Measure and Integral

Measure and integral

Ilkka Holopainen

August 29, 2017 2 Measure and integral

These are lecture notes of the course Measure and integral (Mitta ja integraali).

0 Some background

0.1 Basic operations on sets Let X be an arbitrary set. The power set of X is the set of all subsets of X,

(X)= A: A X , P { ⊂ } and any subset (X) is called a family (or collection) of subsets of X. The union of a family F ⊂ P is F A = x X : x A for some A { ∈ ∈ ∈ F} A[∈F and the intersection (of ) is F A = x X : x A for all A . { ∈ ∈ ∈ F} A\∈F Let be an index set (set of indices) and suppose that for every α there exists a unique A ∈ A subset V X. (In other words, α V is a mapping (X).) Then the collection α ⊂ 7→ α A → P = V : α F { α ∈ A} is an indexed family of X. The union of an indexed family is

V = x X : x V for some α α { ∈ ∈ α ∈ A} α[∈A and the intersection of an indexed family is

V = x X : x V for all α . α { ∈ ∈ α ∈ A} α\∈A We denote also V and V , if is clear from the context. α α A α α [ \ Example. 1. Let (X). We can interpret as an indexed family by using as the index F ⊂ P F F set. That is, if α (thus α is a subset of X), we write V = α. Then = V : α . ∈ F α F { α ∈ F} 2. X = x , x = a singleton. { } { } x[∈X If the index set is N = 1, 2, 3,... , we denote { } ∞ Vn or Vn or Vn, n n n[∈N [ [ and ∞ Vn or Vn or Vn. n n n\∈N \ \ Spring 2017 3

∞ Sequences (of sets) are denoted by (Vn), (Vn)n=1, (Vn)n∈N, or V1,V2,.... The diﬀerence of sets A, B X is ⊂ A B = x X : x A and x B . \ { ∈ ∈ 6∈ } The complement of a set B X (with respect to X) is ⊂ Bc = X B. \ Remark. A B = A Bc. \ ∩

X A Bc = A B ∩ \

A B

Theorem 0.2. Let V : α be a family of X. Then the following de Morgan’s laws hold: { α ∈ A} c c (0.3) Vα = Vα α α [ \ and

c c (0.4) Vα = Vα . α α \ [ Let B X. Then the following distributive laws for union and for intersection hold: ⊂ (0.5) B V = (B V ) ∩ α ∩ α α α [ [ and

(0.6) B V = (B V ). ∪ α ∪ α α α \ \ Proof. (0.3):

x V c x V α: x V α: x V c x V c. ∈ α ⇐⇒ 6∈ α ⇐⇒ ∀ 6∈ α ⇐⇒ ∀ ∈ α ⇐⇒ ∈ α α α α [ [ \ (0.4): Similarly. (0.5):

x B V x B and x V x B and x V for some α ∈ ∩ α ⇐⇒ ∈ ∈ α ⇐⇒ ∈ ∈ α ∈A α α [ [ x B V for some α x B V . ⇐⇒ ∈ ∩ α ∈A ⇐⇒ ∈ ∩ α α [ (0.6): Similarly. 4 Measure and integral

The images and preimages of the union/intersection of a family.

Let X and Y be non-empty sets and f : X Y a mapping. → The image of a set A X under the mapping f is ⊂ f(A)= f(x): x A . ( Y ) { ∈ } ⊂ We usually abbreviate fA. The preimage of a set B Y under the mapping f is ⊂ f −1(B)= x X : f(x) B . { ∈ ∈ } We also abbreviate f −1B and denote

f −1(y)= f −1( y ), { } if y Y. [Note: f need not have an inverse mapping.] ∈ Theorem 0.7. Let f : X Y be a mapping and let V : α be a family of X, and let → { α ∈ A} W : β be a family of Y . Then { β ∈ B}

(0.8) f Vα = fVα α α [ [

−1 −1 (0.9) f Wβ = f Wβ β β [ [

−1 −1 (0.10) f Wβ = f Wβ. β β \ \ Proof. (0.8):

y f V y = f(x) and x V y = f(x) and x V for some α ∈ α ⇐⇒ ∈ α ⇐⇒ ∈ α ∈A α α [ [ y fV for some α y fV . ⇐⇒ ∈ α ∈A ⇐⇒ ∈ α α [ (0.9) and (0.10): Similarly.

Remark. It is always true that f V fV , α ⊂ α α α \ \ but the inclusion can be strict. The equality f( V ) = fV holds, for example, if f os an ∩α α ∩α α injection. Spring 2017 5

Countable and uncountable sets

Countability is a very important notion is measure theory!

Deﬁnition. A set A is countable if A = or there exists an injection f : A N ( a ∅ → ⇐⇒ ∃ surjection g : N A). → A set A is uncountable if A is not countable.

Remark. 1. A countable A finite äärellinen (including ) or countably infinite (when ⇐⇒ ∅ there exists a bijection f : A N). → 2. A countable A = x : n N (repetition allowed, so that A can be finite). ⇐⇒ { n ∈ } 3. A countable, B A B countable. ⊂ ⇒ Theorem 0.11. If the sets A are countable n N, then n ∀ ∈

An is countable. n[∈N (”countable union of countable sets is countable”.)

Proof. We may assume that A = n N. Since A is countable, we may write A = n 6 ∅ ∀ ∈ n n x (n): m N . Define a mapping { m ∈ } g : N N A , g(n,m)= x (n). × →∪n n m Then g is a surjection N N A . Hence it suffices to find a surjection h: N N N, because × →∪n n → × then g h: N A ◦ → n n[∈N is surjective and therefore A is countable. An example of a surjection h: N N N is: ∪n n → × (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) =h(1) =h(3) =h(6) =h(10) =h(15) · · · ր ր ր ր (2, 1) (2, 2) (2, 3) (2, 4) =h(2) =h(5) =h(9) =h(14) ր ր ր (3, 1) (3, 2) (3, 3) =h(4) =h(8) =h(13) ր ր (4, 1) (4, 2) =h(7) =h(12) ր (5, 1) =h(11) . . 6 Measure and integral

Corollary. The set of all rational numbers m Q = n,m Z, n = 0 { n | ∈ 6 } is countable. Reason: The set m A = n,m Z, n = 0, m k, n k k { n | ∈ 6 | |≤ | |≤ } is ﬁnite (and hence countable) k N. Theorem 0.11 Q = A countable. ∀ ∈ ⇒ ∪k∈N k Example. (Uncountable set). The interval [0, 1] (and hence R) is uncountable. Idea: x [0, 1] x has a decimal expansion ∈ ⇒

x = 0, a1a2a3 ..., where a 0, 1, 2,..., 9 . j ∈ { } Contrapositive: [0, 1] is countable, so [0, 1] = x : n N . Points x have decimal expansions { n ∈ } n (1) (1) (1) x1 = 0, a1 a2 a3 . . . (2) (2) (2) x2 = 0, a1 a2 a3 . . . (3) (3) (3) x3 = 0, a1 a2 a3 . . . . . (n) (n) (n) (n) xn = 0, a1 a2 a3 . . . an . . . . .

(1) (2) (3) (n) (n) On the ”diagonal” there is a sequence a1 , a2 , a3 , . . . , an ,..., where an is the nth decimal of x . Let x [0, 1] be defined by x = 0, b b b ..., where n ∈ 1 2 3 (n) (n) an + 2, if an 0, 1, 2,..., 7 , (0.12) bn = (n) (n) ∈ { } (an 2, if an 8, 9 . − ∈ { } The nth decimal of x satisfies b an = 2 n N, and therefore x = x n N. This is a | n − n| ∀ ∈ 6 n ∀ ∈ contradiction, because [0, 1] = x : n N . Hence [0, 1] is uncountable. { n ∈ } [Note: A decimal expansion need not be unique: for instance, 0, 5999 . . . = 0, 6000 . . .. However, (n) this makes no harm, because in (0.12) b = an 2.] n ± Infinite sums. Let = be an arbitrary index set and a 0 α . Question: What does the sum A 6 ∅ α ≥ ∀ ∈A

aα αX∈A mean? Deﬁne a = sup a ﬁnite . α { α |A0 ⊂A } αX∈A αX∈A0 We will return to this a bit later. Spring 2017 7

0.13 Euclidean space Rn

n times Rn = R R Cartesian product ×···× The elements are called points or vectorsz }|. {

x Rn x = (x ,...,x ), x R, j = 1,...,n. ∈ ⇐⇒ 1 n j ∈ Algebraic structure. The sum of points x,y Rn is ∈ x + y = (x + y ,...,x + y ) Rn. 1 1 n x ∈ The product of a real number λ R and a point x Rn is ∈ ∈ λx = (λx , . . . , λx ) Rn. 1 n ∈ Zero vector 0= 0¯ = (0,..., 0). The inverse element (point) of x Rn is ∈ x = ( 1)x = ( x ,..., x ). − − − 1 − n The diﬀerence of x Rn and y Rn is ∈ ∈ x y = x + ( y). − − In Rn the addition and multiplication by a real number satisfy the axioms of a vector space, for example

x + y = y + x, x +0=0+ x = x, λ(x + y)= λx + λy, (λ + µ)x = λx + µx etc x,y Rn, λ, µ R. ∀ ∈ ∈ The inner product of x,y Rn is ∈ n x y = x y R. · i i ∈ Xi=1 Denote n 1/2 x = √x x = xixi norm of x. | | · ! Xi=1 The Euclidean distance in Rn. The distance between x,y Rn is ∈ n 1/2 x y = x y 2 . | − | i − i i=1 ! X 8 Measure and integral

Often we write d(x,y) = x y . Then d is a metric in Rn, i.e. the mapping d: Rn Rn R | − | × → satisﬁes the axioms of a metric:

d(x,y) 0 x,y Rn ≥ ∀ ∈ d(x,y) = 0 x = y ⇐⇒ d(x,y)= d(y,x) x,y Rn ∀ ∈ d(x,y) d(x, z)+ d(z,y) x,y,z Rn (triangle inequality, -ie). ≤ ∀ ∈ △ Open sets and closed sets in Rn. The Euclidean metric d determines open and closed sets of Rn (and hence the topology of Rn) as follows: Let x Rn and r> 0. The set ∈ B(x, r)= y Rn : y x < r { ∈ | − | } is an open ball with the center x and radius r and

S(x, r)= y Rn : y x = r { ∈ | − | } is the sphere (centered at x and with radius r. Similarly,

B¯(x, r)= y Rn : y x r { ∈ | − |≤ } is a closed ball (centered at x with radius r). A set V Rn is open if x V r = r(x) > 0 such that B(x, r) V . ⊂ ∀ ∈ ∃ ⊂ A set V Rn is closed is Rn V is open. ⊂ \ S(x,r) r −|x − y| > 0

r B(x,r) y x

Example. 1. B(x, r) is open x Rn,r > 0 ( -ie, see the picture above). ∀ ∈ △ 2. A closed ball B¯(x, r) is a closed set.

3. Rn and are both open and closed. ∅ 4. A half open interval, e.g. [0, 1), is neither open nor closed. Remark. The closure of a set A Rn is ⊂ A = x Rn : x A or x is an accumulation (or a cluster) point of A . { ∈ ∈ } Recall that x Rn is an accumulation point of A Rn if r> 0 B(x, r) (A x ) = . In Rn it ∈ ⊂ ∀ ∩ \ { } 6 ∅ holds that B¯(x, r)= B(x, r). Remark. If (X, d) is a metric space, i.e. d: X X R satisﬁes the axioms of a metric, we can × → deﬁne open and closed sets of X by using th emetric d as in the case of Rn by replacing y x | − | with the metric d(x,y). Spring 2017 9

The following result holds in general: Theorem 0.14.

(0.15) V Rn open α (arbitrary index set) V open; α ⊂ ∀ ∈A ⇒ α α[∈A

(0.16) V Rn closed α V closed; α ⊂ ∀ ∈A⇒ α α\∈A

k (0.17) V ,...,V Rn open V open; 1 k ⊂ ⇒ j j\=1

k (0.18) V ,...,V Rn closed V closed. 1 k ⊂ ⇒ j j[=1 Proof. (0.15):

x V α s.t. x V , ∈ α ⇒∃ 0 ∈A ∈ α0 α[∈A V open open ball B(x, r) V V . α0 ⇒∃ ⊂ α0 ⊂ α α[∈A (0.16):

V closed α V c open α α ∀ ⇒ α ∀ (0.15) de Morgan = V c = V c open ⇒ α α α α [ \ V closed. ⇒ α α \ (0.17) and (0.18): (Exerc.).

Remark. ∞ V open j N V open, j ∀ ∈ 6⇒ j j\=1 ∞ V closed j N V closed. (Exerc.) j ∀ ∈ 6⇒ j j[=1 1 Lebesgue measure in Rn

1.1 Introduction A geometric starting point: If I = [a, b] R is a bounded interval, its length is ⊂ ℓ(I)= b a. − 10 Measure and integral

(Similarly if I is an open or half open interval.) A set I Rn is an n-interval if it is of the form ⊂ I = I I , 1 ×···× n where each I R is an interval (either open, closed, or half open). j ⊂

a1 b1

An n-interval I is an open (respectively closed) n-interval if each Ij is open (resp. closed). Let Ij has the end points aj, bj; aj < bj. Then the geometric measure of I is n ℓ(I) = (b a )(b a ) (b a )= (b a ) 1 − 1 2 − 2 · · · n − n j − j jY=1 (n = 1 length, n = 2 area, n = 3 volume). Define ℓ( ) = 0. ∅ Our goal would be to define a ”measure” as a mapping m : (Rn) [0, + ], n P → ∞ such that it satisfies the conditions: (1) m (E) is defined E Rn and m (E) 0. n ∀ ⊂ n ≥ (2) If I is an n-interval, then mn(I)= ℓ(I). (3) If (E ) is a sequence of disjoint subsets of Rn (i.e. E E = if j = k), then k j ∩ k ∅ 6 ∞ m ∞ E = m (E ) countably additivity. n ∪k=1 k n k k=1 X (4) mn is translation invariant, i.e.

mn(E + x)= mn(E), where E Rn, x Rn, and E + x = y + x y E . ⊂ ∈ { | ∈ } It turns out that there exists no such mapping that would satisfy all the conditions (1) – (4) simultaneously. In the case of the (n-dimensional) Lebesgue measure mn we drop the condition (1). Hence m : Leb Rn [0, + ], n → ∞ will be a mapping that satisﬁes the conditions (2), (3) and (4), where Leb Rn ( (Rn) P is the family of Lebesgue measurable sets. The family Leb Rn contains, for instance, all open and closed subsets of Rn. Spring 2017 11

1.2 The Lebesgue outer measure in Rn

Convention.

a + = + a = , a = ∞ ∞ ∞ 6 −∞ a = + a = , a = −∞ −∞ −∞ 6 ∞ , + not deﬁned ∞−∞ −∞ ∞ ( )= , ( )= − ∞ −∞ − −∞ ∞

, a> 0 ∞ a = a = , a< 0 ∞ · ·∞ −∞ 0, a = 0 Note! 0 = 0 ·∞  , a> 0 −∞ ( )a = a( )= + , a< 0 −∞ −∞  ∞ 0, a = 0

 = ( )( )= ∞·∞ −∞ −∞ ∞ ( ) = ( )= −∞ ∞ ∞ −∞ −∞

, a> 0 a ∞ = , a< 0 0 −∞ not deﬁned , a = 0 a a =  = 0, a R ∈ ∞ −∞ ±∞ not deﬁned ±∞

Recall: If (a ) N is a sequence such that a 0 j, then either j j∈ j ≥ ∀ ∞ k ∞ aj = lim aj R or aj =+ . k→∞ ∈ ∞ Xj=1 Xj=1 Xj=1 k Reason: partial sums j=1 aj form an increasing sequence. Let A Rn. Consider countable open covers of A (possibly ﬁnite) ⊂ P = I ,I ,... , F { 1 2 } where each I Rn is a bounded open n-interval (or ) and k ⊂ ∅ ∞ A I . ⊂ k k[=1 12 Measure and integral

Then we say that is a Lebesgue cover of A. We form a series F ∞ S( )= ℓ(I ), 0

m∗ (A) = inf S( ): is a Lebesgue cover of A . n { F F } (Later we will prove that closed n-intervals would work as well.) Remark. 1. Denote J = x Rn : x < k j (open n-interval). Clearly k { ∈ | j| ∀ } ∞ n R = Jk, k[=1 and therefore always there exist open covers ∞ I A (and hence inf exists). ∪k=1 k ⊃ 2. I Rn open n-interval 0 ℓ(I ) < the sum is well-deﬁned and k ⊂ ⇒ ≤ k ∞ ⇒ ∞ 0 ℓ(I ) + . ≤ k ≤ ∞ Xk=1

3. The outer measure mn(A) depends (of course) on the dimension n. If n is clear from the ∗ ∗ context, we abbreviate m (A)= mn(A). 4. It follows directly from the deﬁnition that ε > 0 there exists a Lebesgue cover of A ∀ F (usually depending on ε) such that

S( ) m∗(A)+ ε. F ≤ (We allow m∗(A)=+ .) Note that it is usually not possible to ﬁnd a Lebesgue cover of ∞ F A for which m∗ (A)= S( ). n F 5. Thus A m∗(A) is a mapping (Rn) [0, ], in particular, m∗ is deﬁned in the whole 7→ P → ∞ (Rn). P Example. 1. Let n = 2 and let A = (x, 0): a x b R2 (a line segment in the plane). ∗ { ≤ ≤ }⊂ Claim: m2(A) = 0. Proof: Let ε> 0 and I =]a ε, b + ε[ ] ε, ε[ R2 an open 2-interval. ε − × − ⊂ A I 0 m∗(A) ℓ(I ) = 2ε(b a + 2ε) ε→0 0, ⊂ ε ⇒ ≤ 2 ≤ ε − −−−→ ∗ hence m2(A) = 0. Spring 2017 13

2. Let n = 1. Consider the set of rational numbers Q R. ∗ ⊂ Claim: m1(Q) = 0. Proof Since Q is countable, we may write Q = q : j N . Let ε> 0 be arbitrary. For each { j ∈ } j N let ∈ ε ε I = q ,q + R j j − 2j+1 j 2j+1 ⊂ j+1 j be an open interval. Its length is ℓ(Ij) = 2ε/2 = ε/2 .

q I j N Q I j ∈ j ∀ ∈ ⇒ ⊂ j ⇒ [j ∞ ∞ ∞ ε 1 ε→0 0 m∗(Q) ℓ(I )= = ε = ε 0, ≤ 1 ≤ j 2j 2j −−−→ Xj=1 Xj=1 Xj=1

∗ hence m1(Q) = 0. 3. Similarly, A Rn countable m∗ (A) = 0. ⊂ ⇒ n 4. Let A Rn be a bounded set, that is R> 0 such that A B(0,R). Then A I, where ⊂ ∃ ⊂ ⊂ n times I = ] R,R[ ] R,R[ open n-interval. − ×···× − R

R A

We get an estimate m∗(A) ℓ(I) = (2R)n. ≤ Basic properties of the (Lebesgue) outer measure.

Theorem 1.3. (1) m∗ ( ) = 0; n ∅ (2) ”monotonicity”: A B m∗ (A) m∗ (B); ⊂ ⇒ n ≤ n (3) ”subadditivity”: A , A ,... Rn 1 2 ⊂ ⇒ ∞ ∞ m∗ A m∗ (A ). n j ≤ n j j=1 j=1 [ X Remark. (3) holds also for ﬁnite unions k (A ) (choose A = = ). ∪j=1 j k+1 · · · ∅ Proof. (1): Clear. 14 Measure and integral

(2): Let be a Lebesgue cover of B. F A B is also a Lebesgue cover of A deﬁnition= m∗ (A) S( ). ⊂ ⇒ F ⇒ n ≤ F Take the inf over all Lebesgue covers of B m∗ (A) m∗ (B). ⇒ n ≤ n (3): Denote A = A . Let ε > 0. For each j choose a Lebesgue cover = I ,I . . . of A ∪j j Fj { j1 j2 } j such that S( ) m∗ (A )+ ε/2j . Fj ≤ n j Now = = I : j N, k N is a Lebesgue cover of A, hence (by deﬁnition) F j Fj { jk ∈ ∈ } S ∞ ∞ ∞ ∞ m∗ (A) S( )= S( ) m∗ (A )+ ε/2j = m∗ (A )+ ε. n ≤ F Fj ≤ n j n j Xj=1 Xj=1 Xj=1 Xj=1 Letting ε 0 we get the claim. →

Remark. Above we need some facts on ”summing” (more precisely, why S( ) = ∞ S( ))? F j=1 Fj See Lemma 1.7 and 1.8 below. P Theorem 1.4. Let A Rn. Then ⊂ ∗ ∗ (1.5) mn(A + x)= mn(A) for all x Rn, where A + x = y + x: y A ; ∈ { ∈ } ∗ n ∗ (1.6) mn(tA)= t mn(A), whenever t> 0 and tA = ty : y A . { ∈ } Proof (Exerc.) On summing. Let I be an (index) set and a 0 i I. If J I is ﬁnite, we denote i ≥ ∀ ∈ ⊂

SJ = ai , S∅ = 0. Xi∈J Definition. a = sup S : J I finite . i { J ⊂ } Xi∈I Lemma 1.7. n ai = lim ai. n→∞ Xi∈N Xi=1 That is, this ”new” definition coincide with the usual one (for countable sums).

Proof Denote J = 1,...,n , S = a (= sup S : J N ﬁnite ). n { } i∈N i { J ⊂ } P ′ SJ increasing sequence lim SJ = S n ⇒ ∃ n→∞ n S S S′ S. Jn ≤ ⇒ ≤ Spring 2017 15

On the other hand,

J N ﬁnite n N s.t. J J ⊂ ⇒ ∃ ∈ ⊂ n S S S′ ⇒ J ≤ Jn ≤ S S′ (taking sup over J). ⇒ ≤ ∀

Next both I and J are arbitrary index sets (i.e. they may be uncountable). (In addition, we abbreviate aij = a(i,j).) Lemma 1.8. aij = aij = aij. (i,jX)∈I×J Xi∈I Xj∈J Xj∈J Xi∈I Proof Denote by Svas the sum on the left hand side, by Skes the sum in the middle, and by Soik the sum on the right hand side. (a): If I J is finite, then finite I′ I, J ′ J s.t. I′ J ′ A⊂ × ∃ ⊂ ⊂ A⊂ × (∗) SA SI′×J′ = aij aij Skes ⇒ ≤ ′ ′ ≤ ′ ≤ Xi∈I jX∈J Xi∈I Xj∈J S S (taking sup over ). ⇒ vas ≤ kes ∀ A [( ): there is only finitely many terms in S ′ ′ , so the order of summing does not matter.] ∗ I ×J (b): Let I′ I be finite and J ′ J be finite i I′. Denote ⊂ i ⊂ ∀ ∈ = (i, j): i I′, j J ′ . A { ∈ ∈ i} Then S S = a . vas ≥ A ij i∈I′ ′ X jX∈Ji Take ( i I′) the sup over finite J ′ J ∀ ∈ i ⊂

Svas aij ≥ ′ Xi∈I Xj∈J sup over ﬁnite I′ I S S . ⊂ ⇒ vas ≥ kes

Similarly, Svas = Soik. Corollary 1.9. aij = aij = aij . (i,jX)∈N×N Xi∈N Xj∈N Xj∈N Xi∈N Remark. The subadditivity does not (in general) hold in the form

(1.10) m∗ A m∗ (A ), n i ≤ n i i∈I i∈I [ X where A Rn, i I, and I is an uncountable index set. Reason: i ⊂ ∈ n ∗ n R = x , mn( x ) = 0 x R . n{ } { } ∀ ∈ x[∈R 16 Measure and integral

If (1.10) would hold, then

(1.10) 0 m∗ (Rn)= m∗ x m∗ ( x ) = 0. ≤ n n { } ≤ n { } x∈Rn x∈Rn [ X On the other hand, we will prove later that m∗ (Rn)=+ . This is a contradiction, so (1.10) does n ∞ not hold!

1.11 (Lebesgue )measurable sets We will deﬁne the (Lebesgue) measurable sets of Rn, denoted by Leb Rn, by using so-called Carath´eodory’s condition. Recall the subadditivity (Theorem 1.3 (3)): A, B Rn ⊂ ⇒ m∗(A B) m∗(A)+ m∗(B). ∪ ≤ Later we will prove that A, B Rn s.t. A B = , but ∃ ⊂ ∩ ∅ m∗(A B)

A E

Definition. (Carathéodory’s condition, 1914.) A set E Rn is (Lebesgue) measurable if ⊂ m∗(A)= m∗(A E)+ m∗(A E ) for all A Rn. ∩ \ ⊂ =A∩Ec Remark. E Rn measurable | {z } ⊂ ⇐⇒ m∗(A) m∗(A E)+ m∗(A E) for all A Rn, with m∗(A) < . ≥ ∩ \ ⊂ ∞ Reason: follows from the subadditivity and holds always if m∗(A)=+ . ≤ ≥ ∞ Definition. If E Rn is measurable, we denote ⊂ ∗ m(E)= m (E) or mn(E) if needed. m(E) is the (n-dimensional Lebesgue) measure of E. We write Leb Rn = E Rn : E Lebesgue measurable (Rn). { ⊂ } ⊂ P Spring 2017 17

Hence

m = m∗ Leb Rn : Leb Rn [0, ], restriction of the outer measure. | → ∞ Later we will show that Leb Rn ( (Rn). P Theorem 1.12. m∗(E) = 0 E measurable. ⇒ Proof. Let A Rn be an arbitrary test set. ⊂ monotonicity A E E = m∗(A E) = 0 ∩ ⊂ ⇒ ∩ monotonocity A A E = m∗(A) m∗(A E)= m∗(A E) +m∗(A E) ⊃ \ ⇒ ≥ \ ∩ \ =0 E measurable. ⇒ | {z }

Theorem 1.13. E measurable Ec measurable. ⇐⇒ Proof. It s enough to show : Let E be measurable and A Rn. Then ⇒ ⊂ m∗(A)= m∗(A E)+ m∗(A Ec) ∩ ∩ = m∗ A (Ec)c + m∗(A Ec) ∩ ∩ Ec measurable. ⇒

Example.

Ex. 3 E Rn countable = m∗(E) = 0 ⊂ ⇒ Thm.= 1.12 E measurable Thm.= 1.13 Ec measurable. ⇒ ⇒ Special cases:

Leb R, R Leb R, ∅∈ ∈ rational numbers Q Leb R, irrational numbers R Q Leb R. ∈ \ ∈

Let E1, E2,... be measurable. We will prove that

∞ ∞ Ei and Ei are measurable. i[=1 i\=1 To prove these statements we need some auxiliary lemmata. First the case of a ﬁnite union/intersection:

Lemma 1.14. E ,...,E measurable k E and k E measurable. 1 k ⇒ i=1 i i=1 i S T 18 Measure and integral

Proof. (a) union: k k−1 Ei = Ei Ek ! ∪ i[=1 i[=1 we may assume k = 2. ⇒ Suppose E and E are measurable. Let A Rn be a test set. 1 2 ⊂ E measurable 1 ⇒ m∗(A)= m∗(A E )+ m∗(A Ec) ∩ 1 ∩ 1   = c  ⇒ E2 measurable, with test set A E1  ∗ c ∗ c ∗ ∩ c⇒ c m (A E1)= m (A E1 E2)+ m (A E1 E2) ∩ ∩ ∩ ∩ ∩    m∗(A)= m∗(A E )+ m∗(A Ec E ) +m∗(A Ec Ec), ∩ 1 ∩ 1 ∩ 2 ∩ 1 ∩ 2 (subadd. ⇒) ≥ m∗(B) where | {z }

B = (A E ) (A Ec E )= A E (Ec E ) = A E (E E ) ∩ 1 ∪ ∩ 1 ∩ 2 ∩ 1 ∪ 1 ∩ 2 ∩ 1 ∪ 2 \ 1 = A (E E ). ∩ 1 ∪ 2 Hence

m∗(A) m∗(B)+ m∗(A Ec Ec) ≥ ∩ 1 ∩ 2 = m∗ A (E E ) + m∗ A (E E )c ∩ 1 ∪ 2 ∩ 1 ∪ 2 E E measurable. ⇒ 1 ∪ 2 B

c A ∩ E1

(b) intersection: de Morgan, Theorem 1.13 (”measurability of the complement”) and part (a)

⇒ k k c c Ei = Ei measurable. ! i\=1 i[=1

Theorem 1.15. E , E measurable E E measurable. 1 2 ⇒ 1 \ 2 Proof. E E = E Ec. 1 \ 2 1 ∩ 2 Spring 2017 19

Lemma 1.16. Let E ,...,E be disjoint and measurable, and let A Rn be an arbitrary set. 1 k ⊂ Then k k m∗ A E = m∗(A E ). ∩ i ∩ i i=1 i=1 [ X

E4 E1

Proof. (a) The case k = 2 : E measurable, A (E E )= B as the test set 1 ∩ 1 ∪ 2 ⇒ m∗(B)= m∗(B E )+ m∗(B E ) ∩ 1 \ 1 =A∩E1 =A∩E2 | {z } | {z } = m∗(A E )+ m∗(A E ) i.e. the claim. ∩ 1 ∩ 2 (b) general case: By induction: Suppose that the claim holds for 2 k p, that is ≤ ≤

E1,...,Ep measurable p p E E = , i = j m∗ A E = m∗(A E ). i ∩ j ∅ 6  ⇒ ∩ i ∩ i A Rn i=1 i=1 ⊂  [ X Thus we get (for k = p + 1)  p+1 p A i=1 Ei = A i=1 Ei Ep+1 ∩ ∩ ∪ = p S S  ⇒ i=1 Ei, Ep+1 disjoint and measurable  p+1 p S k=2  m∗ A E = m∗ A E + m∗(A E ) ∩ i ∩ i ∩ p+1 i=1 i=1 [ p [ k=p = m∗(A E )+ m∗(A E ) ∩ i ∩ p+1 Xi=1 p+1 = m∗(A E ). ∩ i Xi=1

∞ Lemma 1.17. Let E = i=1 Ei, where the sets Ei are measurable. Then there exist disjoint and measurable sets Fi Ei s.t. ⊂ S ∞ E = Fi. i[=1 20 Measure and integral

Proof. Choose

F1 = E1, [measurable] F = E E , [measurable (Thm. 1.15)] 2 2 \ 1 . . k−1 F = E E , [measurable (Thm. 1.15 and L. 1.14)] k k \ i i[=1 . .

F2 = E1 = F1 E2

E3 F3 =

Then clearly ∞ F E i, E = F and F F = i = j. i ⊂ i ∀ i i ∩ j ∅ ∀ 6 i[=1

The main result of Lebesgue measurable sets

Theorem 1.18. Let E1, E2,... be a sequence (possibly ﬁnite) of measurable sets. Then the sets

Ei and Ei [i \i are measurable. If, in addition, the sets Ei are disjoint, then

(1.19) m Ei = m(Ei). (”countably additivity”) i i [ X Proof. Denote 1.17 S = Ei = Fi, Fi measurable and disjoint, [i [i k S = F , S S. k i k ⊂ [i L. 1.14 (measurability of ﬁnite unions) S measurable. Let A be a test set. Then ⇒ k m∗(A)= m∗(A S )+ m∗(A S ) ∩ k \ k monot. m∗(A S )+ m∗(A S) ≥ ∩ k \ k 1=.16 m∗(A F )+ m∗(A S) k N. ∩ i \ ∀ ∈ Xi=1 Spring 2017 21

Letting k we get →∞ ∞ (1.20) m∗(A) m∗(A F )+ m∗(A S) ≥ ∩ i \ Xi=1 subadd. m∗ ∞ (A F ) + m∗(A S) ≥ ∪i=1 ∩ i \ = m∗(A S)+ m∗(A S) ∩ \ S = E measurable. ⇒ i [i Inequality (1.20), in the case A = S, and the subadditivity ⇒ =0 ∞ ∞ =Fi ∞ subadd. (1.20) m(F ) m(S) m∗(S F )+ m∗(S S)= m(F ). i ≥ ≥ ∩ i \ i Xi Xi=1 z }| { z }| { Xi=1

If Ei are disjoint, we may choose Fi = Ei, and therefore (1.19) holds. c c The ﬁrst part of the proof and Thm. 1.13 imply that i Ei = i Ei is measurable. T S

Example. Let A R2 s.t. ⊂ (1.21) m∗ A B(x, r) x r3 x R2, r> 0. ∩ ≤ | | ∀ ∈ ∀ Claim: m(A) = 0 Proof. (a) Suppose ﬁrst that A is bounded, so A Q = [ a, a] [ a, a] (closed square) for ⊂ − × − some a. Let n N. Devide Q into closed (sub-)squares Q , with side length = 2a/n, j = 1,...,n2. ∈ j Let xj be the center of Qj. Then

x 2a and Q B(x , 2a/n) (rough estimates) | j|≤ j ⊂ j monot. (1.21) m∗(A Q ) m∗ A B(x , 2a/n) x (2a/n)3 (2a)4n−3. ⇒ ∩ j ≤ ∩ j ≤ | j| ≤ n2 subadd. A = (A Q ) = ∩ j ⇒ j[=1 n2 n2 m∗(A)= m∗ (A Q ) m∗(A Q ) ∩ j ≤ ∩ j j=1 j=1 [ X n2(2a)4n−3 = (2a)4n−1 n ≤ ∀ n=→∞ m∗(A) = 0 m(A) = 0. ⇒ ⇒ (b) General case:

A = A , where A = A B(0, j) bounded. j j ∩ j[∈N (a) A A A satisﬁes the assumption (1.21) m(A ) = 0 j j ⊂ ⇒ j ⇒ j ∀ subadd. = m(A) = 0. ⇒ 22 Measure and integral

1.22 Examples of measurable sets So far we know that: m∗(A) = 0 A and Ac measurable. ⇒ Now we will prove that, for example, open sets and closed sets are measurable. First:

I Rn n-interval (open, closed, etc.) I is measurable and m(I)= ℓ(I). ⊂ ⇒ We use (Riemann) integration: Let I = I I Rn n-interval, where I R is an interval, with end points a < b , j = 1 ×···× n ⊂ j ⊂ j j 1,...,n. Let χ : Rn 0, 1 (the characteristic function of I) I → { } 1, x I χI (x)= ∈ (0, x I . 6∈ Choose an n-interval Q I and (Riemann) integrate ⊃

b1 bn χ = 1 dx dx = (b a ) (b a )= ℓ(I). I · · · 1 · · · n 1 − 1 · · · n − n ZQ Za1 Zan Lemma 1.23. Let I and I ,...,I be n-intervals s.t. I k I . Then ℓ(I) k ℓ(I ). If, 1 k ⊂ j=1 j ≤ j=1 j furthermore, the intersections Ii Ij, i = j, do not have interior points (i.e. no Ii Ij, i = j, ∩k 6 k S P∩ 6 contains an open ball) and I = j=1 Ij, then ℓ(I)= j=1 ℓ(Ij). Proof. Deﬁne χ,χ : Rn 0, 1S, P j → { }

1, x I 1, x Ij χ(x)= ∈ and χj(x)= ∈ 0, x I 0, x I . ( 6∈ ( 6∈ j Then it follows from the assumption I k I that χ(x) k χ (x) x Rn. Choose an ⊂ j=1 j ≤ j=1 j ∀ ∈ n-interval Q that contains all the n-intervals mentioned above and (Riemann) integrate over Q S P

ℓ(I)= χ χ = χ = ℓ(I ). ≤ j j j ZQ ZQ j j ZQ j X X X k If the n-intervals Ij do not have common interior points, then χ(x)= j=1 χj(x) except possible on the boundaries of n-intervals that do not contribute to the integrals. P Lemma 1.24. If I is an n-interval, then

m∗(I)= ℓ(I).

Proof. (a): ε> 0 an open n-interval J I s.t. ℓ(J) <ℓ(I)+ ε. ∀ ∃ ⊃ J Leb. cover of I m∗(I) ℓ(I)+ ε { } ⇒ ≤ ε> 0 arbitr. m∗(I) ℓ(I). ⇒ ≤ Spring 2017 23

(b): Suppose ﬁrst that I is closed. Let be a Lebesgue cover of I. Since I is closed and bounded, F I is compact. So a ﬁnite subcover = I ,...,I . Lemma 1.23 ∃ F0 { 1 k} ⊂ F ⇒ ℓ(I) S( ) S( ) ≤ F0 ≤ F inf over ℓ(I) m∗(I). ∀F ⇒ ≤ Hence: ℓ(I) = m∗(I) if I is closed. Suppose then that I need not be closed. Let ε > 0. Now a ∃ closed n-interval I I s.t. ℓ(I ) >ℓ(I) ε. Thus c ⊂ c − monot. m∗(I) m∗(I )= ℓ(I ) >ℓ(I) ε ≥ c c − ε> 0 arbitr. m∗(I) ℓ(I). ⇒ ≥

n Remark. The above holds also for degenerate n-intervals I = I1 In R , where at least def. ∗ ×···× ⊂ one Ij is a singleton. Then ℓ(I) = 0= mn(I). Let A Rn, ε> 0 and let J , J ,... Rn be arbitrary n-intervals s.t. A ∞ J . For each ⊂ 1 2 ⊂ ⊂ i=1 i i open n-interval I J s.t. ℓ(I ) <ℓ(J )+ ε/2i. Now I ,I . . . is a Lebesgue cover of A, and ∃ i ⊃ i i i { 1 2 } S therefore m∗(A) ∞ ℓ(I ) ∞ ℓ(J )+ ε. (Recall a geometric series.) It follows that ≤ i=1 i ≤ i=1 i P P∞ ∞ m∗(A) = inf ℓ(J ): A J , J arbitrary n-interval . i ⊂ i i i=1 i=1 X [ Theorem 1.25. If I is an n-interval, then I is measurable and

m(I)= ℓ(I).

Proof. L. 1.24 it suﬃces to prove that I is measurable. Let A Rn be a test set. Claim: ⇒ ⊂ m∗(A) m∗(A I)+ m∗(A I). ≥ ∩ \ Let ε> 0. Then a Lebesgue cover of A by open n-intervals = I ,I ,... s.t. ∃ F { 1 2 } S( ) m∗(A)+ ε. F ≤

I =∆ ∆ 1 ×···× n  ⇒ I =]a , b [ ]a , b [ j 1 1 ×···× n n  ′  n-interval Ij Ij I = ]a1, b1[ ∆1 ]an, bn[ ∆n = ∩ ∩ ×···× ∩ ( . ∅ I I is not necessarily an n-interval but j \ I I = I′′ j \ j,k [k is a ﬁnite union of n-intervals s.t. the intersections I′ I′′ and I′′ I′′ , k = i, do not have interior j ∩ j,k j,k ∩ j,i 6 points. 24 Measure and integral

A ′′ Ij,1 ′′ Ij,2 Ij ′ Ij

Lemma 1.23 and 1.24 ⇒ 1.23 ′ ′′ 1.24 ∗ ′ ∗ ′′ ℓ(Ij) = ℓ(Ij)+ ℓ(Ij,k) = m (Ij)+ m (Ij,k). Xk Xk Taking the sum over j ⇒ m∗(A)+ ε S( )= ℓ(I )= m∗(I′ )+ m∗(I′′ ) ≥ F j j j,k Xj Xj Xj Xk subadd. m∗ I′ + m∗ I′′ ≥ j j,k [j [j,k ⊃A∩I ⊃A\I monot. m∗(A| {z I})+ m∗(A| {zI).} ≥ ∩ \ Letting ε 0 m∗(A) m∗(A I)+ m∗(A I). → ⇒ ≥ ∩ \ Theorem 1.26. (Lindel¨of’s theorem) Let A Rn be an arbitrary set and ⊂ V A, α ⊃ α[∈A where the sets V Rn, α are open. Then there exists a countable sub-cover α ⊂ ∈A V A. αj ⊃ j[∈N Proof. Exerc.

Theorem 1.27. Open subsets and closed subsets of Rn are measurable. Proof. (a) Let A be open. If x A, an open n-interval I(x) s.t. x I(x) A ( an open ball ∈ ∃ ∈ ⊂ ∃ B(x, r ) A and it contains an open n-interval). x ⊂ I(x): x A is an open cover of A. { ∈ } Lindel¨of countable sub-cover I(x ): j N ⇒ ∃ { j ∈ } A = I(x ) is a countable union of measurable sets ⇒ j j[∈N A is measurable. ⇒ (b) If A is closed, its complement Ac is open and hence measurable A = (Ac)c is measurable. ⇒ Spring 2017 25

Example. Let f : R2 R2 be continuous. Claim: fR2 is measurable. → Proof.

2 R = Aj, where Aj = B¯(0, j) si compact j[∈N f continuous fA compact ⇒ j fA closed fA measurable ⇒ j ⇒ j fR2 = fA fR2 measurable. j ⇒ j[∈N

Recall: Let n,m 1. A mapping f : Rn Rm is continuous f −1U Rn is open open ≥ → ⇐⇒ ⊂ ∀ U Rm. ⊂ Rn Rm f

− U f 1U

If f : Rn Rm is continuous and C Rn is compact, then fC Rm is compact. Reason: → ⊂ ⊂ fC U open cover ⊂ i i[∈I C f −1U open cover ⇒ ⊂ i∈I i ﬁnite sub-cover S  ⇒ ∃ C compact  k k C f −1U fC U . ⊂ ij ⇒ ⊂ ij j[=1 j[=1 More general measurable sets, σ-algebras.

sets F , F closed (e.g. Q, [a, b), (a, b]) Fσ i i i[∈N sets G , G open (e.g. R Q, [a, b), (a, b]) Gδ i i \ i\∈N sets A , A Fσδ j j ∈ Fσ i\∈N sets B , B Gδσ j j ∈ Gδ i[∈N etc.

Deﬁnition. Let X be an arbitrary set. A family Γ (X) is a σ-algebra (”sigma-algebra”) of X ⊂ P if 26 Measure and integral

(a) Γ; ∅∈ (b) A Γ X A Γ; ∈ ⇒ \ ∈ (c) A Γ, i N ∞ A Γ. i ∈ ∈ ⇒ i=1 i ∈ Remark. (1) If Γ is aSσ-algebra and A Γ, i N, then also A Γ since i ∈ ∈ i i ∈ T A = Ac c = Ac c Γ. i i i ∈ i i i=1 \ \ [

(2) We have proved: The family of Lebesgue measurable sets Leb Rn is a σ-algebra of Rn (The- orems 1.12, 1.13, 1.18).

(3) (X) is the largest σ-algebra of X; , X is the smallest σ-algebra of X; A X (ﬁxed) P {∅ } ⊂ , X, A, Ac is a σ-algebra of X. ⇒ {∅ } Deﬁnition. The family of Borel sets Bor Rn is the smallest σ-algebra of Rn that contains all closed sets.

Existence: Denote

= Γ: Γ is a σ-algebra of Rn, Γ contains closed sets . B { } \ (For instance Γ = (Rn) is a σ-algebra of Rn that contains all closed sets.) P is a σ-algebra since: B (a) ; ∅∈B (b) A Ac Γ Γ Ac ; ∈B ⇒ ∈ ∀ ⇒ ∈B (c) A A Γ Γ A . i ∈B ⇒ i∈N i ∈ ∀ ⇒ i∈N i ∈B The construction S is the smallestS σ-algebra of Rn that contains closed sets, and so ⇒ B Bor Rn = . B Open sets, closed sets, sets, sets, etc. are Borel sets. Fσ Gδ Theorem 1.28. Every Borel sets is measurable.

Proof. The family of measurable sets Leb Rn is a σ-algebra and contains closed sets, and therefore

Bor Rn Leb Rn. ⊂ Spring 2017 27

1.29 General measure theory Deﬁnition. Let Γ be a σ-algebra in X. A function µ: Γ [0, + ] is a measure in X if → ∞ (i) µ( ) = 0; ∅ (ii) A Γ, i N, disjoint µ ∞ A = µ(A ). ”countably additivity” i ∈ ∈ ⇒ i=1 i i∈N i The triple (X, Γ,µ) is a measure spaceS . P

Remark. 1. A measure µ is also monotonic:

A, B Γ, A B 0 µ(A) µ(B). ∈ ⊂ ⇒ ≤ ≤ Reason: A, B A Γ disjoint, B = A (B A) \ ∈ ∪ \ µ(B)= µ(A)+ µ(B A) µ(A). ⇒ \ ≥ ≥0 | {z } 2. A, B Γ, A B, µ(A) < µ(B A)= µ(B) µ(A). ∈ ⊂ ∞ ⇒ \ − 3. A measure µ is a probability measure if µ(X) = 1.

Example. (1) n-dimensional Lebesgue measure

m : Leb Rn [0, + ] n → ∞ is a measure. Reason: Leb Rn is a σ-algebra in Rn and m is countably additive.

(2) Let X = be an arbitrary set. Fix x X and deﬁne for all A X 6 ∅ ∈ ⊂ 1, if x A; µ(A)= ∈ (0, if x A. 6∈ Then µ: (X) [0, + ] is a probability measure (so-called Dirac measure at the point P → ∞ x X). ∈ Reason: (a) (X) is σ-algebra. P (b) Let A X, j N, be disjoint. Then j ⊂ ∈ ∞ ∞ µ Aj = µ(Aj) j=1 j=1 [ X since

x ∞ A both sides = 0 6∈ j=1 j ⇒  S ∞ disjoint  x Aj = exactly one j0 N s.t. x Aj0 both sides = 1. ∈ j=1 ⇒ ∃ ∈ ∈ ⇒  S (3) µ: (X) [0, + ], µ(A) = 0 A X, is a measure. P → ∞ ∀ ⊂ 28 Measure and integral

(4) Let a 0, j N, s.t. ∞ a = 1. Define for all A N j ≥ ∈ j=1 j ⊂ P µ(A)= aj. jX∈A Then µ: (N) [0, 1] is a probability measure. P → Definition. Let X be an arbitrary set. A mapping µ∗ : (X) [0, + ] is an outer measure in X P → ∞ if (1) µ∗( ) = 0; ∅ (2) A B µ∗(A) µ∗(B); ⊂ ⇒ ≤ (3) A X, j N µ∗ ∞ A ∞ µ∗(A ). j ⊂ ∈ ⇒ j=1 j ≤ j=1 j Furthermore, a set E X isS (µ∗-)measurable P , if (Carathéodory’s criterion) ⊂ (1.30) µ∗(A)= µ∗(A E)+ µ∗(A E) ∩ \ holds A X. ∀ ⊂ Denote ∗ (X)= E X : E µ∗-measurable Mµ { ⊂ } of (X) is µ∗ is clear from the context. M Remark. (X) (X) is a σ-algebra in X and the restriction M ⊂ P µ∗ (X): (X) [0, + ] |M M → ∞ is a measure. Proof as in the case of Lebesgue measure.

1.31 Convergence of measures Let X = , Γ (X) a σ-algebra, and µ: Γ [0, + ] a measure. 6 ∅ ⊂ P → ∞ Theorem 1.32. Let A Γ, j = 1,..., be an increasing sequence (i.e. A A X j ∈ 1 ⊂ 2 ⊂ ··· ⊂ (µ-)measurable). Then ∞ µ Aj = lim µ(Aj). j→∞ j=1 [ Note: A Γ j N ∞ A Γ. j ∈ ∀ ∈ ⇒ j=1 j ∈ Proof. S ∞ ∞ A = ( A A ), A = (a convention) j j \ j−1 0 ∅ j=1 j=1 [ [ disjoint, measurable A j+2| {z } Aj+1

Aj+1 \ Aj Spring 2017 29

µ countably additive ⇒ ∞ ∞ µ A = µ(A A ) j j \ j−1 j=1 j=1 [ X k = lim µ(Aj Aj−1) k→∞ \ Xj=1 k = lim µ (Aj Aj−1) k→∞ \ j=1 [ =Ak

= lim µ(Ak). k→∞ | {z }

Theorem 1.33. Let A Γ, j = 1,..., be a decreasing sequence (i.e. X A A j ∈ ⊃ 1 ⊃ 2 ⊃ ··· (µ-)measurable). If, in addition, µ(A ) < for some k N, then k ∞ ∈ ∞ µ Aj = lim µ(Aj). j→∞ j=1 \ Note: Γ σ-alg. ∞ A Γ. ⇒ j=1 j ∈ T Proof. We may assume that µ(A ) < . Denote ∞ A = A and B = A A . Then B B 1 ∞ j=1 j j 1 \ j 1 ⊂ 2 ⊂ are measurable. · · · T A1

A3 B2

∞ Theorem 1.32 µ Bj = lim µ(Bj). ⇒ j→∞ j[=1 ∞ ∞ ∞ B = (A A )= A A = A A j 1 \ j 1 \ j 1 \ j[=1 j[=1 j\=1 A = A A A disjoint union µ(A )= µ(A )+ µ(B ) 1 j ∪ 1 \ j ⇒ 1 j j =Bj A = A (A A) disjoint union µ(A )= µ(A)+ µ(A A) 1 ∪ |1 \{z } ⇒ 1 1 \ 30 Measure and integral

µ(A)= µ(A ) µ(A A) (here we need µ(A ) < ) ⇒ 1 − 1 \ 1 ∞ ∞ = µ(A ) µ B 1 − j j=1 [ = µ(A1) lim µ(Bj) − j→∞

= µ(A1) lim µ(A1) µ(Aj) − j→∞ −

= lim µ(Aj). j→∞

Remark. The assumption µ(A ) < for some k N is necessary. Ex. k ∞ ∈ A = (x,y) R2 : x > j j { ∈ } A A A 1 ⊃ 2 ⊃ 3 ⊃··· m (A )= j 2 j ∞ ∀ Aj = m2 Aj = 0 = lim m2(Aj). ∅ ⇒ 6 j→∞ j∈N j∈N \ \ Remark. (An important application for instance in probability theory) Borel-Cantelli lemma: Let (X, Γ,µ) be a measure space, A Γ, j N, and j ∈ ∈ A = x X : x A for inﬁnitely many j N . { ∈ ∈ j ∈ } Then: ∞ µ(A ) < µ(A) = 0. j ∞ ⇒ Xj=1

1.34 Non-(Lebesgue-)measurable set in R Theorem 1.35. (Vitali, 1905) Leb R( (R), P in other words, there exists a subset E R that is not Lebesgue measurable. ⊂ An idea is to ﬁnd a set B R, 0 disjoint sets Ai s.t. m∗(A )= m∗(A ) i. i 1 ∀ Then some Ai must be non measurable. A way to guarantee that the sets Ai have the same outer measure is to choose Ai = A + xi for some (ﬁxed) A R and x R, and use the translation invariance of the outer measure m∗. ⊂ i ∈ Spring 2017 31

Proof. Consider the quotient space R/Q whose elementys are equivalence classes E(x), x R. ∈ E(x)= E(y) x y x y Q. ⇐⇒ ∼ ⇐⇒ − ∈ We may write E(x) = x + Q. Choose from each equivalence class E(x), x R, exactly one ∈ representative that belongs to the unit interval [0, 1]. Let A be the set of such chosen points (representatives). Claim: A Leb R. 6∈ Assume on the contrary: A Leb R. ∈ (i) The sets A + r, r Q, are disjoint since: ∈ x (A + r) (A + s), r,s Q x = a + r and x = a + s, a , a A ∈ ∩ ∈ ⇒ 1 2 1 2 ∈ a a = s r Q ⇒ 1 − 2 − ∈ a a E(a )= E(a ) ⇒ 1 ∼ 2 ⇒ 1 2 a = a (because we choose exactly one representative) ⇒ 1 2 s = r. ⇒ (ii) m(A) = 0 (we use the tranlation invariance: A Leb R A + a Leb R and m(A) = ∈ ⇒ ∈ m(A + a)):

1 A [0, 1] A + [0, 2] n N ⊂ ⇒ n ⊂ ∀ ∈ ∞ ∞ ∞ 1 disjoint 1 2 m (A + ) = m(A + )= m(A) ⇒ ≥ n n n=1 n=1 n=1 [ X X m(A) = 0. ⇒

(iii) R = r∈Q(A + r): S x R a E(x) A x a = r Q, a A ∈ ⇒ ∃ ∈ ∩ ⇒ − ∈ ∈ x = a + r, a A ⇒ ∈ x A + r. ⇒ ∈ (i), (ii) ja (iii) ⇒ + = m(R)= m(A + r)= m(A) = 0. contradiction ∞ rX∈Q rX∈Q =0 | {z }

Remark. 1. Also in Rn, n 1, similar examples, and so ∀ ≥ ∃ Leb Rn ( (Rn). P

2. If A R is an arbitrary set s.t. m∗(A) > 0, then B A s.t. B Leb R. ⊂ ∃ ⊂ 6∈ 32 Measure and integral

2 Measurable mappings

2.1 Measurable mapping Denote R˙ = R + . ∪{−∞}∪{ ∞} Deﬁnition. Let A Rn. A mapping f : A Rm is measurable (w.r.t. σ-algebra Leb Rn) if f −1G ⊂ → is (Lebesgue-)measurable for all open G Rm. A mapping f : A R˙ is measurable if ⊂ → (i) f −1G is measurable for all open G Rm, ⊂ (ii) f −1(+ ) is measurable, and ∞ (iii) f −1( ) is measurable. −∞ Rn Rm

f A

f −1G G

Remark. 1. f : A Rm measurable → ⇒ A = f −1Rm Rn is a measurable set. ⊂ Similarly f : A R˙ measurable → ⇒ A = f −1(R) f −1(+ ) f −1(+ ) Rn is a measurable set. ∪ ∞ ∪ ∞ ⊂ 2. f : A Rm measurable, B A measurable f B : B Rm measurable. → ⊂ ⇒ | → Reason: G Rm open ⊂ ⇒ (f B)−1(G)= B f −1G | ∩ measurable measurable is measurable. |{z} | {z } 3. Let X be an arbitrary set and Γ (X) a σ-algebra. ⊂ P Deﬁne: A mapping f : X R is measurable (w.r.t. σ-algebra Γ) if f −1G Γ for all open → ∈ G R. ⊂ Recall A mapping f : A Rm, A Rn, is continuous at x A if ε> 0 δ = δ(ε) > 0 s.t. → ⊂ ∈ ∀ ∃ f B(x, δ) A B(f(x), ε). ∩ ⊂ f : A Rm is continous if f is continuous at every x A. → ∈ B(f(x),ε) f A

B(x, δ) fB(x, δ) Spring 2017 33

Fact: f : A Rm continuous → ⇐⇒ (2.2) f −1G is open in A poen G Rm, i.e. f −1G = A V, where V Rn is open. ∀ ⊂ ∩ ⊂ Theorem 2.3. A measurable and f : A Rm continuous f measurable. → ⇒ Proof.

(2.2) G Rm open = f −1G open in A open V Rn s.t. ⊂ ⇒ ⇒ ∃ ⊂ f −1G = A V Leb Rn ∩ ∈ measurable measurable f measurable. ⇒|{z} |{z}

Theorem 2.4. If f : A Rm is measurable, then f −1B is measurable for all Borel sets B Rm. → ⊂ Proof. Denote Γ = V Rm : f −1V measurable . Then Γ is a σ-algebra because: { ⊂ } (1) f −1 = measurable Γ, ∅ ∅ ⇒ ∅∈ (2) V Γ f −1V c = A f −1V measurable V c Γ, ∈ ⇒ \ ⇒ ∈ measurable measurable (3) V Γ, i N f −1 |{z} V =| {z } f −1V measurable V Γ. i ∈ ∈ ⇒ i∈N i i∈N i ⇒ i∈N i ∈ S S measurable S Furthermore Γ contains all closed sets because:| {zF }closed F c open f −1F = f −1(F c) c ⇒ ⇒ measurable measurable F Γ. ⇒ ∈ Hence Γ Bor Rm (= the smallest σ-algebra that contains all closed sets). | {z } ⊃ Corollary 2.5. If f is measurable, then the preimage f −1(y) of a point y and the preimage f −1I of an interval are measurable.

Example. Let E Rn amd χ : Rn 0, 1 the characteristic function of E, ⊂ E → { } 1, if x E, χE(x)= ∈ 0, if x E. ( 6∈ Claim: χ measurable function E measurable set. E ⇐⇒ Proof. E = χ−1(1) measurable (Cor. 2.5). ⇒ E Let E be measurable and G R ope. ⇐ ⊂ Rn, if 0, 1 G, { }⊂ −1  , if 0, 1 G = , χE (G)= ∅ { }∩ ∅ E, if 0, 1 G = 1 ,  { }∩ { } Ec, if 0, 1 G = 0 .  { }∩ { }  These sets are measurable χ measurable function. ⇒ E 34 Measure and integral

Theorem 2.6. Let f : A Rm be measurable, A Rn, and g : B Rk continuous, where → ⊂ → fA B Rm. Then g f is measurable. ⊂ ⊂ ◦ Proof.

G Rk open (2.2) ⊂ = g−1G open in B g continuous ⇒ open V Rm s.t. g−1G = B V ⇒ ∃ ⊂ ∩ fA⊂B (g f)−1G = f −1(g−1G)= f −1(B V ) = f −1(V ) measurable. ⇒ ◦ ∩

Warning: f and g measurable g f measurable. 6⇒ ◦ If f : A Rm, then →

f = (f1,...,fm), f(x)= f1(x),...,fm(x) , where

f : A R, f (x) = (P f)(x) and P (y ,...,y )= y (= projection onto j’s coordinate axis). j → j j ◦ j 1 m j Theorem 2.7. f = (f ,...,f ): A Rm is measurable f is measurable j 1,...,m . 1 m → ⇐⇒ j ∀ ∈ { } Proof. If f is measurable, then f = P f is measurable (Thm. 2.6) since P is continuous. ⇒ j j ◦ j Suppose that f is measurable j. Let G Rm be open. ⇐ j ∀ ⊂

−1 P2 I2

−1 −1 I1 × I2 = P1 I1 ∩ P2 I2

−1 P1 I1

Lindel¨of G = I(i), I(i) open m-interval (cf. proof of Thm. 1.27) ⇒ i∈N [ m I(i) = I(i) I(i) = P −1I(i), I(i) R open 1 ×···× m j j j ⊂ j\=1 m m −1 −1 (i) −1 −1 (i) −1 (i) f G = f I = f Pj Ij = fj Ij measurable. i∈N i∈N j=1 i∈N j=1 [ [ \ [ \ measurable | {z }

Theorem 2.8. Let f : A R˙ and g : A R˙ be measurable. Then their sum and product are → → measurable (whenever deﬁned). Furthermore, λf, λ R and f a, a> 0, are measurable. ∈ | | Spring 2017 35

Proof. Sum: Suppose ﬁrst that f, g : A R are measurable. Denote f + g = u v, where → ◦ A v R2 u R, v = (f, g) and u(x,y)= x + y. −→ −→ Thm. 2.7 v measurable ⇒ f + g = u v measurable. u continuous ⇒ ◦ Note: The case f, g : A Rm measurable f + g measurable follows from Theorem 2.7. → ⇒ Suppose then that f, g : A R˙ are measurable. [The sum f +g is deﬁned if there exists no point → x A such that f(x), g(x) = + , .] Denote f + g = h. We know that A is measurable ∈ { } { ∞ −∞} (Remark 1.). On the other hand,

A = h−1(+ ) h−1( ) A , where A = h−1R. ∞ ∪ −∞ ∪ 0 0 h−1(+ )= f −1(+ ) g−1(+ ) is measurable. ∞ ∞ ∪ ∞ h−1( )= f −1( ) g−1( ) is measurable. −∞ −∞ ∪ −∞ A is measurable. ⇒ 0

beginning of proof f A and g A measurable (Remark 2.) = h−1G is measurable G R open | 0 | 0 ⇒ ∀ ⊂ h is measurable. ⇒ Product. Similarly (Exerc.) λf Special case of the product. f a f a = u f, where u(x)= x a continuous if a> 0. Thm. 2.6 f a is measurable. | | | | ◦ | | ⇒ | | From now on we consider only functions f : A R˙ , A Rn. → ⊂ An important basic criterion:

Theorem 2.9. Let A Rn be measurable and f : A R˙ . TFAE (= the following are equivalent) ⊂ → (1) f is measurable;

(2) E = x A: f(x) < a is measurable a R; a { ∈ } ∀ ∈ (3) E′ = x A: f(x) > a is measurable a R; a { ∈ } ∀ ∈ (4) E′′ = x A: f(x) a is measurable a R; a { ∈ ≤ } ∀ ∈ (5) E′′′ = x A: f(x) a is measurable a R. a { ∈ ≥ } ∀ ∈ Proof.

E′′′ = A E hence (2) (5) a \ a ⇐⇒ E′′ = A E′ hence (3) (4) a \ a ⇐⇒ E′′ = E hence (2) Thm.= 1.18 (4) a a+1/j ⇒ j\∈N Thm. 1.18 E = E′′ hence (4) = (2) a a−1/j ⇒ j[∈N E = f −1 ( , a) f −1( ) hence (1) (2) a −∞ ∪ −∞ ⇒ open | {z } 36 Measure and integral

Suppose that (2) holds [and thus also (3),(4),(5)] Claim: (1) holds, that is, f is measurable. Proof: Let G R be open. ⊂

G = Ij, Ij = (aj, bj) open interval (Lindel¨of) j[∈N f −1G = f −1I , f −1I = x: a

Remark. The assumption ”A measurable” is necessary in Theorem 2.9. Example: Let A be non- measurable (Thm. 1.35) and x A. Deﬁne f : A R˙ , 0 ∈ → + if x A x , f(x)= ∞ ∈ \ { 0} ( if x = x0. −∞ Then E = x A: f(x) < a = x is measurable a R, thus (2) holds but f can not be a { ∈ } { 0} ∀ ∈ measurable (since A non-measurable), that is (1) does not hold.

Example. Claim: f : R R measurable → ⇐⇒ (1) f 2 measurable function, ((2) E = x: f(x) > 0 measurable set. { } Proof: Denote E = x: f(x) < a . We must prove E is measurable a R (Theorem 2.9). ⇐ a { } a ∀ ∈ (i) Let a> 0.

f(x) < a f(x)2 < a2 or f(x) 0, hence ⇐⇒ ≤ E = x: f 2(x) < a2 Ec measurable. a { } ∪ measurable (1) measurable (2) | {z } |{z} (ii) Let a 0. ≤ f(x) < a f(x)2 > a2 and f(x) 0, hence ⇐⇒ ≤ E = x: f 2(x) > a2 Ec measurable. a { } ∩ measurable (1) measurable (2) | {z } |{z} Theorem 2.9 f is measurable. ⇒ Thm. 2.8 Thm. 2.9 f measurable f 2 = f f is measurable. Similarly: f measurable E ⇒ ⇒ · ⇒ measurable. Spring 2017 37

Remark. f 2 measurable f measurable. Reason: Let E R be non-measurable and f : R R, 6⇒ ⊂ → 1, if x E, f(x)= ∈ 1, if x Ec. (− ∈ Then f 2 is measurable as a constant function f 2(x) 1 but x: f(x) > 0 = E is non-measurable ≡ { } set. Thm.= 2.9 f non-measurable. ⇒ 2.10 lim sup and lim inf of a sequence

Deﬁnition. Let a1, a2,... be a sequence in R˙ . Denote

bk = sup ai, ck = inf ai. (bk, ck R˙ allowed) i≥k i≥k ∈

Then

b b b b and 1 ≥ 2 ≥···≥ k ≥ k+1 ≥··· c c c c (sup / inf taken over a smaller set) 1 ≤ 2 ≤···≤ k ≤ k+1 ≤··· limits ⇒ ∃

lim bk = inf bk = β and lim ck = sup ck = γ ( allowed). k→∞ k∈N k→∞ k∈N ±∞ Denote

β = limsup ai or lim ai ”upper limit” or ”limes superior” i→∞ i→∞

γ = lim inf ai or lim ai ”lower limit” or ”limes inferior”. i→∞ i→∞

Thus

lim sup ai = lim sup ai = inf sup ai , i→∞ k→∞ i≥k k∈N i≥k lim inf ai = lim inf ai = sup inf ai . i→∞ k→∞ i≥k k∈N i≥k Remark. (a ) a sequence in R˙ lim sup a and lim inf a always exist ( R˙ ) and are i ⇒ i→∞ i i→∞ i ∈ unique.

Example. (1) , , , ,... ; b = k, c = k β = , γ = ∞ −∞ ∞ −∞ k ∞ ∀ k −∞ ∀ ⇒ ∞ −∞ (2) 1, 2, 3, 4,... ; b = k, c = k k β = = γ k ∞ ∀ k ∀ ⇒ ∞ (3) 0, 1, 0, 1, 0, 1,... ; b = 1 k, c = 0 k β = 1, γ = 0 k ∀ k ∀ ⇒ (4) 0, 1, 0, 2, 0, 3,... ; b = 0 k, c = k β = 0, γ = . − − − k ∀ k −∞ ∀ ⇒ −∞ Theorem 2.11. (i) lim inf a lim sup a , i→∞ i ≤ i→∞ i (ii) a M i i lim sup a M, i ≤ ∀ ≥ 0 ⇒ i→∞ i ≤ (iii) a m i i lim inf a m. i ≥ ∀ ≥ 0 ⇒ i→∞ i ≥ 38 Measure and integral

Proof. (i) c b γ = lim c lim b = β, k ≤ k ⇒ k→∞ k ≤ k→∞ k (ii) b M k i β = lim b M, k ≤ ∀ ≥ 0 ⇒ k→∞ k ≤ (iii) c m k i γ = lim c m. k ≥ ∀ ≥ 0 ⇒ k→∞ k ≥

Theorem 2.12. Let (ai) be a sequence in R˙ . Then

lim ai ( R˙ ) lim inf ai = limsup ai ( R˙ ). ∃ i→∞ ∈ ⇐⇒ i→∞ i→∞ ∈

In this case

lim ai = lim inf ai = limsup ai ( allowed). i→∞ i→∞ i→∞ ±∞

Proof. Suppose that α = lim a . ⇒ ∃ i→∞ i (a1) α R ∈ ε> 0 i s.t. α ε < a < α + ε i i ⇒ ∃ 0 − i ∀ ≥ 0 α ε c γ β b α + ε ⇒ − ≤ i0 ≤ ≤ ≤ i0 ≤ ε arbotrary γ = β ⇒ (a2) α = ∞ M R i s.t. a > M i i ∈ ⇒ ∃ 0 i ∀ ≥ 0 M c γ β ⇒ ≤ i0 ≤ ≤ M arbitrary γ = β = ⇒ ∞ (a3) α = similarly. −∞ Suppose that β = γ denote= α. ⇐ (b1) α R ∈ ε> 0 k s.t. b < α + ε k k ⇒ ∃ 1 k ∀ ≥ 1 k s.t. c > α ε k k ∃ 2 k − ∀ ≥ 2 k max k , k α ε < c a b < α + ε ≥ { 1 2}⇒ − k ≤ k ≤ k ε arbitrary α = lim ak ⇒ k→∞

(b2) α = ∞ M R k s.t. c > M k k ∈ ⇒ ∃ 0 k ∀ ≥ 0 a c > M k k ⇒ k ≥ k ∀ ≥ 0 lim ak = ⇒ k→∞ ∞

(b3) α = similarly. −∞ Spring 2017 39

2.13 Measurablity of limit function Theorem 2.14. Let f : A R˙ , j N, be measurable. Then the functions j → ∈

sup fj, inf fj, lim sup fj, lim inf fj j∈N j∈N j→∞ j→∞ are measurable. If f = lim f , then f is measurable. ∃ j→∞ j Remark. These functions are deﬁned pointwise x A. For instance, the value of the function ∀ ∈ sup f at a point x A is sup f (x) R˙ . j∈N j ∈ j∈N j ∈ Proof. Denote g(x) = sup f (x), x A. For all a R: j∈N j ∈ ∈ (2.15) measurable (∗) x A: g(x) a = x A: fj(x) a is measurable g = sup fj is measurable. { ∈ ≤ } { ∈ ≤ } ⇒ j∈N j\∈N z }| { ( ) : g(x) a f (x) a j N ∗ ≤ ⇐⇒ j ≤ ∀ ∈ (2.16) inf fj = sup( fj) is measurable, j∈N − j∈N −

lim sup fj = inf sup fj is measurabel [(2.15), (2.16)], j→∞ k∈N j≥k lim inf fj = sup inf fj is measurable [(2.15), (2.16)]. j→∞ k∈N j≥k Thm. 2.12 If f = lim fj, then lim fj = limsup fj is measurable. ∃ j→∞ j→∞ j→∞

Almost every(where) (abbreviated a.e.) = except a set of measure zero. Example: (a) a.e. real number is irrational, because m(Q) = 0.

2 j→∞ (b) e−jx 0 for a.e. x R since m( 0 ) = 0. −−−→ ∈ { } Theorem 2.17. Let f, g : A R˙ . Suppose that f is measurable and g = f a.e. Then g is measur- → able. Proof. f, g : A R˙ and f(x)= g(x) x A A , where A A, m(A ) = 0. Let a R. Denote → ∀ ∈ \ 0 0 ⊂ 0 ∈ E = x A: f(x) < a and F = x A: g(x) < a . a { ∈ } a { ∈ } measurable F = F A F A , a | a ∩{z0 ∪ a} \ 0 m∗ F A m∗(A ) = 0 F A is measurable. a ∩ 0 ≤ 0 ⇒ a ∩ 0 F A = E A is measurable a \ 0 a \ 0 F Is measurable. ⇒ a 40 Measure and integral

Remark. Hence sets of measure zero do not aﬀect on measurability we may talk about ⇒ measurability of functions that are deﬁned only a.e. Theorem 2.18. Let f : A R˙ , j N, be measurable and f f a.e. Then f is measurable. j → ∈ j → Proof. f = limsupj→∞ fj a.e. Example. Suppose f : R R and f ′(x) x R. → ∃ ∀ ∈ Claim: f ′ is measurable. Proof: Denote f(x + 1/n) f(x) ′ gn(x)= − , hence f (x) = lim gn(x). 1/n n→∞ f ′(x) x R f continuous and therefore measurable g measurable (Thm. 2.8) ∃ ∀ ∈ ⇒ ⇒ n Thm. 2.14 = f ′ measurable. ⇒

3 Lebesgue integral

3.1 Simple functions Deﬁnition. A function f : Rn R is simple if → (1) f is measurable, (2) f 0 (f(x) 0 x Rn), ≥ ≥ ∀ ∈ (3) f takes only ﬁnitely many values. Denote Y = f f : Rn R simple (or Y ). { | → } n Rn 1 A3 R

A1 0 a1 a2 a3

Remark. 1. f Y f(x) = x. ∈ ⇒ 6 ∞ ∀ 2. f Y, E Leb Rn fχ Y. ∈ ∈ ⇒ E ∈ Let f Y and let a , . . . , a [0, + ) be the values of f. Then ∈ 1 k ∈ ∞ k −1 n Ai = f (ai) are measurable and disjoint, R = Ai i[=1 and k f = a χ is the standard representation of f. i · Ai Xi=1 Spring 2017 41

Deﬁnition. Let f Y and f = k a χ its standard representation. Then the integral of f ∈ i=1 i · Ai (over Rn) is P k I(f)= a m(A ). (recall 0 = 0) i i ·∞ Xi=1 If E Rn is measurable, then the integral of f over E is ⊂

I(f, E)= I(fχE).

In particular:

I(f)= I(f, Rn),

0 I(f, E) , ≤ ≤∞ E Leb Rn I(χ )= m(E). ∈ ⇒ E Theorem 3.2. If f Y and k a χ is the standard representation of f, then ∈ i=1 i · Ai P k I(f, E)= a m(A E). i i ∩ Xi=1 Proof. Omitted.

Theorem 3.3. Let E , j N, be measurable and disjoint sets and let E = E . If f Y, then j ∈ j∈N j ∈ S I(f, E)= I(f, Ej). Xj∈N k Proof. Let f = i=1 aiχAi be the standard representation.

P k L. 3.2 I(f, E)= a m(A E). ⇒ i i ∩ Xi=1 Since A E = (A E ), then (by the countable additivity Thm. 1.18) i ∩ j∈N i ∩ j S m(A E)= m(A E ) i = 1,...,k i ∩ i ∩ j ∀ Xj∈N k k I(f; E)= a m(A E )= a m(A E ) ⇒ i i ∩ j i i ∩ j Xi=1 Xj∈N Xj∈N Xi=1 3.2 = I(f, Ej). Xj∈N

Remark. Clearly I(f, )= I(fχ )= I(0) = 0, and therefore by Thm. 3.3 the mapping ∅ ∅ Leb Rn [0, + ], E I(f, E) → ∞ 7→ is a measure for every (ﬁxed) f Y. ∈ 42 Measure and integral

Convergence theorem 1.32 ⇒ Corollary 3.4. If f Y and E E are measurable, then ∈ 1 ⊂ 2 ⊂··· ∞ I f, j=1Ej = lim I(f, Ej). ∪ j→∞ Theorem 3.5. Let f, g Y, E measurable, and a 0 a constant. Then ∈ ≥ (i) f + g Y and I(f + g, E)= I(f, E)+ I(g, E); ∈ (ii) af Y and I(af, E)= aI(f, E). ∈ Proof. (i): Clearly f + g Y . ∈ (a) Let E = Rn and k ℓ

f = ajχAj , g = biχBi Xj=1 Xi=1 the standard representation. Then

3.2 (f + g)χ = (a + b )χ i, j = Ai∩Bj i j Ai∩Bj ∀ ⇒

I(f + g, A B ) = (a + bj)m(A B )= a m(A B )+ b m(A B ) (3.6) i ∩ j i i ∩ j i i ∩ j j i ∩ j = I(f, A B )+ I(g, A B ) i ∩ j i ∩ j Rn = disjoint union of sets A B . Theorem 3.3 i ∩ j ⇒ (3.6) I(f + g) 3=.3 I(f + g, A B ) = I(f, A B )+ I(g, A B ) i ∩ j i ∩ j i ∩ j Xi,j Xi,j Xi,j 3.3 = I(f)+ I(g)

(b) E arbitrary.

I(f + g, E)= I (f + g)χE = I(fχE + gχE)= I(fχE)+ I(gχE) = I(f, E)+ I(g, E). (ii): af Y clear. ∈ a = 0 I(af, E)=0= aI(f, E). ⇒ k Let a> 0 and f = i=1 aiχAi the standard representation.

P k

af = aaiχAi standard representation. Xi=1 k k I(af, E)= aa m(A E)= a a m(A E)= aI(f, E). i i ∩ i i ∩ Xi=1 Xi=1

Monotonicity properties. Spring 2017 43

Theorem 3.7. (1) E measurable and f, g Y, f g (i.e. f(x) g(x) x) I(f, E) ∈ ≤ ≤ ∀ ⇒ ≤ I(g, E);

(2) E F measurable, f Y I(f, E) I(f, F ); ⊂ ∈ ⇒ ≤ (3) f Y, m(E) = 0 I(f, E) = 0. ∈ ⇒ Proof. (1): g = f + (g f), where g f 0 and g f Y. Theorem 3.5 − − ≥ − ∈ ⇒ 3.5 I(g, E) = I(f, E)+ I(g f, E) I(f, E). − ≥ ≥0 | {z } (2):

E F 0 χ χ ⊂ ⇒ ≤ E ≤ F fχ fχ ( Y )  ⇒ E ≤ F ∈ f Y ∈  (1) I(f, E)= I(fχ )  I(fχ )= I(f, F ). ⇒ E ≤ F

k (3): If f = i=1 aiχAi is the standard representation, then

P k I(f, E)= a m(A E) = 0 since A E E and m(E) = 0. i i ∩ i ∩ ⊂ Xi=1 =0 | {z }

3.8 Lebesgue integral, f 0 ≥ Theorem 3.9. Let f : Rn R˙ be measurable and f 0. Then an increasing sequence of simple → ≥ ∃ functions f Y, f f , s.t. f(x) = lim f (x) x Rn. j ∈ 1 ≤ 2 ≤··· j→∞ j ∀ ∈ Proof. Define f : Rn R˙ as follows: Divide [0, j) into disjoint half open intervals I ,...,I , whose j → 1 k length is 1/2j , i.e. I = [(i 1)2−j , i2−j), i = 1,...,k = j2j . i − Define −j −1 −j −j (i 1)2 , if x f Ii, (i.e. (i 1)2 f(x) < i2 ) fj(x)= − ∈ − ≤ j, if x f −1[j, + ] (i.e. f(x) j). ( ∈ ∞ ≥ f measurable f −1(I ) measurable and ⇒ i f −1[j, + ] measurable. ∞  f Y, j = 1, 2,... ⇒ j ∈  f 0, takes only finitely many values  j ≥  Construction f f (see the picture).  ⇒ j ≤ j+1 44 Measure and integral

j + 1 fj+1 j fj

fj+1

fj fj

−1 −1 −1 f Ii f ([j, +∞]) f Ii

Claim: f (x) f(x) x Rn. j → ∀ ∈ (a): f(x) < + j >f(x). If j j , then ∞ ⇒ ∃ 0 ≥ 0 (i 1)2−j f(x) < i2−j for some i 1,...,j2j − ≤ ∈ { } f (x) = (i 1)2−j f(x) < i2−j = f (x) + 2−j f(x) 2−j

(b): f(x)=+ f (x)= j j f (x) + = f(x). ∞ ⇒ j ∀ ⇒ j → ∞ Deﬁnition. Let f : Rn R˙ be measurable and f 0. Then the (Lebesgue) integral of f over Rn → ≥ is f = sup I(ϕ): ϕ Y, ϕ f . { ∈ ≤ } Z If E Rn is measurable, then the integral of f over E is ⊂

(3.10) f = fχE. ZE Z Denote also

f = f dm = f(x) dm(x), m = n-dimensional Lebesgue measure. ZE ZE ZE b b If n = 1 and E = [a, b], we denote E f = a f = a f(x) dx. Convention. If f : A R˙ andRE A,R then weR deﬁne fχ : Rn R˙ , → ⊂ E → f(x), if x E, fχE(x)= ∈ (0, if x E. 6∈ Then (3.10) deﬁnes f for all measurable f : A R˙ and measurable E A. E → ⊂ Theorem 3.11. f R Y and E measurable I(f, E)= f. ∈ ⇒ E Proof. We may assume E = Rn (otherwise replace f by fχR Y ). E ∈ (a) f f I(f) f. ≤ ⇒ ≤ L. 3.7R (1) (b) ϕ Y, ϕ f = I(ϕ) I(f) f I(f). ∈ ≤ ⇒ ≤ ⇒ ≤ R Spring 2017 45

Basic properties of integrals.

Theorem 3.12. Suppose that the functions below are non-negative and measurable and the sets are measurable subsets of Rn.

(1) f g f g ≤ ⇒ E ≤ E (2) A B R f R g ⊂ ⇒ A ≤ B (3) f(x) = 0 xR E R f = 0 ∀ ∈ ⇒ E (4) m(E) = 0 f =R 0 ⇒ E (5) 0 a< R af = a f. ≤ ∞ ⇒ E E Proof. (1): Let E = RRn, ϕ Y,R ϕ f ϕ g ∈ ≤ ⇒ ≤ ⇒ sup I(ϕ) g = f g. ≤ ⇒ ≤ Z Z Z (1) E Leb Rn fχ gχ in Rn = ∈ ⇒ E ≤ E ⇒

f = fχ gχ = g. E ≤ E ZE Z Z ZE (2): fχ fχ ja (1) claim. A ≤ B ⇒ (3): fχ = 0 f = I(0) = 0. E ⇒ E (4): Let ϕ Y, ϕ fχ . Since ϕ Rn E = 0, then ϕ = ϕχ and ∈ R≤ E | \ E 3.7 (3) sup I(ϕ)= I(ϕ, E) = 0 = f = 0. ⇒ ZE (5): If a = 0, both sides are zero. Let a> 0, ϕ Y, ϕ fχ aϕ afχ ∈ ≤ E ⇒ ≤ E ⇒ 3.5 (ii) af I(aϕ) = aI(ϕ) af a f. ≥ ⇒ ≥ ZE ZE ZE 1 1 yll¨a 1 f = (af) f = (af) af a f af. a ⇒ a ≥ a ⇒ ≥ ZE ZE ZE ZE ZE

Relation to the Riemann integral.

Theorem 3.13. Let E Rn be bounded and f : E R measurable, f 0. If f is Riemann ⊂ → ≥ integrable over E, then the

(Riemann integral) (R) f = f (Lebesgue integral). ZE ZE This is the case, for example, when E is a closed n-interval and f continuous. 46 Measure and integral

Proof. Choose a closed n-interval I E. By deﬁnition ⊃

(R) f = (R) fχE and f = fχE = fχE, ZE ZI ZE Z ZI we may assume that E = I (by replacing f with fχ ). Let D = I ,...,I be a partition of I E { 1 k} into half-open disjoint intervals. Denote

gi = inf f(x), g¯i = inf f(x) g¯i gi and x∈Ii x∈I¯i ⇒ ≤

Gi = sup f(x), G¯i = sup f(x) G¯i Gi. x∈Ii x∈I¯i ⇒ ≥

The (Riemann) lower sum is

k k m = g¯ ℓ(I ) g m(I )= I(ϕ), D i i ≤ i i Xi=1 Xi=1 where ϕ = k g χ Y. Similarly the upper sum is i=1 i Ii ∈ P k k M = G¯ ℓ(I ) G m(I )= I(ψ), D i i ≥ i i Xi=1 Xi=1 where ψ = k G χ Y. Clearly ϕ f ψ, and therefore i=1 i Ii ∈ ≤ ≤ P sup f≤ψ (3.14) m I(ϕ) f ψ = I(ψ) M . D ≤ ≤ ≤ ≤ D ZE ZE Suppose that f is Riemann integrable over E. Then ε> 0 a partition D as above s.t. ∀ ∃ (3.15) m (R) f M (always) and 0 M m < ε. D ≤ ≤ D ≤ D − D ZE Letting ε 0 we obtain from (3.14) and (3.15) → ⇒ (R) f = f. ZE ZE

Remark. The case where E is unbounded (improper Riemann integral) is more complicated. A counterpart of Theorem 3.13 holds if f 0, but not in general. ≥ The Lebesgue integral is more general than the Riemann integral: −1 Example. Let f = χQ, Q = rational numbers. Then f is simple because f (1) = Q and f −1(0) = R Q are measurable. \ f = m(E Q) = 0 measurable E R. ∩ ∀ ⊂ ZE On the other hand, f is not Riemann integrable over any interval [a, b], a < b,: Let D = I ,...,I { 1 k} be a partition of [a, b] into subintervals. Every Ii contains both rational and irrational numbers. Hence m = 0 ℓ(I )=0 and M = 1 ℓ(I )= b a. ⇒ D · i D · i − Xi Xi Spring 2017 47

Theorem 3.16. Let f : E R˙ be measurable, f 0 and f < . Then f(x) < for a.e. → ≥ E ∞ ∞ x E. ∈ R Proof. Denote A = x E : f(x)= (measurable set since f is measurable). { ∈ ∞} f(x) j x A, j = 1, 2,... jχ fχ j ≥ ∀ ∈ ⇒ A ≤ E ∀ f I(jχ )= jm(A) j ⇒ ≥ A ∀ ZE 1 j→∞ 0 m(A) f 0 m(A) = 0. ≤ ≤ j −−−→ ⇒ ZE <∞ |{z}

Monotone convergence theorem.

Theorem 3.17. (MCT) Let f : E R˙ be measurable and j → 0 f f f f . ≤ 1 ≤ 2 ≤···≤ j ≤ j+1 ≤··· Then

lim fj = lim fj (+ aloowed). j→∞ j→∞ ∞ ZE ZE Proof. f f f f a limit lim f = a ( [0, ]). Similarly, j ≤ j+1 ⇒ E j ≤ E j+1 ⇒ ∃ j→∞ E j ∈ ∞ f = lim f that is measurable (Thm. 2.14). ∃ j→∞ j R R R

f f f f a f. j ≤ ⇒ j ≤ ⇒ ≤ ZE ZE ZE Need to prove: f a. E ≤ May assume: E = Rn (otherwise replace f , f by functions f χ , fχ (note: f χ fχ )). R j j E E j E ր E Let 0

If ϕ(x) > 0 then bϕ(x) < ϕ(x) f(x) (because 0

f f χ bϕχ j ≥ j Ej ≥ Ej ∞ 3.4 fj bϕχEj = bI(ϕ, Ej ) bI ϕ, Ej = bI(ϕ), as j ⇒ n ≥ n −−→ →∞ ZR ZR j=1 [ =Rn

a = lim fj bI(ϕ) ϕ| {zY,} ϕ f ⇒ j→∞ ≥ ∀ ∈ ≤ ZE sup = a b f 0

Remark. The order of and lim can not be changed in general: Example: R 1 f = jχ , f Y, I(f )= j = 1 j j (0,1/j] j ∈ j j ∀ j→∞ f (x) 0 x R j −−−→ ∀ ∈

lim fj = 0 = 1 = lim fj (the sequence (fj) is not increasing). ⇒ j→∞ 6 j→∞ ZR ZR Example. Find the limit ∞ e−xt lim dt. x→0+ 1+ t2 Z0 Solution: It’s enough to study the limit

∞ e−xnt lim dt n→∞ 1+ t2 Z0 for all sequences (x ) s.t. x x > 0 and x 0. Denote n n ≥ n+1 n ց e−xnt f (t)= , t [0, ) and n = 1, 2,... n 1+ t2 ∈ ∞

x x > 0 and t [0, ) e−xnt e−xn+1t n ≥ n+1 ∈ ∞ ⇒ ≤ e−xnt e−xn+1t 0 f (t)= = f (t), ⇒ ≤ n 1+ t2 ≤ 1+ t2 n+1 that is, the sequence (fn) is increasing. Furthermore,

e−xnt e0·t 1 f (t)= n→∞ = t [0, ). n 1+ t2 −−−→ 1+ t2 1+ t2 ∀ ∈ ∞ MCT ⇒ ∞ ∞ ∞ j 1 (∗) 1 lim fn(t) dt = lim fn(t) dt = dt = lim dt n→∞ n→∞ 1+ t2 j→∞ 1+ t2 Z0 Z0 Z0 Z0 3.13= lim j arctan t = lim arctan j arctan 0 = π/2. j→∞ 0 j→∞ − Spring 2017 49

Reason for ( ): MCT applied to the increasing sequence (g ), ∗ j

χ[0,j](t) g (t)= . j 1+ t2 (Note: In Theorem 3.13 the set E is bounded.)

Theorem 3.18. Let E Rn be measurable and f ,...,f : E R˙ measurable s.t. f 0. Then ⊂ 1 k → j ≥ k k fk = fk. E E Z Xj=1 Xj=1 Z Proof. We may assume: E = Rn and k = 2. Theorem 3.9 increasing sequences (ϕ ), (ψ ) of ⇒ ∃ j j simple functions s.t. ϕ f and ψ f as j . j ր 1 j ր 2 →∞ 3.5 I(ϕ + ψ )= I(ϕ )+ I(ψ ) ⇒ j j j j  MCT I(ϕj)= ϕj f1 and I(ψj ) f2,  ⇒ → →   (f1 + f2)= f1 + f2 . R R R  ⇒ similarly, ϕj + ψj f1 + f2 and MCT  Z Z Z ր ⇒   I(ϕj + ψj) (f1 + f2)  →   R  Beppo Levi Theorem.

Theorem 3.19. Let E Rn be measurable and f : E R˙ measurable s.t. f 0. Then ⊂ j → j ≥

fj = fj. ZE j∈N j∈N ZE X X k Proof. Denote uk = j=1 fj. Then

P ∞ 0 u u and u f =: u. ≤ 1 ≤ 2 ≤··· k → j Xj=1 MCT and Thm. 3.18 ⇒ k ∞ MCT 3.18 u = lim uk = lim uk = lim fj = fj. E E k→∞ k→∞ E k→∞ E E Z Z Z Xj=1 Z Xj=1 Z

The next convergence result is also very important!

Theorem 3.20. (Fatou’s lemma). Let E Rn be measurable and f : E R˙ measurable s.t. ⊂ j → f 0 j N. Then j ≥ ∀ ∈ lim inf fj lim inf fj (+ allowed). j→∞ ≤ j→∞ ∞ ZE ZE 50 Measure and integral

Proof. Denote gk(x) = inf fj(x), x E. j≥k ∈ Then

0 g g k N ≤ k ≤ k+1 ∀ ∈ gk measurable (Thm. 2.14)

gk fk and lim gk = lim inf fj ≤ k→∞ j→∞ MCT MCT lim inf fj = lim gk = lim gk = lim inf gk lim inf fk . ⇒ j→∞ k→∞ k→∞ k→∞ ≤ k→∞ ZE ZE ZE ZE gk≤fk ZE

Example. (1)

fj = jχ(0,1/j]

lim fj(x) = 0 x R lim inf fj = 0 j→∞ ∀ ∈ ⇒ j→∞ f = 1 j j ∀ ZR Fatou’s lemma holds in the form 0 1. ≤ (2)

fj = χ[j,2j]

lim fj(x) = 0 x R lim inf fj = 0 j→∞ ∀ ∈ ⇒ j→∞ f = m([j, 2j]) = j as j j →∞ →∞ ZR Fatou’s lemma holds in the form 0 . ≤∞ Integral as a set function is a measure:

Theorem 3.21. Let f : Rn R˙ be measurable, f 0. Then the mapping → ≥ Leb Rn [0, + ], E f → ∞ 7→ ZE is a measure, i.e.

(i) f = 0 , Z∅ (ii) if E Rn are measurable and disjoint, then j ⊂ ∞ f = f . ∞ Sj=1 Ej Ej Z Xj=1 Z In particular, Spring 2017 51

(iii) E E Rn measurable 1 ⊂ 2 ⊂···⊂ ⇒

f = lim f , ∞ j→∞ ZSj=1 Ej ZEj

(iv) Rn E E measurable and f < ⊃ 1 ⊃ 2 ⊃··· E1 ∞ ⇒ R f = lim f , ∞ j→∞ ZTj=1 Ej ZEj

Proof. (i): Thm. 3.12 (4); (ii): Exerc.; (iii) and (iv): Theorems on convergence of measures 1.32 and 1.33.

Theorem 3.22. (i) Let f, g : E R˙ be measurable and f 0, g 0. If f = g a.e. in E, then → ≥ ≥

f = g . ZE ZE In particular: f 0 measurable and deﬁned a.e. in E f well-deﬁned. ≥ ⇒ E R (ii) Let f : E R˙ be measurable, f 0. If f = 0, then f = 0 a.e. in E. → ≥ E Proof. (i): Denote A = x E : f(x) = g(x) R. By assumption m(A) = 0. { ∈ 6 }

f 3.21= f + f = g + g = g. E E A A E\A A E Z Z \ Z Z Z Z f=g =0 |{z} (ii): Assume on the contrary that m| {zx} E : f(x) > 0 > 0. By Exercise, r> 0 s.t. { ∈ } ∃ m x E : f(x) > r > 0 { ∈ } denote =A (∗) (∗∗) f f | r χ{z= rm(A}) > 0. contradiction ⇒ ≥ ≥ A ZE ZA ZA [( ) : A E, ( ) : fχ rχ ] ∗ ⊂ ∗∗ A ≥ A

Remark: Let (X, Γ,µ) be a measure space, f Γ-measurable function X [0, ]. Deﬁne the → ∞ integral of f

f = sup I(ϕ): ϕ: X R simple, ϕ f , { → ≤ } ZX f = fχ if E Γ. E ∈ ZE ZX The results in Section 3.8 (except Theorem 3.13 (Riemann int.)) hold. 52 Measure and integral

3.23 Lebesgue integral: general case Let f : E R˙ be measurable and E Rn. Denote → ⊂ 1 f +(x) = max f(x), 0 (= f + f measurable) { } 2 | | 1 f −(x)= min f(x), 0 (= f f measurable). − { } 2 | |−

f +

f −

+ f f −

Then

f +(x) 0, f −(x) 0 ≥ ≥

f(x)= f +(x) f −(x), f(x) = f +(x)+ f −(x). − | | (Note: above the case does not occur because either f +(x)=0 or f −(x) = 0.) ∞−∞ Section 3.8 ⇒ f + and f − deﬁned ( [0, + ]). ∈ ∞ ZE ZE Can we always deﬁne

f = f + f − (cf. f = f + f −)? − − ZE ZE ZE No(!) since now the (undeﬁned) case may occur! ∞−∞ Deﬁnition. A function f : E R˙ is integrable in E if f is measurable and f + < and → E ∞ f − < . Then the integral of f over E is E ∞ R R f = f + f − ( R). − ∈ ZE ZE ZE Theorem 3.24. A function f : E R˙ is integrable in E f measurable and → ⇐⇒

f < . | | ∞ ZE Then f f . E ≤ E| | Z Z

Spring 2017 53

Proof. Measurability is included in the deﬁnition of integrability. Furthermore, ⇒ f = f + + f − =3.18 f = f + + f − < . | | ⇒ E| | E E ∞ ≥0 ≥0 Z Z Z <∞ <∞ |{z} |{z} | {z } | {z } ⇐ 0 f + f f + f < ≤ ≤ | | ⇒ E ≤ E| | ∞ f integrable in E. R R  ⇒ 0 f − f f − f < ≤ ≤ | | ⇒ E ≤ E| | ∞  Furthermore, R R 

f = f + f − f + + f − = f + + f − − ≤ ZE ZE ZE ZE ZE ZE ZE

≥0 ≥0

3.18= f + + f − = |f{z. } | {z } | | ZE ZE

3.16, 3.24 Remark. f integrable in E = f(x) < a.e. x E. ⇒ | | ∞ ∈ Theorem 3.25. If f : E R˙ is measurable, f g and g integrable in E, then f is integrable in → | |≤ E.

Proof. f g < . | |≤ ∞ ZE ZE

Remark. It suﬃces that f g a.e. in E, i.e. | |≤ m x E : f(x) > g(x) = 0, then f = f + f < . { ∈ | | } | | | | | | ∞ ZE ZE\A ZA =A <∞ =0 | {z } Theorem 3.26. If f : E R is measurable and Riemann integrable,| {z then} |f{zis} Lebesgue integrable → in E and f = (R) f . ZE ZE Proof. 1 1 f + = f + f , f − = f f Riemann integrable 2 | | 2 | |− 3.13 = f + ja f + Leb. integrable and Riem./Leb.-integrals are same ⇒ f = f + f − = (R) f + (R) f − = (R) f . ⇒ − − ZE ZE ZE ZE ZE ZE

Theorem 3.27. Let E Rn be measurable, f, g : E R˙ integrable in E and λ R. Then ⊂ → ∈ 54 Measure and integral

(i) f + g integrable in E and E(f + g)= E f + E g;

(ii) λf integrable in E and E Rλf = λ E f;R R (iii) f g f g; R R ≤ ⇒ E ≤ E (iv) m(E) = 0 R fR = 0; ⇒ E (v) f = g a.e. in ER f = g. ⇒ E E Remark. f, g integrable inR E R f(x), g(x) R a.e. x E f + g deﬁned a.e. in E. ⇒ ∈ ∈ ⇒ Proof. (i): Let h = f + g. Then h deﬁned a.e. and measurable

h f + h h f + g < h integrable | | ≤ | | | | ⇒ | |≤ | | | | ∞ ⇒ ZE ZE ZE In general, h+ = f + + g+, but a.e. in E: 6 h+ h− = h = f + g = f + f − + g+ g− − − −

h+ + f − + g− = h− + f + + g+ (functions 0, integrate both sides (Thm. 3.18)) ⇒ ≥

h+ + f − + g− = h− + f + + g+ (integraalit < ) ⇒ ∞ ZE ZE ZE ZE ZE ZE

h = h+ h− = f + f − + g+ g− ⇒ − − − ZE ZE ZE ZE ZE ZE ZE = f + g. ZE ZE (ii): (a) λ 0 ≥ (λf)+ = λf + ja (λf)− = λf −

(λf)+ = λ f + ja (λf)− = λ f − ⇒ ZE ZE ZE ZE claim ⇒ (b) λ< 0 (λf)+ = ( λ)f − ja (λf)− = ( λ)f +, and the claim follows as above − − (iii): (i) and (ii) g f integrable and ⇒ − g = f + (g f) f E E E − ≥ E Z Z Z ≥0 Z (iv): m(E) = 0 f + = 0 and f − = 0 | {z } f = 0 ⇒ E E ⇒ E (v): f = g a.e. in E f + = g+, f − = g− a.e. in E R⇒ R R f + = g+ ja f − = g− claim. ⇒ ⇒ ZE ZE ZE ZE Spring 2017 55

Convergence theorems

Theorem 3.28. (Dominated convergence theorem, DCT) Let E Rn be measurable and ⊂ (f ), j N, a sequence of measurable functions s.t. j ∈

f(x) = lim fj(x) a.e. x E . j→∞ ∈

If g : E R˙ s.t. g is integrable in E and ∃ → f (x) g(x), j N, and a.e. x E , | j |≤ ∀ ∈ ∈ then f is integrable in E and

f = lim fj . (Note f R) j→∞ ∈ ZE ZE ZE Proof. By redeﬁning fj, f and g in a set of measure zero, we may assume

j→∞ f (x) f(x) x E and j −−−→ ∀ ∈ f (x) g(x) x E | j |≤ ∀ ∈

f(x) g(x) x E . ⇒ | | ≤ | | ∀ ∈ g integrable in E, Thm. 3.25) f integrable in E. ⇒ Fatou g + f 0 and g + f g + f = j ≥ j → ⇒ Fatou g + f = (g + f) lim inf (g + fj) = lim inf g + fj E E E ≤ j→∞ E j→∞ E E Z Z Z Z Z Z = g + lim inf fj j→∞ ZE ZE

f lim inf fj (note g < ) ⇒ ≤ j→∞ ∞ ZE ZE ZE

g f 0 , therefore − j ≥ Fatou g f = (g f) lim inf (g fj) = lim inf g fj E − E E − ≤ j→∞ E − j→∞ E − E Z Z Z Z Z Z = g lim sup f − j ZE j→∞ ZE

f lim sup f . ⇒ ≥ j ZE j→∞ ZE Hence

f lim inf fj lim sup fj f claim ≤ j→∞ ≤ ≤ ⇒ ZE ZE j→∞ ZE ZE 56 Measure and integral

Example. Find the limit 1 x lim n x−3/2 sin dx . n→∞ n Z0 Let f (x)= nx−3/2 sin x = (n/x) sin(x/n) x−1/2 n→∞ x−1/2 def.= f(x), then n n −−−→ →1, as n→∞

| {z 1} 1 f = 2√x = 2. 0 0 Z .

Here we just present Fubini’s theorems without proofs. We identify Rp+q = Rp Rq, p,q N. × ∈ z Rp+q z = (x ,...,x ,y ,...,x ) = (x,y). ∈ ⇐⇒ 1 p 1 q =x∈Rp =y∈Rq

q R {x}× Rq | {z } | {z }

y 7→ f(x,y)

Rp × {y} R y x 7→ f(x,y) x Rp

Theorem 4.1. (Fubini’s 1. theorem, f 0) Let f : Rp+q R˙ be measurable and f 0. Then ≥ → ≥ (1)

y f(x,y) is measurable for a.e. x Rp ; 7→ ∈ [i.e. m x Rp : y f(x,y) non-measurable = 0] p { ∈ 7→ }

(2) x f(x,y) is measurable for a.e. y Rq ; 7→ ∈ (3)

x f(x,y) dmq(y) is measurable; 7→ q ZR Spring 2017 57

(4)

y f(x,y) dmp(x) measurable; 7→ p ZR (5)

f = f(x,y) dmq(y) dmp(x) p+q p q ZR ZR ZR

= f(x,y) dmp(x) dmq(y) . (+ allowed) q p ∞ ZR ZR

Theorem 4.2. (Fubini’s 2. theorem, general case) Let f : Rp+q R˙ be measurable and suppose → that at least one of the integrals

f , f(x,y) dmq(y) dmp(x) , or p+q | | p q | | ZR ZR ZR

f(x,y) dmp(x) dmq(y) q p | | ZR ZR is ﬁnite. Then

(1) y f(x,y) is integrable over Rq for a.e. x Rp; 7→ ∈ (2) x f(x,y) is integrable over Rp for a.e. y Rq; 7→ ∈ (3) x f(x,y) dm (y) is integrable over Rp, i.e. 7→ Rq q R f(x,y) dmq(y) dmp(x) < ; Rp Rq | | ∞ Z Z

(4) y f(x,y) dm (x) is integrable over Rq; 7→ Rp p (5) f is integrableR over Rp+q, and

f = f(x,y) dmq(y) dmp(x)= f(x,y) dmp(x) dmq(y) . ( R) p+q p q q p ∈ ZR ZR ZR ZR ZR Below is a list of (some) books that can be used as an additional material.

References

[EG] Evans, Lawrence ja Gariepy Ronald. Measure theory and ﬁne properties of functions, CRC Press, 1992.

[Fr] Friedman, Avner. Foundations of modern analysis, Dover Publications Inc., 1982.

[GZ] Gariepy, Ronald ja Ziemer, William. Modern real analysis, PWS Publishing Company, 1994. 58 Measure and integral

[HS] Hewitt, Edwin ja Stromberg, Karl. Real and abstract analysis, Springer-Verlag, 1975.

[Jo] Jones, Frank. Lebesgue integration on Euclidean space, Jones and Bartlett Publishers, 1993.

[Mat] Mattila, Pertti. Geometry of sets and measures in Euclidean spaces, Cambridge University Press, 1995.

[MW] McDonald, John N. ja Weiss, Neil A. A course in real analysis, Academic Press Inc., 1999.

[Ro] Royden, H. L. Real analysis, Macmillan Publishing Company, 1988.

[Ru] Rudin, Walter. Real and complex analysis, McGraw-Hill Book Co., 1987.