Math 511 Notes
Mathematical Analysis II

Riemann Integral: A function f:[a,b]->R is Riemann integrable if the infinum of all upper sums equals the supremum of all lower sums. In particular, S(x) is a step function if S(x) is piecewise constant on intervals. For a given partition of the interval [a,b], for upper sums take S_u(x)>f(x) such that on each subinterval S_u(x) = max f(x), for all x's in the subinterval. Similarly for lower sums.

Notation: If E is any set, then the characteristic function of E is given by

Thus, a step function is of the form s(x) = S s_k c_interval(k)(x), where s_k is a value of the function on interval k.

Further, the integral of this step function is given by S s_k Dx_k, where Dx_k = length of interval k.

Question: To integrate a general function f(x), take a sequence of step functions on a finer and finer partition of [a,b] and define the integral of f(x) to be the limit of the integrals of the step functions. Is the function f integrable over [0,1] when f is given by:

1. f(x) = x; YES (Easy function)
2. f(x) = sin(x); YES (Transcendental)
3. f(x) = (x - 0.5) / |x - 0.5|; YES (Note, this is discontinuous.)
4. f(x) = c_{{1/n | n=1,2,3,...}}(x); YES (Prove this for homework)
5. f(x) = c_Q(x), where Q=rationals; NO (So, there are non-Riemann integrable functions.)

Problem: There exist Riemann integrable functions f_n(x) such that f_n->c_Q. Indeed, take the set

Note, as n gets larger, E_n gets closer to the rationals Q. Also, notice the sequence of functions

are all Riemann integrable but they approach c_Q(x) which isn't Riemann integrable. This problem arises because the Riemann integral is not "complete". Thus, the need for the Lebesgue integral.

Generalization: Instead of using interval partitions of [a,b] and the resulting step functions to define the integral (like Riemann), use a collection of disjoint sets E_k (not necessarily intervals) such that the union of the E_k = [a,b] and create a simple function

Then, the integral of this simple function is given by S s_k m(E_k), where the measure of the set E_k = m(E_k), generalizes length.

Defn: For given sets E and F:

1. complement E^c = {x | x is not an element of E }
2. set difference E - F = { x | x is an element of E and x is not an element of F }
3. Two sets are disjoint if their intersection is the empty set.
4. The general union or intersection of a collection of sets can be taken and DeMorgan's Laws hold.
5. The superior limit of {E_n}, denoted lim sup E_n, is the set consisting of those points which belong to infinitely many of the E_n.
6. The inferior limit of {E_n}, denoted lim inf E_n, is the set consisting of those points which belong to all but a finite number of the E_n.
7. If E^* = lim sup E_n = lim inf E_n, then we say the sequence {E_n} has a limit E*.

Defn: Ring, algebra, s-ring and s-algebra ... see defns 1.1.1 and 1.1.2.

Note: The definitions above require closure with respect to set union and set difference. It is easy to show that in a ring and s-ring, we also get closure with respect to set intersection by considering E^F = E - (E-F). Also, in an algebra and s-algebra, we get closure with respect to complements by considering X-E.

HOMEWORK: page 3, #3 (assuming s-algebra), 4

Borel's Conditions: Properties that a good measure should have.

• B0: m([a,b]) = length([a,b]) = b-a.
• B1: m(E) > 0, for any set E
• B2: m( U E_k) = S m(E_k), for mutually disjoint sets E_k.
• B3: m(F - E) = m(F) - m(E), where E is contained in F
• B4: m(E) > 0 implies E is uncountable

HOMEWORK: Show the following, assuming the Borel conditions:

• A. B2 itself implies B3
• B. E countable implies m(E) = 0
• C. E contained in F implies m(E) < m(F)

Defn 1.2.1: A measure is an extended real-valued set function m having the following properties:

1. The domain A of m is a s-algebra.
2. m is nonnegative on A
3. m is completely additive on A (see fomula 1.2.1, page 4)
4. m(empty set) = 0

m is said to be additive if its domain is a ring R and if

where EeR, FeR and E and F are disjoint. A general set function m is said to be finitely additive if

where all of the sets are mutually disjoint.

Result: A measure is finitely additive.

Pf: Since a measure is completely additive, it must be finitely additive by taking all the sets E_m to be empty, m > n.

Defn: If X is the entire space under consideration and m(X) < oo, then we say the set function m is a finite measure. If X can be written as the infinite union of sets E_n such that for all n, m(E_n) < oo, then the measure is a s-finite measure.

Theorem 1.2.1. Let m be a measure with domain A. Then:

(i) If EeA and FeA with E C F, then m(E) < m(F).

Pf: Write F = E U (F-E), which is a disjoint union. Since a measure is additive, then m(F) = m(E) + m(F-E). But a measure is also nonnegative and so m(F-E) > 0. Hence, the result follows.

(ii) If EeA and FeA with ECF and m(F)<oo , then m(F-E) = m(F) - m(E).

Pf: Use formula in (i).

(iii) If {E_n} is a monotone-increasing sequence of sets of A, then lim m(E_n) = m( lim E_n).

Pf: Problem 1.1.3 implies that lim E_n is in the domain A.

If E₀ = empty set, notice E_n = (E_n - E_n-1) U (E_n-1 - E_n-2) U ... U (E₂ - E₁) U (E₁ - E₀) = U (E_k - E_k-1). As n-->oo, this becomes an infinite union of mutually disjoint sets. So,

m(lim E_n) = m( U (E_k - E_k-1) ) = S _{k=1... m} m(E_k - E_k-1) , since a measure is completely additive

= lim S _{k=1... m} m(E_k - E_k-1), going to the definition of an infinite summation

= lim m( U (E_k - E_k-1) ), using finitely additive, noting the union is for k=1...n

= lim m( E_n ), noting the union collapses.

(iv) Result (iii) above also holds if the sequence is monotone-decreasing and the measure of one of the sets E_M is finite.

Pf: Notice that the sequence E_M - E_k is now monotone-increasing. Apply (iii) and (ii).

Theorem 1.2.2: Let m be a measure with domain A. Then, for the infinite collection {E_n}, n=1..., of sets of A,

Pf: Create the mutually disjoint, monotone sequence of sets F_n using

F₁ = E₁ and F_n = E_n - [E₁ U E₂ U ... U E_n-1 ].

Hence,

m( U E_n) = m( U F_n) = S m(F_n) < S m(E_n).

HOMEWORK: page 7, #1, 2, 3

Defns:

- m is said to be subadditive if its domain is a ring R and if

m(E U F) < m(E) + m(F),

where EeR, FeR and E ^ F=empty set.

- m is said to be finitely subadditive if

m(E₁ U E₂ U E₃ U ...E_n) < m(E₁) + m(E₂) + m(E₃) + ... + m(E_n),

where E_j ^ E_k=empty set, when j <> k.

- m is said to be countably subadditive if

m(E₁ U E₂ U E₃ U ...) < m(E₁) + m(E₂) + m(E₃) + ... + m(E_n)+ ... ,

where E_j ^ E_k =empty set, when j<>k.

- m is said to be monotone if m(E) < m(F), where ECF.

Defn 1.3.1 Outer measure - page 8

Defn 1.3.2 Given an outer measure m*, then the set E is m* measurable if

(m* is subadditive being an outer measure. So always m*(A) < m*(A ^ E) + m*(A-E). To show equality, it suffices to verify m*(A) > m*(A ^ E) + m*(A-E)).

Theorem 1.3.1: Let m* be an outer measure and denote by A the class of all m*-measurable sets. Then, A is a s-algebra and the restriction m of m* to A is a measure.

Pf: We will show first that A is an algebra, then that m satisfies the measure properties on A and then that A is a s-algebra.

(i) (This will also be used later when discussing completeness)

If m*(E) = 0, for any set A, m*(A ^ E) + m*(A - E) < m*(E) + m*(A) = m*(A), since m* is monotone and (A ^ E) C E and A - E C A

Hence, defn 1.3.2 is satisfied and so, E is measurable.

(ii) (Show empty set is in A.)

Since m*(empty set)=0 by definition, then (i) implies that empty seteA.

(iii) (Show complements are in A.)

Suppose EeA. Then, m*(A) = m*(A ^ E) + m*(A-E), by definition 1.3.2. But A ^ E = A - E^c and A-E = A ^ E^c. Hence, E^ceA.

(iv) (Show unions are in A.)

Let E₁ eA and E₂ eA. We will show E₁ U E₂ satisfies defn. 1.3.2.

From defn 1.3.2, m*(A) = m*(A ^ E₁) + m*(A-E₁) and m*(A-E₁) = m*((A-E₁) ^ E₂) + m*((A-E₁)-E₂).

Note, (A-E₁) - E₂ = A - (E₁ U E₂). Also, [ (A-E₁) ^ E₂ ] U [ A ^ E₁ ] = [(A-E₁) U (A ^ E₁) ] ^ [ E₂ U (A ^ E₁) ] = A ^ [ E₂ U E₁ ].

So, m*(A ^ (E₁ U E₂)) + m*(A-(E₁ U E₂)) = m*([ (A-E₁) ^ E₂ ] U [ A ^ E₁ ]) + m*((A-E₁) - E₂), and by using subadditivity,

< m*([ (A-E₁) ^ E₂ ] ) + m*( A ^ E₁ ) + m*((A-E₁) - E₂)

= m*([ (A-E₁) ^ E₂ ] ) + m*((A-E₁) - E₂) + m*( A ^ E₁ ), and since E₂eA

= m*(A-E₁) + m*( A ^ E₁ ), and finally since E₁eA

= m*(A)

Hence, we have m*(A) > m*(A ^ (E₁ U E₂)) + m*(A-(E₁ U E₂)), which is sufficient to show E₁ U E₂ is measurable.

(v) (Show differences are in the algebra.)

Let E₁ eA and E₂ eA. We will show E₁ -E₂ satisfies defn. 1.3.2.

Indeed, notice E₁ -E₂ = E₁ ^ E₂^c = (E₁^c U E₂)^c. However, complements and unions belong by using (iii) and (iv).

(vi) (Show m* is additive when applied to sets in A.)

Let {E_k}_(k=1..n) be a sequence of mutually disjoint sets in A. Let the union of these be denoted by S_n.

We must show that m*(A ^ S_n) = S_(k=1..n) m*(A ^ E_k). Use induction:

(Basic Step: n=1)

We must prove m*(A ^ S₁) = S_(k=1..1)m*(A ^ E_k) = m*(A ^ E₁), which is true since S₁=E₁.

(Induction step)

Assume m*(A ^ Sm) = S_(k=1..m)m*(A ^ E_k) is true for some m>1.

Then, show m*(A ^ S_m+1) = S_(k=1..m+1) m*(A ^ E_k).

However, by defn 1.3.2, m*(A ^ S_m+1) = m*((A ^ S_m+1) ^ S_m) + m*((A ^ S_m+1) - S_m).

Since, S_m+1 ^ S_m = S_m and (A ^ S_m+1) - S_m = A ^ E_m+1

= m*(A ^ S_m) + m*(A ^ E_m+1), and by using the induction hypothesis on the first term,

= S_(k=1..m) m*(A ^ E_k) + m*(A ^ E_m+1)

= S_(k=1..m+1) m*(A ^ E_k), as desired.

(vii) (Show m* is completely additive when applied to sets in A.)

Let {E_n} be an infinite sequence of mutually disjoint sets in A and let the union of these be denoted by S. Since m* is monotone and using (vi) on the smaller set S_nCS,

m*(A ^ S) > m*(A ^ S_n) = S_(k=1..oo) m*(A ^ E_k)

Letting n-->oo yields m*(A ^ S) > S_(k=1..oo) m*(A ^ E_k).

However, the countable subadditivity of m* gives the reverse inequality and hence equality must hold.

(viii) (Show A is a s-algebra)

Since A - S C A - S_n, then m*(A ^ S) < m*(A ^ S_n).

Hence, m*(A) = m*(A ^ S_n) + m*(A - S_n) > m*(A ^ S_n) + m*(A - S) = S_(k=1..n) m*(A ^ E_k) + m*(A - S).

Letting n-->oo and using (vii) yields

m*(A) > S_(k=1..oo) m*(A ^ E_k) + m*(A - S) = m*(A ^ S) + m*(A - S),

which suffices.

(ix) (Show m is a measure)

Easily, noting (viii) gives A is a s-algebra, the restriction m of m* to A satisfies all measure properties but complete additivity. To show this, simply use (vii) above with A=S.

HOMEWORK: page 10, #1, 2, 3, 6

Defn: The class of sets K is a sequential convering class if empty seteK and if for any set AeK, there is a sequence of sets E_neK such that A C U E_n.

Construction of Outer Measures: The previous work shows that if one can construct an outer measure, then restricting the domain of the outer measure to measurable sets yields a measure for that collection of sets. However, this assumes that one is able to start with some given outer measure. Such an outer measure can be constructed from any collection of sets by using the following:

Suppose we have any nonnegative set function lwith domain K such that l(empty set)=0. For each set A of X, create the set function

Theorem 1.4.1: (Show that the method of constructing an outer measure above indeed yields an outer measure.)

Pf:

(i) Obviously the domain consists of all subsets of X by the way m* is defined.

(ii) m* is nonnegative since l is nonnegative

(iv) m* is monotone since if ACB, then any covering of B also covers A. Hence, there could be a smaller infinum for A and thus a smaller value for m*(A).

(iii) We must show m* is countably additive:

Let A_n be any sequence of sets and take any e > 0.

Since K is a convering class, for each A_n, there is a sequence of covering sets E_nk such that

m*(A) + e/2n > S l(E_nk).

So, U A_n C U E_nk and thus by monotonicity, m*( U A_n) < m*( U E_nk).

Hence, m*( U A_n) < m*( U E_nk) < m*(A) + e/2n = m*(A) + e. Since e is arbitrary, we have m*( U A_n) < m*(A) as desired.

(v) Since m* is nonnegative, the the smallest it can be is zero. Since K is a convering class, empty seteK and so A=empty set can be convered most simply by itself with l(empty set)=0.

HOMEWORK: page 12, #2, 3

Defn: A measure m with domain A is said to be complete if N C EeA and m(E)=0 implies NeA.

Result: The measure constructed in Theorem 1.3.1 is complete.

Pf: By (i) of proof, any set with outer measure zero is measurable and thus in A.

Theorem 1.5.1: Any measure can be extended to be a complete measure.

Lebesgue Measure: Let X = R_n = { (x₁, x₂ x₃ ... x_n) where each component is a real number }.The collection K of open intervals forms a sequential convering class of X. Define the set function l by l(empty set)=0 and for each non-null interval I (see defn on page 13)

From Theorem 1.4.1, this set function yields an outer measure which we call Lebesgue outer measure. By Theorem 1.3.1, the restriction of this outer measure yields a measure (which is complete) called Lebesgue measure. The measurable sets are called Lebesgue measurable sets.

HOMEWORK: page 14, #1, 2, 3

Defn: A metric r is a function defined on a set of points X satisfying positive definiteness, symmetry and the triangle inequality. The set of points X together with the metric r is called a metric space. Given a metric r, the distance between two sets A and B is defined by r(A,B) = inf { r(x,y) } xeA and yeB }. If either set is a single point (say A={x}), we write this distance as r(x,B). For a given set A, its diameter is given by d(A) = sup { r(x,y) } xeA and yeA } and the set is bounded if d(A) is finite.

Defns: For any element x and e>0, an open ball is the set B(x,e) = { y | r(x,y) < e } with x called the center and e called the radius. A closed ball allows equality in above and is denoted `B(x,e). A sequence x_n is said to be convergent to y if r(x_n,y)-->0 as n-->oo. Open sets, closure, closed set, interior, Cauchy sequences, complete metric space....see page 17.

Defn: Denote by B the s-ring generated by the class of all open sets of X. The sets of B are called Borel Sets.

HOMEWORK: page 17, #2, 3, 4, 6