > 1 − (this can be done explicitly in this case: put G = ∪q∈Q∩[0,1]

q − 2 n , q + 2 n

and F = [0, 1] − G contains no rational numbers.) Put g ≡ 1 everywhere on [0, 1]. Now since F contains no rationals, g and f agree on F (they are identically 1 there) and m([0, 1] − F ) < . So in this case, Lusin's theorem gives us a very useless approximation ot χ[0,1]−Q, namely the constant function 1. This example nicely illustrates that the type of approximation there exists a continuous function that agree except for a small measure set is actually not so useful.

The Lebesgue Integral

These are notes from Chapter 6 of [ 1 ]. Definition. (6.1.) Let X, A, μ be a measure space. If s =

∑n i=1 aiχEi^ is a simple function, we dene the integral of a simple function to be: ˆ sdμ =

∑^ n

i=

aiμ(Ei)

We dene the integral of a non-negative measurable function to be: ˆ f dμ = sup

sdμ : s ≤ f and s is simple

For arbitary functions we dene

f dμ =

f +dμ −

f −dμ Definition. (6.2.) We say f is integrable when

|f | dμ < ∞ Proposition. (6.3.) Some basic conseqeunces of the denitions i) If a ≤ f (x) ≤ b for all x ∈ X then aμ(X) ≤

f dμ ≤ bμ(X) ii) If f ≤ g then

f dμ ≤

gdμ iii) If f is integrable, then

cf dμ = c

f dμ for all c iv) If μ(A) = 0 and f is integrable then

f χAdμ = 0 Proof. These are pretty straightforward, most of them are proven by rst proving it for simple functions, and then using the denition of the integral as the sup over simple functions to get the result (sometimes the trick you have to do is to do sup

sdμ : s ≤ f and s is simple

= sup

sdμ : s ≤ f and s is simple and has a certain property

where the certain property helps you get the result. (For example in part i) the property is s ≥ a) 

Proposition. (6.4.) If f is integrable, then: ∣ ∣∣ ∣

f

|f |

Proof. For the real case its easy using f ≤ |f | and −f ≤ |f | 

20