Lecture 2. Measure and Integration

There are several ways of presenting the definition of integration with respect toa measure. We will follow, more or less, the approach of Rudin’s ‘Real and ComplexAnalysis’ - this is probably the fastest route.

Conventions concerning ±∞. In measure theory it is often essential to use ∞ and−∞. We shall view these as two objects and define the following relationship with realnumbers (the set of all real numbers will be denoted by R) :

−∞ < c < ∞ for every c ∈ R

c +∞ = ∞ if −∞ < c ≤ ∞, and c + (−∞) = −∞ if −∞ ≤ c < ∞

0 · ∞ = 0

c(±∞) =

{

±∞, if 0 < c ≤ ∞;∓∞, if −∞ ≤ c < 0;

Note that ∞−∞ is not defined! The convention 0·∞ = 0 might seem like an arbitrarychoice, but it is the really the right one : it will be useful in integration and there it willappear as the natural choice.

These operations have the usual properties on [0,∞] : commutative, associative anddistributive.

Throughout our discussion (X,B, µ) will be a measure space. This means that X is aset, B is a collection of subsets of X forming a σ−algebra, and µ is a measure.

Recall that to say that B is a σ−algebra means that :(S1) ∅ ∈ B;(S2) if A ∈ B then the complement Ac ∈ B;(S3) if A1, A2, A3, ... is a countable collection of sets with each Ai ∈ B then ∪∞i=1Ai ∈ B.

Sets in B will be called B−measurable sets, or simply measurable sets.

Observations. (i) The intersection of a countable collection A1, A2, ... of measurablesets is also measurable (because ∩∞i=1Ai = (∪iA

ci )

c is measurable by (S2) and (S3).)(ii) If A1, ..., Am ∈ B then ∪m

i=1Ai ∈ B; this follows from (S1) and (S3) by taking An = ∅for all n bigger. Using the complements we obtain also that ∩m

i=1Ai ∈ B.(iii) Notice also that if A, B ∈ B then A \B, being equal to A ∩ Bc, is also in B.

To say that µ is a measure on (X,B) means that µ is a mapping B → [0,∞] suchthat :

(M1) µ(∅) = 0(M2) if A1, A2, ... is a countable collection of mutually disjoint sets in B (i.e. each Ai ∈ B

and Ai ∩ Aj = ∅ for every i 6= j) then:

µ(∪∞i=1Ai) =

∞∑

i=1

µ(Ai).

Taking Ai = ∅ for i bigger than some m, we see (from (M1) and (M2)) that µ isfinitely-additive :

1

µ(∪mi=1Ai) =

m∑

i=1

µ(Ai)

provided, of course, that the Ai are disjoint measurable sets.(Comment : It might seem that finite additivity implies the condition (M1): we have

µ(∅) = µ(∅ ∪ ∅) = µ(∅) + µ(∅), “and so” µ(∅) = 0. What is wrong with this argument?Under what additional condition on µ would finite additivity imply (M1) ?)

Exercise. If A and B are measurable sets with A ⊂ B, show that µ(A) ≤ µ(B). (Hint: write B as the union of A and B \A; check that B \A (which is B ∩ Ac) is measurableand use additivity.)

The simplest meaningful example of a measure is counting measure:

µ(A) = number of elements in A (taken to be ∞ if A is infinite)

2.1. Definition. A function f : X → [−∞,∞] is measurable with respect to theσ−algebra B if the set f−1[a, b] (i.e. the set {x ∈ X : f(x) ∈ [a, b]} ) is in B, for everya, b ∈ [−∞,∞].

Note that, in particular, if f is measurable then (taking b = a) every set of the typef−1(a) is measurable (a ∈ [−∞,∞]).

2.2. Terminology. (i) If A ⊂ X, then the indicator function of A is the function1A : X → R, defined by :

1A(x) =

{

1, if x ∈ A;0, otherwise.

(ii) A simple function is a function s : X → R whose range consists of a finite setof points (i.e. the image s(X) is a finite subset of R).

If a1, ..., an are all the distinct values of the simple function s then (verify that)

s =

n∑

i=1

ai1Ai

where Ai = {x : s(x) = ai}. Notice (verify) that the Ai are disjoint sets and ∪ni=1Ai =

X.Notice also that a simple function can be written in different ways in the form

∑

cj1Cj. For instance, 1X is also equal to 1A + 1Ac , for any A ⊂ X. The represen-

tation described above in terms of the distinct values of s is, of course, unique; which is

why we use it in the definition (2.4) of

∫

s dµ below.

2.3. Exercise. Verify that s is measurable if and only if each Ai is.

2.4. Definition. (Integrating simple functions) If s : X → [0,∞) is a measurablesimple function of the form

s =n

∑

i=1

ai1Ai

2

where a1, ..., an are all the distinct values of s (and Ai = s−1(ai)), then

∫

s dµ =

n∑

i=1

aiµ(Ai).

Note that a term of the form 0 · ∞ might occur in the sum on the right side of theabove equation; recall, for this purpose, that 0 · ∞ = 0.

2.5. Question. What is the relevance of Problem 2.3 to Definition 2.4 ?

2.6. Note. If s =

m∑

j=1

cj1Cjis a non-negative measurable simple function then a fact

that is not immediately obvious from Definition 2.4 is that

∫

s dµ =m

∑

j=1

cjµ(Cj) (provided

this sum makes sense). Definition 2.4 says that this is true only when the ci are all thedistinct values of s.

2.7. Proposition. If s and t are nonnegative measurable simple functions then so is

s + t and∫

(s + t) dµ =

∫

s dµ +

∫

t dµ.

If c ∈ [0,∞) is a constant then :

∫

cs dµ = c

∫

s dµ

Proof. Write s =n

∑

i=1

ai1Aiand t =

m∑

j=1

bj1Bj, where a1, ..., an are all the distinct

values of s and b1, ..., bm are all the distinct values of t. The function s + t takes thevalue ai + bj on the set Ai ∩ Bj (because on this set s takes the value ai and t takesthe value bj). Moreover, (verify that) the sets Ai ∩Bj are disjoint and cover all of X (i.e.(Ai∩Bj)∩(Ai′∩Bj′) = ∅ unless i = i′ and j = j′; and ∪i,j(Ai∩Bj) = X.) This is becausethe Ai’s are disjoint and cover X and the Bj ’s are disjoint and cover X. Therefore,

s + t =∑

1≤i≤n,1≤j≤m

(ai + bj)1Ai∩Bj.

(To see this, consider any point x ∈ X. It must belong to exactly one of the setsAi ∩ Bj ; say x ∈ A1 ∩ B2. Then the function defined by the sum on the right hand sideof the above equation takes the value a1 + b2 at x. But this also the value of s(x) + t(x).)By Problem 2.3 we see that s + t is measurable. It is also clear that s + t is simple andnon-negative.

We would like to conclude, by Definition 2.4, that it follows that :

∫

(s + t) dµ =∑

1≤i≤n,1≤j≤m

(ai + bj)µ(Ai ∩ Bj) (2.1)

3

However, there is no guarantee that the values ai + bj are distinct and thereforeDefinition 2.4 cannot be applied directly. Fortunately, equation (2.1) is still true becausethe sets Ai ∩ Bj are disjoint (this is the content of Lemma 2.8 below). So we acceptequation (2.1). The right side of equation (2.1) can be split up into the sum of two sums :

∑

1≤i≤n,1≤j≤m

(ai + bj)µ(Ai ∩ Bj) =n

∑

i=1

ai

(

m∑

j=1

µ(Ai ∩ Bj))

+m

∑

j=1

bj

(

n∑

i=1

µ(Ai ∩Bj))

.

Now the sets Ai ∩Bj are disjoint and, for fixed i and variable j, they cover Ai ( for

∪j(Ai ∩ Bj) = Ai ∩(

∪jBj

)

= Ai ∩X = Ai.)

Therefore :m

∑

j=1

µ(Ai ∩ Bj) = µ(Ai).

Similarly,n

∑

i=1

µ(Ai ∩ Bj) = µ(Bj).

Putting all this together we obtain :

∫

(s + t) dµ =

n∑

i=1

aiµ(Ai) +

m∑

j=1

bjµ(Bj)

and we recognize the the two sums on the right hand side to be

∫

s dµ and

∫

t dµ.

Thus :

∫

(s + t) dµ =

∫

s dµ +

∫

t dµ.

That

∫

cs dµ = c

∫

s dµ, for c any constant in [0,∞), is verified easily by writing out

s as a linear combination of indicator functions and using Definition 2.4.

2.8. Lemma. If C1, ..., Cn are disjoint measurable sets and c1, ..., cn non-negative real

numbers then∫

(

n∑

i=1

ci1Ci) dµ =

n∑

i=1

ciµ(Ci).

Proof. We shall assume that ∪ni=1Ci = X; for if this is not so then we can always

throw in another set Cn+1 = X \n⋃

i=1

Ci, and use cn+1 = 0, and this would not alter the

4

value of either side of the above equation (note : here we need to use 0 · ∞ = 0 in case

µ(Cn+1) = ∞.) Let s =

n∑

i=1

ci1Ci, and let a1, ...am be all the distinct values of s. Note that

s is a non-negative measurable simple function. Moreover, (verify that) c1, ..., cn constituteall the values of s (i.e. {c1, ..., cn} = {a1, ..., am}). The trouble is that the ci may not allbe distinct.

For each i ∈ {1, ..., m}, set

Ji = {all j for which cj = ai}

Then ∪j∈JiCj = Ai, where Ai = s−1(ai) is the set of all points where s takes the

value ai.Then (verify that) the Ji are disjoint and their union is {1, ..., n}. Therefore,

n∑

i=1

ciµ(Ci) =

m∑

i=1

∑

j∈Ji

cjµ(Cj)

=

m∑

i=1

ai

(

∑

j∈Ji

µ(Cj))

=

m∑

i=1

aiµ(∪j∈JiCj)

=

m∑

i=1

aiµ(Ai)

=

∫

s dµ.

This completes the proof.

2.9. Comment. Suppose C1, ..., Cm are measurable sets and c1, ..., cm non-negativereal numbers. Then Proposition 2.7 implies that

∫ m∑

j=1

cj1Cjdµ =

m∑

j=1

∫

cj1Cjdµ =

m∑

j=1

cjµ(Cj).

(Check that Definition 2.4 does imply what we have used in the last equality above :∫

cj1Cjdµ = cjµ(Cj).) Compare this with Note 2.6.

We are now familiar with integrals

∫

s dµ where s is simple. The first order of business

now is to extend this to define the integral

∫

f dµ for a general measurable function f . First

5

a note of caution. We can’t expect

∫

f dµ to be meaningful for all measurable functions

f ; for if f is a simple function of the form 1A− 1B where both A and B are sets of infinite

measure then

∫

f dµ has the nonsense value ∞−∞. In general we should expect that∫

f dµ should be defined as

∫

f+ dµ−

∫

f− dµ (here f+ = f1f≥0 is the function f where

f is non-negative and zero elsewhere, while f− = −f1f≤0) and the integral

∫

f dµ being

meaningful only when we don’t have an ∞−∞ situation. Thus if we can define

∫

f+ dµ

and

∫

f− dµ we should be done, i.e we should focus on defining

∫

g dµ for non-negative

measurable functions g. Since we know how to integrate simple functions, the first idea is

to pick a sequence of simple functions sn converging to g and then defining

∫

g dµ to be

the limit of

∫

sn dµ. Of course there are a lot of things to check. As it turns out, it is a

faster route if we define

∫

g dµ to be the supremum of

∫

s dµ for all non-negative simple

s ≤ g, and this is what we do:

2.10. Definition. We know what

∫

s dµ is when s is a simple function. For a more

general function f , it is reasonable to figure that

∫

f dµ can be obtained by taking the

supremum of all

∫

s dµ for simple functions s ≤ f . Polishing this up, we define for any

measurable function f : X → [0,∞],

∫

f dµ = sup

∫

s dµ

where the supremum is over all measurable simple functions s satisfying 0 ≤ s ≤ f . (Suchsimple functions exist; s = 0 is one such function. Moreover, if f is itself a simple function

then taking s = f we see that

∫

f dµ equals

∫

s dµ - in the sense of Definition 2.4 - and

so there is no conflict between this new definition for

∫

f dµ and the one in Definition 2.4

in the case f is simple.)2.11. Basic observations about integration. Some simple facts that follow immediately

from Definition 2.10 are :

(a) if 0 ≤ f ≤ g then

∫

f dµ ≤

∫

g dµ;

(b) if f ≥ 0 and c is a constant with 0 ≤ c < ∞, then

∫

(cf) dµ = c

∫

f dµ;

6

(c) if f = 0 except on a set of measure 0 then

∫

f dµ = 0 (verify this as follows: let E be

the set of all points of X where f 6= 0; then check that any s with 0 ≤ s ≤ f satisfies

s = s1E - write out s as a linear combination of indicator functions and show

∫

s dµ

is 0 by using the fact that µ(E) is 0.)

Lecture 3.

As usual (X,B, µ) is a measure space.

3.1. Notation. If E ⊂ X, we write

∫

E

f dµ for

∫

f1E dµ. In particular,

∫

X

f dµ =∫

f dµ.

3.2. Exercise. (a) If (1) A and B are measurable sets, (2) A ⊂ B, and if (3) µ(A) < ∞,then

µ(B \A) = µ(B)− µ(A).

(b)Show that if E is a measurable subset of X, and f and g are measurable functions

with 0 ≤ f ≤ g on E, then

∫

E

f dµ ≤

∫

E

g dµ.

Now assume (1) and (2) but instead of (3) assume that µ(A) = ∞. Show that thenµ(B) = ∞. (Consult the Problem preceding Definition 2.1.)

(Hint : Write B as the union of the disjoint measurable sets A and B \ A; thenµ(B) = µ(A) + µ(B \ A). Proceed from here. Do you see why the hypothesis that A hasfinite measure is important (for the first part above)?)

3.3. Proposition. If A1, A2, ... is a countable sequence of measurable subsets of X such

that A1 ⊂ A2 ⊂ A3 ⊂ · · ·, then

µ(∪∞i=1Ai) = limn→∞

µ(An)

If B1, B2, ... is a countable sequence of measurable subsets of X such that B1 ⊃ B2 ⊃B3 ⊃ ..., and if, moreover, some µ(Bj) < ∞, then

µ(∩∞n=1Bn) = limn→∞

µ(Bn)

Proof. Since we know how to calculate the measure of the (countable) union ofdisjoint sets we manufacture a sequence of disjoint sets C1, C2, ..., whose union is still∪∞i=1Ai, according to the following prescription : set Cj = Aj \ Aj−1 for j > 1, andset C1 = A1. Then C1, C2, ... are measurable and mutually disjoint (verify). Note thatµ(Ci) = µ(Ai)−µ(Ai−1), if Ai−1 has finite measure. Moreover, (verify) ∪∞j=1Cj = ∪∞j=1Aj.Therefore,

7

µ(∪∞i=1Ai) = µ(∪∞i=1Ci)

=

∞∑

i=1

µ(Ci)

= limn→∞

(

µ(C1) + µ(C2) + · · ·+ µ(Cn))

= limn→∞

(

µ(A1) + {µ(A2)− µ(A1)}+ · · ·+ {µ(An)− µ(An−1)})

= limn→∞

µ(An).

This would conclude the proof except for the fact the fourth ‘=’ above is valid under theadditional assumption that each µ(Ai) is finite. Thus the above chain of equalities provethe result if we also assume that each Ai has finite measure. If this assumption is not validthen some Ai has infinite measure. In that case (by Exercise 3.2 above) all the Aj , withj bigger than i, also have infinite measure. Therefore lim

n→∞µ(An) would equal ∞. On the

other hand, the set ∪∞n=1An must also have infinite measure since it contains the set Ai as asubset. Thus, even if some Ai has infinite measure, the equality µ(∪∞n=1An) = lim

n→∞µ(An)

still holds.The second part is left as an exercise.

Do Problem 2 in Problem Set 1 before attempting (3) of the following exercise.3.4. Exercise (Some useful facts concerning inverse images) (1) Let f : X → Y be a

mapping. Show that f−1 preserves all set-theoretic operations. That is, (a) f−1(∅) = ∅;(b) if A is any set of subsets of Y then f−1(∪E∈AE) = ∪E∈Af−1(E); (c) for any subsetsE, and H of Y , f−1(H \E) = f−1(H)\f−1(E); (d) prove the analog of (b) with ∪ replacedby ∩.

(2) Verify that every interval [a, b] ⊂ [−∞,∞] can be obtained by using countableunions, intersections, and complements of sets of the form (x,∞] ⊂ [−∞,∞]. (Hint : if

−∞ < a < ∞ then [a,∞] = ∩∞n=1(a−1

n,∞]; on the other hand [−∞, b] = (b,∞]c. Note

that [a, b] = [a,∞] ∩ [−∞, b]. What about the case when a = ∞ or a = −∞?)(3) Show that the σ−algebra of subsets of R generated by the intervals of the type (a, b)

coincide with the one generated by those of the type [a, b] (and also with the σ−algebragenerated by intervals of the type (a,∞)). This σ−algebra is called the Borel σ−algebraof R. (Hint : Show that every interval of each type can be obtained by countable unions,complements, and countable intersections of intervals of any other type.)

(4) Show that the σ−algebra of subsets of R generated by the intervals coincides withthe σ−algebra generated by the open sets. (Hint : You will need the important fact thatevery open subset of R is the union of a countable collection of open intervals.)

(5) If f : X → Y and g : Y → Z show that (g ◦ f)−1 = f−1 ◦ g−1 in the sense that forany subset E ⊂ Z, (g ◦ f)−1(Z) = f−1

(

g−1(E))

.

3.5. Proposition (Manufacturing measurable functions I).

8

(1) A function f : X → [−∞,∞] is measurable if and only if f−1(a,∞] is a measurableset for every a ∈ [−∞,∞] (one may replace (a,∞] with any of the sets [a,∞], [−∞, a),[−∞, a]). In case f : X → R, it is measurable if and only if f−1(a, b) is measurablefor every a, b ∈ R.

(2) A function f : X → R is measurable if and only if f−1(U) is measurable for everyopen set U ⊂ R.

(3) If f : X → R is measurable and g : R → R is any continuous function then g ◦ f :X → R is also measurable.

(4) If f, g : X → R are measurable functions and φ : R2 → R is continuous, then thefunction φ(f, g) : X → R : x 7→ φ

(

f(x), g(x))

is measurable. In particular, both f + g

and fg are measurable (by taking φ to be (p, q) 7→ p+q and (p, q) 7→ pq, respectively.)Proof. (1), (2) : Use Exercise 3.4.(3) Recall that g is continuous if and only if g−1(U) is open for every open set U ⊂ R,

and use part (2) above in combination with Ex 3.4(5).(4) Let h : X → R2 : x →

(

f(x), g(x))

. Then φ ◦ (f, g) = φ ◦ h. By part (3) it will

suffice to show that (φ◦h)−1(U) is measurable for every open set U ⊂ R. Let U be an opensubset of R. Since φ is continuous, φ−1(U) is an open subset of R2. Then, since everyopen subset of R2 is a countable union of open boxes of the form (a, b)× (a′, b′), it followsthat φ−1(U) is a countable union of such boxes. Therefore, by Ex 3.4(1,5), (φ ◦ h)−1(U)is a countable union of sets of the form h−1

(

(a, b)× (a′, b′))

. Thus it will suffice to show

that this set h−1(

(a, b)× (a′, b′))

is measurable. Examination of the definition of h shows

that this set is equal to f−1(a, b)∩ g−1(a′, b′). Since f and g are measurable these inverseimages under f and g, and hence also their intersection, are measurable. This completesthe proof.

3.6. Proposition (Manufacturing measurable functions II). (1) Constant functions aremeasurable. The sum and product of two measurable functions are measurable functions(assuming that the sum/product is defined; i.e. we don’t have sums like ∞ + (−∞) orsimilar undefined products).

(2) If f1, f2, ... is a sequence of measurable functions X → [−∞,∞] then the functionsx 7→ sup

n≥1fn(x) (and hence also the function x 7→ inf

n≥1fn(x)) and x 7→ lim sup

n≥1fn(x) (and

hence also x 7→ lim infn≥1

fn(x)) are measurable. In particular, if limn→∞

fn(x) exists for every

x ∈ X, then the function x 7→ limn

fn(x) is measurable.

(3) If f and g are measurable functions X → [−∞,∞] then so are the functionsmax{f, g} and min{f, g}. In particular, the functions

f+ = max{f, 0}

andf− = −min{f, 0}

are measurable. As a consequence (of this and (1) ) the absolute value |f |, being the sumf+ + f−, is measurable.

Proof. (1) It is easy to see (verify) that a constant function is measurable. Nowsuppose f and g are measurable functions and that f(x)+ g(x) is defined for every x ∈ X.

9

It will suffice (because of Ex 3.5(1) ) to show that that for every a < ∞, the set E(a) ={x ∈ X : f(x) + g(x) > a} is measurable. Now this set is the union of the measurablesets g−1{∞}, f−1{∞}, and the set F (a) = E(a)∩ f−1(R)∩ g−1(R). (The reason why webring in F (a) is that we want to avoid dealing with points where f and g have infinitevalues.) Thus it will suffice to show that the set F (a) is measurable. However, for F (a) weare essentially dealing with finite-valued (i.e. real-valued) functions f and g and thereforethe argument in Proposition 3.5 (4) works. The measurability of fg follows in exactly thesame way.

(2) If f = supn

fn, then it may be verified that f−1(a,∞] = ∪∞n=1f−1n (a,∞] (verify

this). This, being a countable union of measurable sets, is measurable. Similarly (orby using negatives), inf

n≥1fn is also measurable. To deal with lim sup

n

fn note first that

lim supn

fn is equal to infn≥1

{supk≥n

fk}. Therefore, by combining our preceding two statements,

we see that lim supn

fn is also measurable. If limn

fn exists then it is equal to lim supn

fn and

is therefore measurable.(3) This is immediate from (2) by taking f1 = f and f2 = f3 = ... = 0.

Problem Set 1.1. If (a, b) and (c, d) are two points in R2, say that they are equivalent if both c− a

and d− b are rational numbers. Verify that this is indeed an equivalence relation. Verifyalso that every point in R2 is equivalent to at least one point (in fact infinitely manypoints) in the unit square [0, 1]× [0, 1]. Form a set E ⊂ R2 by picking one point in thisunit square from each equivalence class. (The formal existence of such a set is guaranteedby the Axiom of Choice.) The set of all points of [0, 1]× [0, 1] with rational coordinates isa countable set {r1, r2, ...}. Show that the sets E + rn are mutually disjoint and that

∪∞n=1

(

E + rn

)

⊂ [0, 2]× [0, 2].

Use this to show that there is no measure µ defined on the set of all subsets of R2 whichhas the following properties : (1) the measure of the translate of a set is equal to themeasure of the set (i.e. µ(E + r) = µ(E)) , and (2) the measure of the square [0, 2]× [0, 2]is a finite positive (> 0) number.

2. Let E be a collection of subsets of X, and let Σ denote the family of all σ−algebras(of subsets of X) which contain the collection E . First show that Σ is not empty by verifyingthat the set P(X) of all subsets of X is a σ−algebra. Then show that ∩Σ (i.e. the collectionof all sets which belong to each of the σ−algebras in the family Σ) is a σ−algebra and thatit is the smallest σ−algebra containing E (i.e. if B is a σ−algebra containg all the sets ofE then B ⊃ ∩Σ. The σ−algebra ∩Σ is called the σ−algebra generated by E and is oftendenoted by σ(E).)

3. Let X be a set, and P(X) the σ−algebra of all subsets of X. Show that everyfunction f : X → [−∞,∞] is measurable with respect to P(X). Conversely, suppose B

10

is a σ−algebra of subsets of X such that every function f : X → [−∞,∞] is measurable.Show that B = P(X).

4. Let (X,B) be a measurable space and b a fixed point in X. Define µ : B → [0,∞]by :

µ(A) =

{

1, if b ∈ A;0, otherwise.

Show that µ is a measure. This measure is called the ‘delta function’ at b.5. (Hard) Let B be a σ−algebra of subsets of {1, ..., n}. Show that B has 2k elements,

for some integer k ∈ {1, ..., n}.6. Show that every continuous function R → R is measurable with respect to the

Borel σ−algebra of R. Give some examples of discontinuous Borel measurable functions.Can you prove that there exist functions R → R which are not Borel measurable ?

7. If s and t are non-negative simple functions and t ≤ s then show that

∫

t dµ ≤∫

s dµ (with the integrals as defined in Definition 2.4) ? (Hint : show that t − s is a

non-negative simple function and use Proposition 2.7 on s = t + (s − t).) What is therelevance of this to the comments in Definition 2.10 ?

8. The set of all rotations in R3 is denoted SO(3); it is a group under composition of

rotations. Let a denote the rotation around the z−axis by the angle cos−1(1

3), and b the

rotation by the same angle around the x−axis. It can be shown by a clever computationalargument that a and b are ‘independent’ rotations in the following sense : no product ofthe form ar1bs1 ...arnbsn with the ri and si being non-zero integers, or similar products(of non-zero powers of a and b, successively) starting/ending with non-zero powers of a orb, can be equal to the identity rotation. Rotations satisfying this independence propertywere constructed by Hausdorff in 1914. Let G denote the set of all rotations which canexpressed as (finite) products of powers (positive and negative) of a and b. Check that G

is a subgroup of SO(3). Let A+ denote the set of all finite products ar1bs1 ... with r1 > 0(i.e the product starts with a positive power of a), let A− be the set of all finite productsar1bs1 ... with r1 < 0; and let B+ be the set of all finite products bs1ar2 ... with s1 > 0, andB− the set of all finite products bs1ar2 ... with s1 < 0. Thus the sets A± and B± form apartition of G \ {I} into four disjoint subsets.

Let S2 = {(x1, x2, x3) ∈ R3 : x21 + x2

2 + x23 = 1}, the unit sphere in R3.

(i) Show that there is a countable set D ⊂ S2 such that G acts on S2 \ D withoutfixed points; i.e. if T ∈ G, and T 6= I, and x ∈ S2 \ D then Tx ∈ S2 \ D and Tx 6= x.(Hint : how many points on S2 does a rotation keep fixed ?) We shall denote S2 \D byX.

The orbit of a point x ∈ X under G is the set {Tx : T ∈ G}. Distinct orbits aredisjoint sets. Picking one element from each such orbit (this is legal under the Axiom ofChoice) we form a set M ⊂ X such that for every x ∈ X there is a unique element m ∈ M

for which Tx = m for some T ∈ G.Let XA+ = {Tm : m ∈ M, T ∈ A+}, and define sets XA− , XB+ , XB− analogously.(ii) Show that the four sets XA± and XB± are disjoint and their union is X.

11

(iii) Show that X =(

a−1XA+

)

∪XA− , and X =(

b−1XB+

)

∪XB− .

(iv) Suppose µ is a measure on the σ−algebra of all subsets of S2 which has theproperties that : (1) it is invariant under rotations (i.e. µ

(

T (E))

= µ(E) for every

T ∈ SO(3) and every E ⊂ S2); and (2) 0 < µ(S2) < ∞. Use (ii) and (iii) above toshow that such a measure µ cannot exist ! Is it necessary to use the countable additivityof µ ?

(v) Show that there is no measure ν defined on the σ−algebra of all subsets of R3 whichsatisfies the following conditions : (1) ν is invariant under rotations; (2) 0 < ν(Br) < ∞ forsome ball Br, centered at the origin and of radius r(< ∞). (Hint : Suppose such a measureν exists, and assume, for convenience and without loss of generality, that the ν−measure ofthe closed unit ball is finite and positive. Define a measure µ on the set of all subsets of S2

as follows. Consider any Y ⊂ S2, and form the ‘cone’ Y ′ = {ry : y ∈ Y, r ∈ [0, 1]} ⊂ R3.Define µ(Y ) to be ν(Y ′). Now use part (iv) above.)

(vi) Let D be a countable subset of S2. Choose a line through the origin which doesnot pass through any point in D (why does such a line exist ?). Verify that the set ofall rotations around the chosen line which carry some point of D into some point of D

is countable (Hint : how many rotations around the line take a given point to anothergiven point ?); conclude that there is a rotation T such that T (D) ∩ D = ∅. Let µ

be a finitely additive (but not necessarily countably additive) measure on the set of allsubset of S2 which is invariant under rotations and for which 0 < µ(S2) < ∞. Show that

µ(D) ≤1

2µ(S2) (Hint : consider D ∩ T (D).) Apply this to D ∪ T (D) in place of D and

thus show that µ(D) ≤1

4µ(S2). What can you say about µ(D) ? Can you now prove a

stronger form of the results in (iv) and (v) above ?Definition. A set F ⊂ R3 is ‘equidecomposable’ with a set H ⊂ R3 if F can be

partitioned into sets F1, ..., Fm (for some integer m ≥ 1) and H can be partitioned intosets H1, ..., Hm such that each Fi can be rotated to yield Hi; in other words, F can bebroken down and rearranged (by using rotations) and reassembled to form H.

Verify that if E is equidecomposable with F , and F is equidecomposable with H, thenE is equidecomposable with H.

(vii) Let D be a countable subset of S2, and let T be the rotation described in (vi).Let D′ = D ∪ T (D) ∪ T 2(D) ∪ · · ·. Check that T j(D) ∩ T i(D) = ∅, unless j = i. Considerthe partition of S2 into S2 \ D′ and D′. Consider, on the other hand, the partition ofS2 \ D into the sets S2 \ D′ and T (D) ∪ T 2(D) ∪ · · ·. Using these partitions, show thatS2 \D is equidecomposable with S2.

(viii) Recall the sets XA± and XB± . Let XA = XA+ ∪XA− , and XB = XB+ ∪XB− .Combine some of the preceding results to show that XA is equidecomposable with S2, andthat XB is also equidecomposable with S2.

(ix) Let B denote the unit ball, centered at the origin, minus the origin itself; i.e.B = {(x1, x2, x3) ∈ R3 : 0 < x2

1 + x22 + x2

3 ≤ 1}. Use (viii) and the idea used in (v),to show that there are two disjoint subsets of B each of which can be broken down andrearranged to form the whole of B. In particular, by doing this we can create two copies

of B by taking two (disjoint) pieces from B and rearranging (by cutting, rotating, andreassembling) each piece ! (It is possible to include the origin as well but we won’t pursue

12

this any further). This is one (somewhat weak) version of the Banach-Tarski Paradox.For more on this, see the book with this title by Stan Wagon, Cambridge University Press(1985), based on which the above exercises have been composed.

Lecture 4. Integration and the Convergence Theorems.

The main results here are the Monotone Convergence Theorem (4.6), Fatou’s Lemma(4.10), and the Dominated Convergence Theorem (4.15).

As usual, (X,B, µ) is a measure space.

We have already defined

∫

f dµ for non-negative measurable functions f . In the

following we extend to more general f :4.1. Definition. (Integration) Let f : X → [−∞,∞] be a measurable function. As

we have seen before, f = f+ − f− and both f+ and f− are non-negative, measurable

functions, and so

∫

f+ dµ and

∫

f− dµ make sense. If at least one of these integrals is

finite then we define :∫

f dµ =

∫

f+ dµ−

∫

f− dµ

(Note that we have excluded the troublesome case ∞−∞ by not defining

∫

f dµ for those

f for which both

∫

f+ dµ and

∫

f− dµ are ∞.)

Note . Let f : X → [−∞,∞] be measurable. Then

∫

f dµ is finite if and only if∫

|f | dµ < ∞. (Recall that |f | is measurable and that |f | = f+ + f−. The reason why

this fact is useful is that |f | is non-negative and so

∫

|f | dµ always makes sense - in some

situations it is possible to show that this is finite without having to deal with f+ and f−.)

Notation.

∫

E

f dµ means

∫

f1E dµ.

Check that in case f is non-negative measurable then

∫

f dµ as defined by Definition

2.10 equals

∫

f dµ as defined in Definition 4.1 above. (Hint : what is f+ and what is f−

for non-negative f ?)Exercise. Here is something we forgot to verify earlier. In case f is a non-negative

simple function we had two definitions of

∫

f dµ: one according to Definition 2.4 (integral

of non-negative measurable simple functions) and one according to Definition 2.10 (fornon-negative measurable functions). Show that there is no conflict between these two

13

definitions. (Hint : if f is non-negative measurable simple then in Definition 2.10 takes = f ; you need also the fact that if s and t are non-negative simple functions and t ≤ s

then

∫

t dµ ≤

∫

s dµ - which follows by writing s = t + (s − t), checking that t − s is a

non-negative measurable simple function, and using Proposition 2.7).

4.2. Exercise. (a) Let f = 1A − 1B , where A and B are disjoint measurable subsets of X.

Under what conditions on µ(A) and µ(B) is

∫

f dµ well-defined ? What is its value

in this case ?

(b) Check that if f is any non-negative measurable function and E is any measurable set

then

∫

E

f dµ ≤

∫

f dµ. (Is f1E ≤ f?)

4.3. Exercise. (Sets of measure zero aren’t important) Suppose f and g are measurablefunctions X → [−∞,∞].

(a) Show that the set N = {x ∈ X : f(x) 6= g(x)} is measurable. (Hint : N is the unionof Ns = {x ∈ X : f(x) < g(x)} and Nb = {x ∈ X : f(x) > g(x)}. A point x lies inNs if and only if there is a rational number r such that f(x) < r and r < g(x). ThusNs is the union, as r runs over the countable set of all rational numbers, of the sets{x ∈ X : f(x) < r} ∩ {x ∈ X : r < g(x)}. Verify that each of the two sets involvedthis intersection here is measurable. Proceed. Note that this argument shows thatthe sets {f > g} (which is short-hand for {x ∈ X : f(x) > g(x)}) and {f < g} ismeasurable.) Then show that the sets {f ≥ g} and {f ≤ g} are also measurable.

(b) Suppose µ(N) = 0. That is, f and g are equal everywhere except on a set N ofmeasure zero : we say then that f and g are equal almost everywhere (of course, withrespect to the measure µ under consideration) - in short, f = g a.e. Show that thenf+ equals g+ almost everywhere, and f− = g− a.e. (Hint : if f(x) = g(x) then bothf+(x) = g+(x) and f−(x) = g−(x); thus if at x, either f+ doesn’t equal f−, or g+

doesn’t equal g− then f(x) 6= g(x) - i.e. {x : f+(x) 6= g+(x)} and {x : f−(x) 6= g−(x)}are subsets of N . Why are these sets measurable ?)

(c) Show that if s is a non-negative simple function which is equal to 0 almost everywhere

then

∫

s dµ = 0. (Hint : write s =∑

aj1Ajand verify that whenever aj 6= 0 the set

Aj has measure zero; then look at the formula defining

∫

s dµ.)

(d) If s is a non-negative simple function and N is a (measurable) set of measure zero,

show that

∫

s dµ =

∫

s1Nc dµ. (Hint : you can do this directly from the Definition

of

∫

s dµ or by writing s = s1Nc + s1N and using (c) on

∫

s1N dµ.)

(e) Let h be a non-negative measurable function, and N a set of measure 0. Show that∫

h dµ =

∫

h1Nc dµ. (Hint : Consult Definition 2.10, replace each s in that definition

by s1Nc and use part (d). Check : (i) if 0 ≤ s ≤ h then 0 ≤ s1Nc ≤ h1Nc , and (ii) if0 ≤ t ≤ h1Nc then 0 ≤ t ≤ h.)

(f) Let f, g : X → [−∞,∞] be measurable, and suppose f = g a.e.; i.e. that the set N =

14

{f 6= g} has measure zero. Show by combining (b) and (e) that

∫

f+ dµ =

∫

g+ dµ

and that

∫

f− dµ =

∫

g− dµ. From this show that

∫

f dµ is defined if and only if∫

g dµ is defined, and this case the two integrals are equal. Thus, if f and g are equal

almost everywhere then

∫

f dµ =

∫

g dµ whenever either side is defined. medskip

4.4. Excercise. (to be used ) Let s be a non-negative measurable simple function onX. Define, for every measurable set E,

ν(E) =

∫

E

s dµ

Show that ν is a measure in the following way. First check quickly that ν(E) makes senseand is non-negative for every measurable set E. Check that ν(∅) = 0 (hint : What is s1∅?)The non-trivial part is checking that ν is countably additive. For this write out s in the

form

n∑

i=1

ai1Ai, where a1, ..., an are all the distinct values of s. Consider now a countable

collection of disjoint measurable sets E1, E2, .... Then examine carefully each step of thefollowing chain of equalities and justify any step that is not obvious :

ν(∪jEj) =

∫

s1∪jEjdµ

=

∫ n∑

i=1

ai1Ai∩∪jEjdµ (Hint : 1A1B = 1A∩B)

=n

∑

i=1

aiµ(Ai ∩ ∪jEj)

=n

∑

i=1

ai

{

∞∑

j=1

µ(Ai ∩ Ej)}

=

∞∑

j=1

n∑

i=1

aiµ(Ai ∩Ej)

=

∞∑

j=1

ν(Ej).

The following result as well as the construction given in its proof is very useful.

4.5. Proposition. Let f : X → [0,∞] be measurable. Then there exist measurable

simple functions sn on X such that :(1) 0 ≤ s1(x) ≤ s2(x) ≤ · · · ≤ f(x) for every x ∈ X;(2) sn(x) → f(x) as n →∞, for every x ∈ X.

15

Proof. The idea is to ‘discretize’ the values of f and in this way approximate f bysimple functions. For example, suppose we want a simple function sn which doesn’t differ

from f by more than 12n . For this we cut up [0,∞) into pieces of length 1

2n ; the k-th piece

would be [k − 1

2n,

k

2n). Let En

k denote the set of all point x ∈ X at which f(x) ∈ [k − 1

2n,

k

2n].

We set sn(x) =k − 1

2n, for x ∈ En

k . Thus for x ∈ Enk , both sn(x) and f(x) belong to an

interval of length 12n , and therefore f(x) and sn(x) differ by at most 1

2n . Defining sn inthis way for each En

k , we obtain the function sn approximating f pretty closely, and sn

should converge to f everywhere. Notice also that the way we have defined sn insuresthat sn(x) ≤ f(x) for every x. While this is the key strategy there are a number of small

flaws. First, you can’t cut up [0,∞] into bits of length 12n , for in this way you can never

capture ∞. Secondly, the way we defined sn above, the set of values of sn looks more like acountable set than a finite set. These problems can be fixed in one stroke. Define sn(x) asabove for those points x where f(x) < n; while for points x where f(x) ≥ n, set sn(x) = n.By doing this we give up our initial goal of approximating f closely everywhere (we giveup in the region where f is bigger than n - but as n →∞ this works out all right). To beprecise, define :

sn =n2n

∑

k=1

k − 1

2n1{ k−1

2n ≤f< k2n }

+ n1f−1[n,∞].

The sets f−1[k − 1

2n,

k

2n) and f−1[n,∞] are measurable and mutually disjoint (for fixed n

and variable k). Thus sn is a measurable simple function, and is clearly non-negative. If x

is a point where f(x) < ∞, then for large enough integer n we will have f(x) < n and for

all such integers n the values f(x) and sn(x) will differ by at most 12n . Thus sn(x) → f(x),

as n →∞, for such x. If, instead, f(x) = ∞, then x belongs to f−1[n,∞] for every integern, and so sn(x) = n for every n; as a result, sn(x) → ∞ as n → ∞. Thus in either casesn(x) → f(x) as n → ∞. Finally, it is easy to check that we still have sn(x) ≤ f(x) forevery x ∈ X.

One of the questions one meets over and over again in analysis is when, given a

sequence of functions fn converging to a function f ,

∫

fn dµ converges to

∫

f dµ. There

three important and versatile tools to handle such situations :(a) the Montone Convergence theorem(b) the Lebesgue Dominated Convergence theorem(c) Fatou’s Lemma.

Of these, Fatou’s lemma is used less frequently but is nevertheless just the right toolto use in certain situations. We will use all three results over and over again.

We turn first to the monotone convergence theorem.4.6. Monotone Convergence Theorem. Let (fn) be a sequence of functions on X

such that :(1) each fn is measurable and non-negative;(2) 0 ≤ f1(x) ≤ f2(x) ≤ · · · ≤ ∞ for every x ∈ X;

16

(3) fn(x) → f(x) as n →∞, for every x ∈ X.Then f is measurable and

∫

fn dµ →

∫

f dµ as n →∞.

Proof. We shall go through a very descriptive account : after all the ideas have beendigested, it is possible to condense them into a fairly short efficient proof. Note first that

(1) guarantees that

∫

fn dµ exists for each n ≥ 1. Moreover, f being the limit of a sequence

of non-negative measurable functions is itself a non-negative measurable function. Thus∫

f dµ is also defined.

Condition (2) implies that

∫

fn dµ ≤

∫

fn+1 dµ, and so the sequence of numbers∫

fn dµ → α as n →∞, for some α ∈ [0,∞]. Clearly, fn(x) ≤ f(x), for every x ∈ X, and

therefore

∫

fn dµ ≤

∫

f dµ. Since this holds for every n, we can take n →∞ to see that

α ≤

∫

f dµ.

Now we need to prove the opposite inequality, that α is greater or equal to

∫

f dµ.

This is the hard part.

Recall that

∫

f dµ is the supremum of

∫

s dµ as s runs over measurable simple

functions with 0 ≤ s ≤ f . So we want to show that α is greater or equal to this supremum.

That is, we want to show that α is greater or equal to

∫

s dµ for every s of the type

mentioned above. Since α is the limit of the sequence of

∫

fn dµ we would be done if some

fn is greater or equal to s. But, although fn → f , and s ≤ f , we can’t conclude that somefn is greater or equal to s. All we can say is that if s(x) < f(x) (at some point x) thenfor n large enough fn(x) > s(x). So let us see if we can get anywhere with the set :

En = {x : fn(x) ≥ s(x)}.

This set is measurable (see Excercise 4.3 (a) ). Moreover, E1 ⊂ E2 ⊂ E3 ⊂ ... (becausef1 ≤ f2 ≤ f3 ≤ · · ·.) The comments above show that {x : s(x) < f(x)} ⊂ ∪n≥1En. Also :

∫

En

fn dµ ≥

∫

En

s dµ

because - and check this - fn1En≥ s1En

. Now

∫

En

fn dµ is less or equal to the whole

integral

∫

fn dµ, since fn is non-negative (Excercise 4.2(b)). Therefore,

∫

fn dµ ≥

∫

En

s dµ.

17

Now we take n →∞ to obtain :

α ≥ limn→∞

∫

En

s dµ (4.1)

Now since (by Excercise 4.4) E 7→

∫

E

s dµ is a measure, and since E1 ⊂ E2 ⊂ · · ·, we

have

∫

En

s dµ →

∫

∪nEn

s dµ, as n →∞.

Thus inequality (4.1) reads :

α ≥

∫

∪nEn

s dµ (4.2)

If ∪nEn were equal to X then we would be done. But in general, ∪nEn is not the wholespace X. Let us see how big it is. As we have already seen it contains the set {x : s(x) <

f(x)}. Now this set clearly excludes the points where f(x) = 0 (because s ≥ 0). Butwe can see that the set {f = 0} is contained in each En (because of the requirements0 ≤ fn ≤ f , and 0 ≤ s ≤ f). Thus A = {f = 0} ∪ {s < f} ⊂ ∪nEn. So inequality (4.2)implies :

α ≥

∫

A

s dµ (4.3)

But still A might not equal all of X. Recalling that 0 ≤ s ≤ f , we see upon examiningA that what we are missing is the set on which s = f 6= 0. The following trick fixes theproblem. Pick a fraction c close to 1 (any c ∈ (0, 1) will do) and replace s by cs. Now cs

is still a measurable simple function satisfying 0 ≤ cs ≤ f . The nice thing about cs is thatcs(x) < f(x) unless f(x) = 0. Thus if we redefine A by replacing s in the definition of A

by cs we actually obtain all of X; i.e. {f = 0}∪{cs < f} = X. Therefore, inequality (4.3)reads (with s replaced by cs) :

α ≥

∫

X

cs dµ

which is the same as :

α ≥ c

∫

s dµ (4.4)

Now we want to get rid of the c. This is easy : inequality (4.4) is valid for every c ∈ (0, 1)and so we can take c → 1, to obtain :

α ≥

∫

s dµ (4.5)

Recall that here s is any measurable simple function with 0 ≤ s ≤ f . Taking the supremumover all such s gives the desired inequality :

α ≥

∫

f dµ

18

and this completes the proof.

Remarks : The hypothesis that each fn is non-negative is important. For consider anexample where µ(X) = ∞ (for example, µ is counting measure on the set of all subsets of

an infinite set X), and set fn = −1

n. Then all the hypotheses, except for non-negativity,

are satisfied but each

∫

fn dµ = −∞ while

∫

( limn→∞

fn) dµ =

∫

0 dµ = 0.

4.7. Fact. Let f be a function X → [−∞,∞] which can be written in the form g− h,where h and g are functions X → [0,∞] (i.e. g, h ≥ 0). Then f+ ≤ g and f− ≤ h. (Thus,the representation f = f+ − f− is ‘minimal’ in a sense.)

Proof . Consider any x ∈ X. If f(x) ≥ 0, then f+(x) = f(x) and so :

f+(x) = f(x) = g(x)− h(x) ≤ g(x).

If f(x) < 0, then f+(x) = 0 and so certainly f+(x) ≤ g(x). Thus f+ ≤ g. Similarly,f− ≤ h.

4.8. Proposition. Let f and g be measurable functions X → [−∞,∞]. Suppose that :

(1)

∫

f dµ and

∫

g dµ are defined,

(2) the sum

∫

f dµ +

∫

g dµ is defined, and

(3) f(x) + g(x) is defined for every x ∈ X.

Then

∫

(f + g) dµ is defined and :

∫

(f + g) dµ =

∫

f dµ +

∫

g dµ (4.6)

If c ∈ R is a constant then

∫

cf dµ is defined if and only if

∫

f dµ is defined, and in this

case :∫

cf dµ = c

∫

f dµ (4.7)

Proof. First suppose that f and g are non-negative measurable functions. Then byProposition 4.5, there exist sequences of measurable simple functions (sn) and (tn) suchthat 0 ≤ sn ≤ f , 0 ≤ tn ≤ g, and sn(x) → f(x), and tn(x) → g(x) for every x ∈ X. Thensn + tn is a sequence of measurable simple functions satisfying 0 ≤ sn + tn ≤ f + g andsn(x)+ tn(x) → f(x)+g(x), as n →∞, for every x ∈ X. Then the monotone convergencetheorem gives the following limits (as n →∞):

∫

(sn + tn) dµ →

∫

(f + g) dµ

∫

sn dµ →

∫

f dµ

19

∫

tn dµ →

∫

g dµ.

Now

∫

sn dµ +

∫

tn dµ =

∫

(sn + tn) dµ. So letting n →∞ we obtain :

∫

(f + g) dµ =

∫

f dµ +

∫

g dµ.

A similar argument (with sn and csn) proves that if f is non-negative measurable andc ∈ (0,∞) then:

∫

cf dµ = c

∫

f dµ.

Now we deal with general measurable f, g : X → [−∞,∞], for which the hypotheses (1),(2)and (3) hold. Let h = f + g. Now f = f+ − f− and g = g+ − g−. So :

h+ − h− = h = f + g = f+ − f− + g+ − g−.

(this is okay even if ±∞’s are involved.)We then have (the case when ±∞ are involved requires special checking):

h+ + f− + g− = f+ + g+ + h− (4.8)

Since these are all non-negative measurable functions we can conclude that :

∫

h+ dµ +

∫

f− dµ +

∫

g− dµ =

∫

f+ dµ +

∫

g+ dµ +

∫

h− dµ (4.9)

From which we would like to conclude that :

∫

h+dµ−

∫

h− dµ =

∫

f+ dµ−

∫

f− dµ +

∫

g+ dµ−

∫

g− dµ

and hence that

∫

h dµ =

∫

h+dµ−

∫

h− dµ

=

∫

f+ dµ−

∫

f− dµ +

∫

g+ dµ−

∫

g− dµ

=

∫

f dµ +

∫

g dµ

which is what we wanted. The only problem with this argument is that the second linewhich is obtained by a rearrangement of equation (4.9) is a valid conclusion from (4.9)

only if

∫

h+ dµ and

∫

h− dµ are not both ∞. Now, by Fact 4.7, h+ ≤ f+ + g+, and

20

h− ≤ f− + g−. So if both

∫

h+ dµ and

∫

h− dµ were ∞ then both

∫

f+ dµ +

∫

g+ dµ

and

∫

f− dµ +

∫

g− dµ would be ∞ (here we have again used the fact, proved above,

that the integral of the sum of two non-negative measurable functions is the sum of theintegrals). But, as may be verified by checking each of the four possible cases here, this

would contradict the hypotheses that

∫

f dµ,

∫

g dµ and their sum are defined.

This takes care of all the details in proving that

∫

(f + g) dµ =

∫

f dµ +

∫

g dµ.

The proof of

∫

(cf) dµ = c

∫

f dµ for general f (i.e. not necessarily non-negative) is

done similarly.

Exercise. Show that

∫ n∑

j=1

cj1Cjdµ =

n∑

j=1

cjµ(Cj) if c1, ..., cn are real numbers and

C1, ..., Cn are measurable sets of finite measure. Is it possible for

∫ n∑

j=1

cj1Cjdµ to exist

even if

n∑

j=1

cjµ(Cj) is undefined ?

4.9. Proposition. If fn : X → [0,∞] (note fn ≥ 0) is measurable, for n = 1, 2, ..., and

f(x) =

∞∑

n=1

fn(x) for every x ∈ X

then f is measurable and∫

f dµ =∞∑

n=1

∫

fn dµ

(i.e. the sum and integral can be interchanged).Proof. Let gn = f1 + · · · + fn. Then each gn is measurable, 0 ≤ g1 ≤ g2 ≤ · · ·,

and gn(x) → f(x) as n → ∞, for every x ∈ X. Therefore by the monotone convergencetheorem, f is measurable and

∫

gn dµ →

∫

f dµ as n →∞.

Now by Proposition 4.8 :

∫

gn dµ =n

∑

i=1

∫

fi dµ.

Letting n →∞ in this formula and comparing with the earlier stated limit of

∫

gn dµ we

obtain the desired result.

21

4.10. Fatou’s Lemma. If fn : X → [0,∞] (note : fn ≥ 0) is measurable, for

n = 1, 2, ..., then∫

(lim infn→∞

fn) dµ ≤ lim infn→∞

∫

fn dµ.

Proof. This is obtained easily from the monotone convergence theorem by realizingthat lim inf

n→∞fn(x) equals lim

n→∞gn(x), where gn(x) = inf

k≥nfk(x), and that then g1, g2, ... are

measurable functions with 0 ≤ g1 ≤ g2 ≤ · · ·. For, with this notation, we have first (since,by definition of gn, we have gn ≤ fn) :

∫

gn dµ ≤

∫

fn dµ (4.10)

By the monotone convergence theorem (and since lim infn→∞

fn = limn→∞

gn):

∫

gn dµ →

∫

lim infn→∞

fn dµ (4.11)

We might be tempted to let n → ∞ in equation (4.10) to conclude (using (4.11)) that∫

lim infn→∞

fn dµ is less or equal to limn→∞

∫

fn dµ. However, there is no guarantee that

limn→∞

∫

fn dµ exists. So the best we can do is to take lim infn→∞

in equation (4.10) and this

gives the desired inequality.4.11. Standard Convention. Consider a function f : E → [−∞,∞] where E ⊂ X is

measurable and µ(Ec) = 0. Suppose, moreover, that f is measurable in the sense thatf−1[a, b] is a measurable subset of X (of course, it is in fact a subset of E) for everya, b ∈ [−∞,∞]. Choose any function f ′ : X → [−∞,∞] which is measurable and agreeswith f on the set E. (For example, the function h defined by h(x) = f(x) for x ∈ E andh(x) = 0 for x ∈ Ec is one such function - verify that h is measurable.) Then we define∫

f dµ to be

∫

f ′ dµ. This value is independent of the choice of f ′, for if we use a different

choice say f ′′ then f ′ and f ′′ are equal (on the set E) almost everywhere and hence havethe same integral.

4.12. Chebyshev’s Inequality. Let f : X → [0,∞] be a measurable function and

let c ∈ (0,∞) a positive number. Then :

µ({x ∈ X : f(x) ≥ c}) ≤1

c

∫

f dµ.

Proof. We have :

∫

f dµ ≥

∫

{f≥c}

f dµ ≥

∫

{f≥c}

c dµ = cµ({f ≥ c})

where the second inequality follows from the simple observation that on the set {f ≥ c}the function f is ≥ c ! The desired result now follows by dividing by c.

22

The following observation will be used constantly.

4.13. Exercise. If f : X → [−∞,∞] is measurable and

∫

|f | dµ < ∞ then f is

finite-valued almost everywhere.

4.14. A Useful Basic Fact. If f : X → [−∞,∞] is measurable and

∫

f dµ is defined

then :

|

∫

f dµ| ≤

∫

|f | dµ.

Proof. Since f ≤ |f |, we have

∫

f dµ ≤

∫

|f | dµ. Replacing f by −f this reads :

−

∫

f dµ ≤

∫

|f | dµ. These two inequalities are summarized in one in :

|

∫

f dµ| ≤

∫

|f | dµ.

We come now to one of the most frequently used convergence theorem for integrals :4.15. Lebesgue’s Dominated Convergence Theorem. Suppose (fn) is a sequence

of measurable functions X → [−∞,∞] (note that this time we don’t require that fn be

non-negative) such that

f(x) = limn→∞

fn(x)

exists for every x ∈ X. Suppose, moreover, that there is a measurable function g : X →[0,∞] (the ‘dominating function) such that

|fn(x)| ≤ g(x) for every x ∈ X

and that∫

g dµ < ∞ (4.12)

Then∫

|f | dµ < ∞,

limn→∞

∫

|fn − f | dµ = 0 (4.13)

and

limn→∞

∫

fn dµ =

∫

f dµ (4.14)

(Note : There might be points where fn−f is of the form ∞−∞ and hence undefined.

The set of such points has measure zero because

∫

f dµ (as will be shown) and

∫

fn dµ

are finite and we can use the result 4.13 above. Therefore Convention 4.11 is applicable.)

23

Proof. Each

∫

|fn| dµ and

∫

|f | dµ (note that |f | ≤ g and

∫

g dµ < ∞) are finite.

Therefore we may and will assume that each fn and f are finite valued; in particular,fn − f will then make sense everywhere.

The following shows that equation (4.13) implies equation (4.14) :

|

∫

fn dµ−

∫

f dµ| = |

∫

(fn − f) dµ| ≤

∫

|fn − f | dµ.

Thus it will suffice to prove equation (4.13). Since there is no monotone sequence insight we try to see if Fatou’s Lemma does anything for us here. The first try would bewith lim inf

n→∞|fn− f | but as you can readily this doesn’t give anything useful. The key idea

is is to notice that equation (4.13) can be read as saying that lim supn→∞

∫

|fn − f | dµ ≤ 0

(for this would be mean that zero is the lim sup of the sequence of non-negative numbers∫

|fn − f | dµ - and this is the same as saying that the limit of this sequence is zero). To

get lim inf the obvious thing to do is to switch signs (keeping in mind that lim infn

an =

− lim supn

(−an)). But then we would get into trouble with the requirement in Fatou’s

Lemma that the functions involved be non-negative. We can get around this problem byworking with h−|fn−f |, for some suitable function h which is greater or equal to |fn−f |,and hopefully we can throw it out at the end of the argument. A choice for h is obtainedeasily by noting that :

|fn − f | ≤ |fn|+ |f | ≤ 2g.

Thus our candidate for h can be 2g. So we apply Fatou’s Lemma to the sequence offunctions 2g − |fn − f |. The lim inf of this sequence, as n →∞, is the same as its limit -lim

n→∞(2g − |fn − f |) = 2g − 0 = 2g.

Thus Fatou’s Lemma gives :

∫

(2g) dµ ≤ lim infn→∞

∫

(2g − |fn − f |) dµ

=

∫

2g dµ + lim infm→∞

(−

∫

|fn − f | dµ)

=

∫

2g dµ− lim supn→∞

∫

|fn − f | dµ.

(Notice that we have used the simple fact that lim infn→∞

(c + an) = c + lim infn→∞

an valid for

any sequence of real numbers an, and any fixed real number c). Now, because

∫

(2g) dµ is

finite, we can subtract it off to obtain :

0 ≤ − lim supn→∞

∫

|fn − f | dµ

which is the same as :

24

lim supn→∞

∫

|fn − f | dµ ≤ 0

and this, as noted earlier, is equivalent to

limn→∞

∫

|fn − f | dµ = 0.

The following result is of some interest but we probably won’t need it.Egorov’s Theorem. If (fn) is a sequence of measurable functions which converges to a

measurable function f on a set E of finite measure, then for any ε > 0 there is a measurable

set F ⊂ E with µ(E − F ) < ε (i.e. F can be as close to E in size as we want), such that

(fn) converges uniformly to f on F .Proof. Try to prove it yourself. Or look up Folland’s ‘Real Analysis’ (or Halmos’

‘Measure Theory’).

Problem Set 2.

1. (This requires knowledge of algebra) If A, B ⊂ X, then the symmetric difference

A∆B is the set of all points of X which belong to either A or B but not to both; i.eA∆B = (A∪B)− (A∩B) = (A−B)∪ (B−A). Let P(X) denote the set of all subsets ofX. For this exercise we denote the elements of P(X) by lower case letters a, b, c, ... Also,for this exercise only, we will write a+b for a∆b, and a ·b for a∩b. Note that ∅, X ∈ P(X).

(1) Check that under + and ·, P(X) is a commutative ring with X as identity elementunder multiplication (‘·’). (That is : (i) a+∅ = a, (ii) a+b = b+a, (iii) a+(b+c) = (a+b)+c,(iv) a + a = ∅ - so each element is its own ‘negative’ -(v) a ·X = a, (vi)a(bc) = (ab)c, (vii)ab = ba, (viii) a(b + c) = ab + ac.)

(2) Verify that {∅, X} forms a subring of P(X) which is isomorphic to the field Z2 ={0, 1} (∅ 7→ 0, X 7→ 1). Recall that Z2 is the set of integers modulo 2.

(3) Verify that P(X) is a vector space over the field Z2 under the following naturaldefinition of ‘multiplication by scalars ’ : (i) 0.a = 0 and 1.a = a for every a ∈ P(X). (Youwill need to use the fact - item (iv) in part (1) - that a + a = ∅ for every a ∈ P(X).)

(Because of (1),(2),(3) one says that P(X), or any subset of P(X) which contains ∅and X and is closed under ‘+’ and ‘·’, is an algebra over Z2.)

(4) Suppose B ⊂ P(X) contains ∅ and is closed under ∪ (i.e. A ∪ B ∈ B wheneverA, B ∈ B) and complements (i.e. Ac ∈ B for every A ∈ B). Show that then B is closedunder ‘+’ and ‘·’ (i.e. a + b ∈ B and ab ∈ B whenever a, b ∈ B). Check that B is a vectorsubspace of the vector space P(X).

(5) Let B be as described in (4). From the theory of vector spaces we know that everyvector space has a basis (this uses the Axiom of Choice in case B is infinite). Let A ⊂ Bbe a basis of B. Thus every element x ∈ B can be written in a unique way as a linearcombination (with coefficients 0 and 1) of elements of A. Show that if A is finite andcontains n elements then B contains 2n elements.

25

(6) Use the ideas in (5) to show that no σ−algebra can have a countably infinitenumber of elements. (Hint : Use the fact that if A is an infinite set then the set of allmappings A → {0, 1} is uncountable.)

(7) Now let µ be a measure on a σ−algebra B of subsets of X. Let N denote thecollection of all sets N ∈ B for which µ(N) = 0. Show that N is an ideal in B : i.e. (i)∅ ∈ N , (ii) if a, b ∈ N then a + b ∈ N , amd (iii) if a ∈ N and c ∈ B then ac ∈ N .

2. Let (X,B, µ) be a measure space. Let A1, A2, ... ∈ B. Show that

µ(∪∞i=1Ai) ≤∞∑

i=1

µ(Ai).

(Hint: Let A = ∪∞i=1Ai. Verify that 1A(x) ≤∑

i≥1

1Ai(x) for every x ∈ X. Then apply

Proposition 4.9 (which relies on the monotone convergence theorem). This Problem canalso be done directly without use of the monotone convergence theorem.) Show that ifeach Ai has measure zero then so does their union.

3. Let (X,B, µ) be a measure space. Intuitively, a subset of a set of measure zeroshould also have measure zero. However, a subset of a set of measure zero might notbe measurable; i.e. might not be in B. This suggests that we remedy the situation byenlarging B to include all subsets of sets in B which have measure zero. But then we willalso have to include all sets formed by the union of these newly measurable sets with theold ones along with complements also. With this in mind, we let N ′ denote the set of allN ⊂ X for which there exists M ∈ B such that N ⊂ M and µ(M) = 0, and we define B′

to be the collection of all sets of the form E∆N where E ∈ B and N ∈ N ′. Verify that :(i) B′ ⊃ B;(ii) F ∈ B′ if and only if there exist A, B ∈ B with A ⊂ F ⊂ B and µ(B − A) = 0.

(Hint : If F does lie between two such sets A and B, then check that E = A∆(F −A) andthat F −A ⊂ B−A. Next suppose F = E∆N where E ∈ B and N ∈ N ′. Then there is aset M ∈ B with N ⊂ M and µ(M) = 0. Put A = E −M and B = E ∪M , and verify thatthese two sets satisfy all the necessary requirements.)

(iii) If E ∈ B′ then Ec ∈ B′. (Hint : by (ii), there are A, B ∈ B with A ⊂ E ⊂ B

with µ(B − A) = 0. Try Ac and Bc for the set Ec. You will find it useful to verify thatAc − Bc = B −A - think of what it means for a point x to belong to Ac −Bc.)

(iv) Show that if N1, N2, ... ∈ B is a countable collection of sets of measure zero thentheir union ∪iNi also has measure 0.

(v) Show that B′ is closed under countable unions. (Hint : Suppose E1, E2, ... ∈ B′.

Then there are sets A1, B1, A2, B2, ... ∈ B such that Ai ⊂ Ei ⊂ Bi and µ(Bi −Ai) = 0 foreach i. Set E = ∪iEi, A = ∪iAi, and B = ∪iBi. To work with B − A check first thatB − A = ∪i(Bi −A) and next that Bi − A ⊂ Bi − Ai; thus (B −A) ⊂ ∪i(Bi − Ai) - nowuse (iv) to show that B − A has measure zero.)

Summary : Parts (i) -(v) show that B′ is a σ−algebra with B ⊂ B′.(vi) Suppose that E ∈ B′ and A, A′, B, B′ ∈ B are such that A ⊂ E ⊂ B and A′ ⊂

E ⊂ B′ with µ(B−A) = 0 and µ(B′−A′) = 0. Show that µ(A) = µ(B) = µ(A′) = µ(B′).

26

(Hint : That µ(A) = µ(B) and µ(A′) = µ(B′) is easy to see and is not the substantial partof this problem. To show that µ(A) = µ(A′), verify first that A− (A ∩A′) ⊂ B′ −A′ andnext, as a consequence, that µ(A) = µ(A ∩ A′); switching the roles of A and A′, obtainµ(A′) = µ(A′ ∩A). Proceed.)

(vii) Define µ′ : B′ → [0,∞] by µ′(E) = µ(A) where A, B ∈ B, A ⊂ E ⊂ B andµ(B − A) = 0. By (vi) the value µ′(E) does not depend on the specific choice of A andB (any other choice A′, B′ would lead to the same value for µ′(E).) Check first thatµ′(E) = µ(E) in case E ∈ B. (What would be your choice for A and B be in this case?)Next show that µ′ is a measure. (Hint : Let E1, E2, ... be disjoint sets in B′, and write

E = ∪iEi. The goal is to show that µ(E) equals∑

i≥1

µ(Ei). Choose Ai,Bi,as before, for

each Ei. By (v), the sets A = ∪iAi and B = ∪iBi work for E. Now µ(Ei) = µ(Ai) bydefinition. What is µ(E)? Are the sets Ai disjoint ? Proceed.)

(viii) Show that if E ∈ B′, µ′(E) = 0, and F ⊂ E then F ∈ B′.Terminology : The measure µ′ is called the completion of µ.4. Let X = {a, b}, and let µ be counting measure on the σ−algebra P(X) of all subsets

of X. Define a sequence of P(X)−measurable functions fn (recall that all functions on X

are P(X)−measurable) as follows : for n odd, set fn(a) = 1 and fn(b) = 0; while for n

even, set fn(a) = 0 and fn(b) = 1. Apply Fatou’s Lemma to this sequence and check thatstrict inequality ‘<’ holds.

5. Let X be the set of positive integers {1, 2, 3, ...} with counting measure µ on theσ−algebra of all subsets of X. Let fn = 1{n}; i.e. fn(x) is equal to 1 at x = n and iszero at all other values of x. Note that each fn is measurable (with respect to P(X))and 0 ≤ fn ≤ 1; i.e. they are uniformly bounded. What is lim

n→∞fn(x) at any x ∈ X?

Calculate limn→∞

∫

fn dµ and

∫

limn→∞

fn dµ. Why is the dominated convergence theorem

not applicable here?

6. Let µ be counting measure on the σ−algebra P(X) of all subsets of the set X ={1, 2, 3, ...}. Recall that this means that if A ⊂ X, then µ(A) is the number of elementsin A, taken to be ∞ if A is infinite. Show that for any function f : X → [0,∞] : j 7→ fj,

the integral

∫

f dµ equals the sum

∞∑

j=1

fj. (Hint : first verify this in the case fj is zero

for all j bigger than some n; next, for arbitrary f , consider the sequence of functions

f (n) = f1{1,...,n} and use monotone convergence to deal with

∫

f dµ, and also the definition

of the infinite sum, i.e. that∞∑

j=1

fj = limn→∞

n∑

j=1

fj .)

7. If aij ≥ 0, for every i, j ∈ {1, 2, 3, ...}, show that :

∑

i≥1

∑

j≥1

aij =∑

j≥1

∑

i≥1

aij.

(Hint : Use Proposition 4.9 and Problem 5.)

27

8. Let fn be a measurable function X → R for each n = 1, 2, 3, .... Suppose that there

is a number c ∈ R such that

∫

|fn| dµ ≤ c for every n. Suppose also that fn(x) → f(x),

as n →∞, for every x ∈ X. Show that

∫

|f | dµ ≤ c.

9. Let (X,B, µ) be a measure space. Verify that if f, g, f ′, g′ are measurable functionson X such that f = f ′ a.e and g = g′ a.e. then : f + g = f ′ + g′ a.e, fg = f ′g′ a.e. andf

g=

f ′

g′a.e. (provided, of course, that these ratios are defined almost everywhere). If (fn)

and (f ′n) are sequences of measurable functions with fn = f ′n a.e. for every n = 1, 2, ..., thenlim infn→∞

fn = lim infn→∞

f ′n, lim supn→∞

fn = lim supn→∞

f ′n, and limn→∞

fn = limn→∞

f ′n, all a.e. (provided

these two limts exist almost everywhere). (Hint : it is convenient to prove the last of theseequalities first and use these in proving the first two.)

10. Suppose f1, f2, ... is a sequence of real-valued measurable functions and f : X → R

is measurable. Suppose limn→∞

∫

|fn − f | dµ = 0. Show that, for every ε > 0, limn→∞

µ{|fn −

f | > ε} = 0 (this condition is stated as ‘fn converges to f in measure’.)(i) Consider a sequence of sets A1, A2, .... ⊂ X, and write Em = ∪n≥mAn. Verify that

E1 ⊃ E2 ⊃ · · ·. Define lim supn

An = ∩nEn; i.e. lim supn

An = ∩n ∪m≥n Am. Verify that

x ∈ lim supn

An if and only if x belongs to An for infinitely many values of n. (Hint : for

each m, there is an integer n ≥ m such that x ∈ An.)(For part (ii) below you will need the simple observation that if a1, a2, ... is a sequence

of real numbers for which

∞∑

i=1

ai < ∞, then∑

i≥n

ai → 0 as n → ∞. Verify this (let

sn =n

∑

i=1

ai, and s =∞∑

i=1

ai; then, by definition, sn → s, as n → ∞; what does this say

about s− sn -assuming s < ∞ - as n →∞?)(ii) Let A1, A2, ... be measurable subsets of X, where (X,B, µ) is a measure space.

Prove the (first) Borel-Cantelli Lemma : If∑

n

µ(An) < ∞ then µ(lim supn

An) = 0.

(Hint : lim supn

An ⊂ ∪n≥mAn; so µ(lim supn

An) ≤ µ(∪n≥mAn). Recall that µ(∪iBi) ≤∑

i

µ(Bi) and use the fact mentioned above about convergent sums.) Thus if the sum of

the measures of the sets An is finite, then almost every point belongs to at most finitelymany An.

28