MAT205a, Fall 2019 Part II: Integration Lecture 4, Following Folland, ch 2.1, part of 2.4

1. Measurable functions 1.1. Definition. Suppose that (X, M) and (Y, N ) are two measurable spaces, i.e., M and N are σ-algebras of subsets of X and Y respectively. Consider a f : X → Y and subsets of X of the form {f −1(E): E ∈ N }. These subsets form a σ-algebra.

Definition 1.1. A function f : X → Y is called (M, N )-measurable if the pre- image of any measurable set is measurable, i.e., f −1(E) ∈ M for any E ∈ N . A function f : X → R is called M-measurable if for any Borel set E ⊂ R its preimage, f −1(E), is in M.

It is easy to see that {E ⊂ Y : f −1(E) ∈ M} is a σ-algebra. Thus f : X → R is measurable if and only if {x ∈ X : f(x) > a} are measurable for all a ∈ R. We will denote this set by {f > a}.

Suppose that A ∈ M, we define the characteristic function of A by χA(x) = 1 when x ∈ A and χA(x) = 0 when x 6∈ A. Then χA is measurable. Clearly for any −1 c E ⊂ R we have χA (E) ∈ {∅, A, A ,X}. We remark that if X is a topological space with the Borel σ-algebra BX and f : X → R is continuous then f is BX -measurable. By the definition of continuity, the preimage of any open interval I ⊂ R is open. Thus the preimage of any Borel set is a Borel set in X.

1.2. Properties of measurable functions. We show that the class of measurable functions is closed under the standard operation.

Proposition 1.1. If f is a measurable function, then for any c ∈ R the function cf is also measurable. If f, g ∈ X → R are measurable, then max{f, g}, min{f, g}, f + g and fg are mea- surable.

−1 −1 −1 Proof. First, clearly if c 6= 0 then (cf) (E) = f (c E) and if E ∈ BR then −1 c E ∈ BR, thus cf is measurable. Now, {max{f, g} > a} = {f > a} ∩ {g > a} and, similarly, {min{f, g} > a} = {f > a} ∪ {g > a}. Thus max{f, g}, min{f, g} are measurable. For the sum we have {x : f + g > a} = ∪r({f > r} ∩ {g > a − r}), where the union is taken over all rational numbers r. It is a countable union of measurable sets, thus it is measurable. To show that fg is measurable it is enough to show that for any measurable function f the function f 2 is also measurable and 1 2

1 2 2 2 2 then write fg = 2 ((f + g) − f − g ). But {f > a} = X when a < 0 and 2 √ √ 2 {f > a} = {f > a} ∪ {f < − a} when a ≥ 0. Therefore f is measurable.  For any measurable function f we define f + = max{f, 0} and f − = max{−f, 0}, those functions are non-negative and measurable, f = f + − f −. Definition 1.2. A finite linear combination of characteristic functions of measurable sets is called a simple function. It follows from the previous proposition that simple functions are measurable. We will show that any measurable function can be approximated by simple ones.

Proposition 1.2. If {fn} is a sequence of measurable functions then the follow- ing functions are measurable: g1(x) = supn fn(x), g2(x) = infn fn(x), g3(x) = lim supn fn(x), and g4 = lim infn fn(x).

Proof. Clearly, {g1 > a} = ∪n{fn > a} is measurable as the countable union of measurable sets. Similarly, for g2, {g2 ≥ a} = ∩n{fn ≥ a} is measurable. Further- more, g3 = lim supn fn = infn supk≥n fk is measurable, and so is g4 = lim infn = supn infk≥n fk.  The proposition implies that if there is a limit, then it is equal to both the upper and the lower limits and is measurable.

Corollary 1.1. Suppose that {fn} is a sequence of measurable functions on (X, M), and for any x ∈ X there is the limit f(x) = limn→∞ fn(x) (i.e., f is the point-wise limit of fn). Then f is also measurable. 1.3. Approximation of measurable functions. We denote by L+(M) the family of all non-negative measurable functions on (X, M). As we have seen above, for any measurable function f there are functions f +, f − ∈ L+ such that f = f + − f −. Proposition 1.3. Let (X, M) be a measurable space and f ∈ L+(M) be a non- negative measurable function. Then there exists a sequence of simple functions {φj} such that φ1 ≤ φ2 ≤ ... and f(x) = limj→∞ φj(x) for any x ∈ X. Proof. For each n define the sets

−n −n n En,k = {f ∈ [k2 , (k + 1)2 )},En,∞ = {f ≥ 2 } for 0 ≤ k ≤ 4n. Clearly, those are measurable sets, now we set

X −n n φn = k2 χEn,k + 2 χEn,∞ . k

Then it is easy to check that φn ≤ φn+1 ≤ f and f(x) = limn→∞ φn(x) for each x ∈ X.  3

Now assume that X = R and µ is a Lebesgue-Stieltjes on R. We want to approximate a measurable function by continuous ones almost everywhere.

Proposition 1.4. If µ is a Lebesgue-Stieltjes measure on (R, M) and f is M- measurable, then there is a sequence of continuous functions {gn} which converges to f almost everywhere.

Proof. It is enough to prove the proposition for the case of a non-negative function.

Let En,k be sets defined in the proof of the previous proposition. Since µ is a Lebesgue-

Stieltjes measure there exists an open set Un,k such hat En,k ⊂ Un,k and µ(Un,k \ −n−k En,k) ≤ 2 . Let also ψn,k be a with support on Un,k such that −n−k 0 ≤ ψn,k ≤ 1 and µ(Un,k \{ψn,k < 1}) < 2 . Now let

Cn = ∪k(Un,k \ En,k) ∪ (Un,k \{ψn,k 6= 1)}

∞ and C = ∩j ∪n+j Cn. Then C is the set of x which belong to infinitely many sets Cn and µ(C) = 0. We define X −n gn = k2 ψn,k. k Clearly gn is a continuous function. Now if x ∈ R\C then there exists n(x) such that n(x) x 6∈ Cn for n > n(x) and f(x) < 2 . Then for n > n(x) we have |gn(x) − f(x)| < n c 1/2 . Therefore gn converges to f on C . 

2. Modes of convergence 2.1. Point-wise and . Let (X, M, µ) be a measure space.

The statement ”the sequence {fn} converges to f” can be understood in a number of ways. First we discuss the following three types of convergence.

(1) We say that {fn} converges to f uniformly on X if for any ε > 0 there exists m such that |f(x) − fn(x)| < ε for every x ∈ X and n ≥ m.

(2) We say that {fn} converges to f point-wise if for any x ∈ X and ε > 0 there exists m = m(x) such that |f(x) − fn(x)| < ε for n ≥ m. (3) If the property described in (2) holds for x ∈ E such that µ(Ec) = 0, we say that

{fn} converges to f almost everywhere. Clearly, uniform convergence implies point-wise convergence, which implies con- vergence almost everywhere. It is easy to construct examples showing that inverse implications are wrong. If a sequence of measurable functions converges to a func- tion almost everywhere and the measure is complete, then the limit function is also measurable.

Theorem 2.1 (Egorov). Suppose that (X, M, µ) is a measure space with a finite measure, µ(X) < ∞. If a sequence of measurable functions {fn} converges to a 4 measurable function f almost everywhere then for any ε > 0 there exists E ∈ M such that µ(X \ E) < ε and fn converges to f uniformly on E.

Proof. First, let Y be the set where {fn} converges to f, µ(X \ Y ) = 0. For each n and k we define

An,k = ∪m≥n{|f − fm| > 1/k}.

c Then A1,k ⊃ A2,k ⊃ .. for each k and ∩nAn,k ⊂ Y . Since the measure is finite, −k we have limn→∞ µ(An,k) = 0. We choose n(k) such that µ(An(k),k) < ε2 . And let

A = ∪kAn(k),k, so that µ(A) ≤ ε. On the complement of A we have |fn − f| < 1/k c when n ≥ n(k). Thus fn converges to f uniformly on A . 

2.2. Lusin theorem. Now we show that any measurable function is continuous on some compact subset of almost full measure.

Theorem 2.2 (Lunsin). Let µ be a Lebesgue-Stieltjes measure on (R, M). Suppose that f : R → R is a M-measurable function and A ⊂ R is such that µ(A) < ∞. Then for any ε > 0 there exists a compact set K ⊂ A such that µ(A \ K) < ε and f is continuous on K.

Proof. By Proposition 1.4 there is a sequence of continuous functions gn that con- verges to f almost everywhere on A. By Egorov’s theorem there exists E ⊂ A such that µ(A\E) < ε/2 and gn converges to f uniformly on E. We know that µ(E) < ∞ then there exists a compact set K ⊂ E such that µ(E \ K) < ε/2. But the uni- form limit of continuous functions on a compact set is continuous. Hence f|K is continuous. 

2.3. Convergence in measure. Another notion of convergence that appears often in is convergence is measure.

Definition 2.1. We say that a sequence of measurable functions {fn}n converges in measure to a measurable function f if for any ε > 0 the measure µ({|f − fn| > ε}) goes to zero as n → ∞.

We also say that the sequence {fn}n is Cauchy in measure if for each ε > 0 we have µ({|fn − fm| > ε} → 0 as n, m → ∞.

First, we show that the limit in measure is unique in the following sense.

Proposition 2.1. If fn → f in measure and fn → g in measure then f = g almost everywhere. 5

Proof. We have {f 6= g} = ∪k{|f −g| > 1/k}, thus it suffices to show that µ({|f −g| > 1/k}) = 0 for any k. In fact,

{|f − g| > 1/k} ⊂ {|f − fn| > 1/(2k)} ∪ {|g − fn| > 1/(2k)}, for each n. But as n → 0 the sum in the right-hand side goes to zero and therefore f = g a.e. 

Proposition 2.2. Suppose that {fn} converges to f almost everywhere, functions fn are measurable, f is measurable and µ(X) < ∞. Then {fn} converges to f in measure.

∞ Proof. We fix ε > 0 and consider the sets En = {|fn − f| > ε} and Fn = ∪j=nEj. Then F1 ⊃ F2 ⊃ ... and µ(∩nFn) = 0, since ∩nFn ⊂ {fn does not converge to f} and

µ(X) < ∞. Then limn→∞ µ(Fn) = 0 and since En ⊂ Fn, limn→∞ µ(En) = 0. Thus fn converges to f in measure.  It is easy to see that the condition µ(X) < +∞ is important.

Proposition 2.3. Is a sequence of measurable functions {fn} is Cauchy in measure, then there exists a measurable function f such that fn → f in measure.

−j Proof. First, we can find a subsequence {gk} of {fn} such that µ{|gj − gk| > 2 } < −j −j ∞ 1−k 2 for k > j. Let Ej = {|gj − gj+1| > 2 } and Fk = ∪j=kEj. Then µ(Fk) ≤ 2 and for F = ∩kFk we have µ(F ) = 0. On the other hand, if x 6∈ F then x 6∈ Fk −j for some k and for j ≥ k we have |gj(x) − gj+1(x)| < 2 , thus {gj(x)} is a Cauchy sequence of real numbers, it has a limit, we define f(x) = limj→∞ gj(x) when x 6∈ F and f(x) = 0 otherwise. We know that f is a measurable function and gj converges 1−j 1−j to f on X \ F and also in measure µ({|f − gj| > 2 } ≤ µ(Fj) ≤ 2 . Then it is easy to see that {fn} converges to f in measure as

{|fn − f| > ε} ⊂ {|fn − gj| > ε} ∪ {|gj − f| > ε}, foe any j. 

Corollary 2.1. If {fn} converges to f in measure, then there exists a subsequence

{fnk } that converges to f almost everywhere. It follows from the proof of the proposition above. 1 MAT205a, Fall 2019 Part II: Integration Lecture 5, Following Folland, ch 2.2

3. Integration of non-negative functions 3.1. Integral of simple function. Let (X, M, µ) be a measure space. We defined simple functions as linear combinations or the characteristic functions of measurable P sets. Thus if φ is simple, then φ = k ckχEk and φ takes only finitely many values (the values are sums of some of the ck’s). Suppose that φ(X) = {a1, ..., an}, then we −1 denote Aj = φ ({aj}). Then the standard representation of a simple function φ is X φ = ajχAj . j

Definition 3.1. Suppose that φ ∈ L+(M) is a simple function with the standard P representation φ = ajχAj . Then the integral of φ (over X with respect to the measure µ) is defined by Z X φ dµ = ajµ(Aj). j

Clearly all aj are non-negative and the sum is non-negative, it could be infinite. Before we extend the integral to all non-negative measurable functions, we examine the properties of the integral.

Proposition 3.1. Let φ and ψ be simple non-negative functions on (X, M) then (a) if c ≥ 0 then R cφ dµ = c R φ dµ, (b) R (φ + ψ) dµ = R φ dµ + R ψ dµ, (c) If φ ≤ ψ then R φ dµ ≤ R ψ dµ. P Proof. First, (a) follows from the definition, if φ = ajχAj is the standard repre- P sentation and c > 0 then the standard representation of cφ = j(caj)χAj . If c = 0 both sides are clearly zeros. P P For (b) let φ = ajχAj and ψ = l blχBl be the standard representations. Then X φ + ψ = (aj + bl)χAj ∩Bl , j,l where the sets Aj ∩ Bl are disjoint. For the standard representation of the sum, we have X φ + ψ = ckχCk , k 2 where Ck is the union of the sets of the form Aj ∩ Bl with aj + bl = ck. Then by the additivity of the measure and the property µ(∅) = 0, we have Z X X (φ + ψ) dµ = ckµ(Ck) = (aj + bl)µ(Aj ∩ Bl) = k j,l X X Z Z ajµ(Aj) + blµ(Bl) = φ dµ + ψ dµ. j l For (c) we see that ψ − φ is a non-negative simple function and by (b) we obtain R R R R ψ dµ = φ dµ + (ψ − φ) dµ ≥ φ dµ.  Now we prove the following lemma that will be useful later.

Lemma 3.1. Let φ be a simple non-negative function. For each E ∈ M define R ν(E) = φχE dµ. Then ν is a measure on (X, M). P Proof. First of all if φ = j ajχAj is the standard representation, then φχE is a P simple function with the standard representation φχE = j ajχAj ∩E, where some of the sets Aj ∩ e may be empty. Clearly, ν(E) ≥ 0 an ν(∅) = 0. We need to show that

ν is countably additive. Suppose that En ∈ M are disjoint and E = ∪jEn. Since µ P is countably additive, for each Aj we have µ(Aj ∩ E) = n µ(Aj ∩ En). Thus Z X X X ν(E) = φχE dµ = ajµ(Aj ∩ E) = ajµ(Aj ∩ En) = j n j X Z X φχEn dµ = ν(En). n n  R R We will also use the notation E φ dµ := φχE dµ. 3.2. Monotone Convergence Theorem. We now extend the definition to all non- negative measurable function.

Definition 3.2. For f ∈ L+(M) we define the integral of f by Z Z  f dµ = sup φ dµ : φ is simple, 0 ≤ φ ≤ f .

The previous proposition (part (c)) implies that for the simple functions the def- inition of the integral agrees with the one given before. By the definition we have also that if f, g ∈ L+(M) and f ≤ g then R f dµ ≤ R g dµ. While the definition and part (a) of the previous proposition imply that R (cf) dµ = c R f dµ for c ≥ 0 and f ∈ L+(M). Before we prove that the integral is linear, we establish the following powerful convergence result, called the Monotone Convergence Theorem (MCT). 3

Theorem 3.1 (Monotone Convergence). Suppose that {fn} is an increasing sequence + of functions in L (M), fn ≤ fn+1 and f = limn→∞ fn. Then Z Z f dµ = lim fn dµ. n→∞ R R R Proof. By the remark above, fn dµ ≤ fn+1 dµ ≤ f dµ and therefore the limit, R R R limn→∞ fn dµ, exists and limn→∞ fn dµ ≤ f dµ. Now we want to show the reverse inequality. Fix some α ∈ (0, 1) and let φ be a simple non-negative function such that φ ≤ f. Define the sets En = {fn ≥ αφ}.

Then En ⊂ En+1 and ∪nEn = X since {fn} increases and converge to f. We have Z Z Z

fn dµ ≥ fnχEn dµ ≥ α φχEn dµ = αν(En).

By Lemma 3.1 and properties of measures, we know that the right-hand side converges to αν(X), then, letting n → ∞, we obtain Z Z lim fn dµ ≥ α φ µ. n→∞ R Thus, by the definition of the integral, limn→∞ fn dµ ≥ α f dµ for each α < 1 and we obtain the reverse inequality.  Now we are ready to prove that integration is a linear operation.

Proposition 3.2. If f, g ∈ L+(M) then R (f + g) dµ = R f dµ + R g dµ.

Proof. As we know form Proposition 1.3 there exist increasing sequences {φn} and

{ψn} of simple functions that converge to f and g respectively. Then {φn + ψn} is an increasing sequences that converges to f + g and by the monotone convergence theorem and the additivity of the integral on simple functions, we conclude that Z Z Z Z Z Z f dµ+ g dµ = lim φn dµ+ lim ψn dµ = lim (φn +ψn) dµ = (f +g) dµ. n→∞ n→∞ n→∞



+ P Corollary 3.1. If {fn} is a sequence of function in L (M) and f = n fn then R P R f dµ = fn dµ.

The corollary follows from the monotone convergence theorem and the finite addi- tivity of the integral.

Proposition 3.3. Let f ∈ L+(M). Then R f dµ = 0 if and only if f = 0 µ-almost everywhere. 4

Proof. Suppose that f = 0 a.e. and 0 ≤ φ ≤ f where φ is a simple function. If

φ = ak > 0 on some set Ak, then f ≥ ak > 0 on Ak and thus µ(Ak) = 0. Then by the definition of the integral R φ dµ = 0 and therefore R f dµ = 0. R On the other hand, suppose that f dµ = 0. Let Ek = {f ≥ 1/k}, then we have −1 R −1 f ≥ k χEk and 0 = f dµ ≥ k µ(Ek) . Therefore µ(Ek) = 0 and {f 6= 0} = ∪kEk is a set of zero measure.  3.3. Fatou’s Lemma. It is important in the assumptions of the monotone conver- gence theorem that the sequence is increasing, in general for non-negative functions the integral of the limit is not equal to the limit of integrals (for example consider the sequence of simple functions φn = nχ(0,n−1) on the real line, it converges to zero R everywhere and φn dm = 1. The following inequality is true in general.

+ Lemma 3.2 (Fatou). If {fn} is a sequence of functions in L (M) then Z Z lim inf fn dµ ≤ lim inf fn dµ. n→∞ n→∞

Proof. Let gk = infn≥k fn, then lim infn→∞ fn = limk→∞ gk and gk is an increasing R R sequence. Moreover gk dµ ≤ fk dµ. Applying the monotone convergence theorem to this increasing sequence, we get Z Z Z lim inf fn dµ = lim gk dµ ≤ lim inf fn dµ. n→∞ k→infty n→∞  1 MAT205a, Fall 2019 Part II: Integration Lecture 6, Following Folland, ch 2.3, 2.5

4. Integrable functions 4.1. Definition and properties of the integral. We extend now the definition of the integral from the family of non-negative functions to measurable function.

Definition 4.1. Let f = f + − f − is a measurable function. If at least one of the numbers R f + dµ and R f − dµ is finite, then we define Z Z Z f dµ = f +dµ − f −dµ.

If both integrals are finite, we say that f is integrable.

Clearly, a measurable function f is integrable if and only if R |f| dµ < ∞. It follows from the fact that|f| = F + + f − and, since the integration is linear on non-negative functions, R |f| dµ = R f + dµ + R f − dµ. We denote the class of integrable functions by L1(µ).

Proposition 4.1. The set of integrable functions is a and for any a, b ∈ R and f, g ∈ L1(µ) we have Z Z Z (af + bg) dµ = a f dµ + b g dµ.

Proof. If f, g ∈ L1(µ) and a, b ∈ R then |af + bg| ≤ |a||f| + |b||g| and therefore R |af + bg| dµ < ∞. Hence L1(µ) is a vector space. We want to check that the integration is a linear operator. First, we know that R af dµ = a R f dµ for a non-negative function f and a ≥ 0. Then writing f = f + − f − and noticing that R (−f) dµ = − R f dµ, we conclude that R R af dµ = a f dµ holds for all f (for which integral makes sense) and all a ∈ R. It remains to show that R (f + g) dµ = R f dµ + R g dµ when f, g ∈ L1(µ). We denote h = f + g, then we need to show that Z Z Z Z Z Z h+ dµ − h− dµ = f + dµ − f − dµ + g+ dµ − g− dµ, where all terms are non-negative. We know that h+ − h− = f + − f − + g+ − g−, it implies that h+ + f − + g− = h− + f + + g+. We know that the integral is linear on non-negative functions, therefore Z Z Z Z Z Z h+ dµ + f − dµ + g− dµ = h− dµ + f + dµ + g+ dµ.

And the required identity follows.  2

4.2. Lebesgue space. As we saw in Proposition 3.3 the integral of a function which is zero almost everywhere is zero. By the linearity it implies that if two functions f, g are integrable and f = g a.e. then R f dµ = R g dµ. We want to refine the definition of the space L1(µ) by identifying two integrable functions when they are equal a.e. (clearly, it is an equivalence relation, so we consider equivalence classes of functions). With this refinement of the definition, we introduce the norm on L1(µ) by Z kfkL1(µ) = |f| dµ.

We see that kafkL1(µ) = |a|kfkL1(µ), where a ∈ R, and

kf + gkL1(µ) ≤ kfkL1(µ) + kgkL1(µ).

The metric, induced by this norm, defines another notion of convergence. We say that a sequence of integrable functions {fn} converges to an integrable function f in 1 L (µ) if kf − fnkL1(µ) tends to zero as n → ∞.

1 Proposition 4.2. If {fn} and f are integrable functions and fn → f in L (µ) then fn converges to f in measure.

−1 Proof. We fix ε > 0 and define En = {|f − fn| > ε} Then µ(En) > ε kf − fnkL1(µ). Hence µ(En) → 0 when n → ∞. 

We now prove our next convergence result.

Theorem 4.1 (Dominated Convergence). Let {fn} be a sequence of integrable func- 1 tions such that fn → f almost everywhere, and suppose that there exists g ∈ L (µ) 1 R R such that |fn| ≤ g a.e. for each n. Then f ∈ L (µ) and limn→∞ fn dµ = f dµ. S Proof. Let Y = {fn 6→ f} ∪ n{|fn| > g}, then Y is a set of measure zero. We redefine fn and f on Y by letting them to be 0 there. Then fn → f pointwise and |fn| ≤ g, while the integrals have not changed. We apply the Fatou lemma to sequences of non-negative functions g − fn and g + fn to obtain Z Z Z Z (g − f) dµ ≤ lim inf (g − fn) dµ, and (g + f) dµ ≤ lim inf (g + fn) dµ. n→∞ n→∞

Using the linearity of the integral, we get Z Z Z lim sup fn dµ ≤ f dµ ≤ lim inf fn dµ. n→∞ n→∞

The conclusion of the Theorem follows.  3

4.3. Lebesgue integral. A particular case of the integral we defined is the inte- gral on the measure space (R, L, m). We want to compare this integral, called the Lebesgue integral, with the on a bounded interval. Clearly, a Rie- mann integrable function is bounded.

Theorem 4.2. Suppose that f is Riemann integrable on a bounded interval [a, b]. R R b Then f is Lebesgue measurable and f dm = a f(x) dx. Proof. For each integer n we divide the interval [a, b] onto N = 2n equal subinter- vals I1, ..., IN and denote mj = infIj f and Mj = supIj f. Now we consider simple P P functions φn = j ajχIj and ψn = j AjχIj , clearly, φn ≤ f ≤ ψn. We also have φn ≤ φn+1 and ψn ≥ ψn+1, let φ = limn φn and ψ = limn ψn, then φ ≤ f ≤ ψ. We know that f is bounded on [a, b], thus all the functions are bounded by Mχ[a,b]. By the dominated convergence theorem (we could have used monotone convergence as well), R R R R we have limn φn dm = φ dm and limn ψn dm = ψ dm. The assumption that f is Riemann integrable implies that the limits are equal, and thus R (ψ − φ) dm = 0. We know that φ = ψ a.e. and f = φ almost everywhere. Then f is measurable (it is measurable after we change it on a set of zero , and Lebesgue measure is complete). Moreover, Z Z Z X X Z b f dm = φ dm = lim ajχI dm = lim ajm(Ij) = f(x)dx. n→∞ j n→∞ j j a  We have the following characterization of Riemann integrable functions.

Theorem 4.3. A bounded function f on a bounded interval [a, b] is Riemann inte- grable if and only if it is continuous m-almost everywhere.

Proof. We define functions

h(x) = lim inf f(y) and g(x) = lim sup f(y). δ→0 |x−y|<δ δ→0 |x−y|<δ Then h(x) ≤ g(x) and h(x) = g(x) if and only if f is continuous at x. We note that for any point which is not of the form k2−n we have h(x) = φ(x) and g(x) = ψ(x), where φ and ψ are functions define in the proof of the previous theorem. Suppose that f is Riemann integrable, then as we know from the previous proof, φ = ψ almost everywhere and f is continuous almost everywhere. Now assume that f is continuous almost everywhere, then we know that h = g almost everywhere and thus for the sequence of dyadic partitions of the interval the upper and lower Darboux integrals are equal. This implies that f is Riemann integrable.  4

5. Integration with respect to product measures 5.1. Monotone Class Lemma. Before we discuss integration on product spaces, we need one technical and useful result.

Definition 5.1. A family C of subsets of X is called a monotone class if for any increasing sequence of sets Ej in C their union E = ∪jEj is in C and for any decreasing sequence of sets Fj in C their intersection F = ∩jFj is in C.

Clearly any σ-algebra is a monotone class. Furthermore, an intersection of mono- tone classes is a monotone class, so for any E ∈ P(X) we can define the monotone class generated by E as the smallest monotone class that contains E.

Lemma 5.1. Suppose that A is an algebra of subsets of X, then the monotone class C generated by A coincides with the σ-algebra M generated by A.

Proof. Clearly C ⊂ M, we want to show that C is a σ-algebra. It is enough to show that C is an algebra. Then we write the countable union of sets in C as ∪jEj = ∪jFj, where Fj = ∪k≤jEk is an increasing sequence of sets and if each finite union Fj is in

C, which is a monotone class, we have that ∪Ej is in C. We note that X is in A and thus X ∈ C. So it suffices to prove that if E,F ∈ C then E ∩ F,E \ F,F \ E ∈ C. For each E ∈ C define D(E) = {F ∈ C : E ∩ F,E \ F,F \ E ∈ C}. We want to show first that D(E) is a monotone class. Suppose that Fj is an increasing sequence of sets in C then F = ∪jFj ∈ C and E ∩ F = ∩j(E ∩ Fj), F \ E = ∪J (Fj \ E) and

E \ F = ∩j(E \ Fj) are in C. Thus F is in D(E). Similarly, D(E) is closed under the intersections of increasing sequences. If A ∈ A then clearly A ⊂ D(A), and since D(A) is a monotone class, we have C ⊂ D(A). Therefore for any E ∈ C we have A ⊂ D(E) and once again C ⊂ D(E).  5.2. Fubini-Tonelli theorem. In this section we assume that (X, M, µ) and (Y, N , ν) are two measure spaces with σ-finite measures and denote by M × N the product σ-algebra and by µ × ν the product measure. We remind that for a set L ⊂ X × Y y we define Lx = {y ∈ Y :(x, y) ∈ L} and L = {x ∈ X :(x, y) ∈ L}.

y Theorem 5.1. Let E ⊂ M × N then the functions x 7→ ν(Ex) and y 7→ µ(E ) are well-defined on X and Y respectively, are measurable, and Z Z y µ × ν(E) = ν(Ex) dµ(x) = µ(E ) dν(y).

Proof. We assume first that the measures are finite and define by D the family of sets E for which the conclusion of the the theorem holds. Clearly all rectangular 5 sets A × B, where A ∈ M and B ∈ N , are in D. Moreover, finite disjoint unions of such sets are also in D. Those sets form an algebra, so it is enough to show that D is a monotone class. Suppose that En is an increasing sequence of sets in D and E = ∪nEn. Then (En)x are measurable and so is Ex = ∪n(En)x, we have also

ν(Ex) = limn ν((En)x), by the monotone convergence theorem Z Z ν(Ex) dµ = lim ν((En)x) dµ = lim (µ × ν)(En) = µ × ν(E). n→∞ n→∞ Similarly for µ(Ey). Since the measures are finite, we can do the same for decreasing sequence of sets Fn and F = ∩Fn. We consider functions ν((Fn)x) which converge to ν(Fx) and are bounded by the constant function ν(Y ) that is integrable, since µ(X) < ∞.

For the general case of σ-finite measures, we decompose X × Y = ∪j,kXj × Yk into disjoint sets, where µ(Xj) < ∞ and ν(Yk) < ∞. For any E ∈ M × N we write E = ∪j,k(E ∩ (Xj × Yk)) and apply the theorem on each piece. Applying the monotone convergence theorem twice, we get

X X Z µ × ν(E) = µ(E ∩ (Xj × Yk)) = ν(Ex ∩ Yk)χXj (x) dµ = j,k j,k X Z Z ν(Ex ∩ Yk) dµ = ν(Ex) dµ. k  Now we are ready to prove the main theorems of this section.

Theorem 5.2 (Tonelli). If f ∈ L+(M × N ) then g(x) = R f(x, ·) dν and h(y) = R f(·, y) dµ are in L+(M) and L+(N ) and Z Z Z  Z Z  f d(µ × ν) = f(x, y) dν(y) dµ(x) = f(x, y) dµ(x) dν(y).

Proof. If f is a characteristic function of a set in M × N , the theorem follows from the previous one. By linearity of the integral, the conclusion holds for any simple function f. In general, let f be the limit of an increasing sequence of simple functions R φn and let gn(x) = φn(x, ·) dν. Then, by the monotone convergence theorem, g(x) = limn→∞ gn(x). Moreover, gn(x) is an increasing sequence. Then once again by the monotone convergence theorem Z Z Z Z g dµ = lim gn dµ = lim φn d(µ × ν) = f d(µ × ν). n→∞ n→∞

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Theorem 5.3 (Fubini). Suppose that f ∈ L1(µ × ν) then f(x, ·) ∈ L1(ν) for µ-a.e. x, f(·, y) ∈ L1(µ) for ν-a.e. y and the functions g(x) = R f(x, ·)dν and h(y) = R f(·, y)dµ are in L1(µ) and L1(ν) respectively. Moreover, Z Z Z  Z Z  f d(µ × ν) = f(x, y) dν(y) dµ = f(x, y) dµ(x) dν(y).

Proof. We apply the Tonelli theorem for f + and f − and corresponding functions g+, h+ and g−, h−. Then R g+ dµ + R g− dµ = R |f| d(µ × ν) < ∞ and therefore g+(x) + g−(x) < ∞ for µ-a.e. x ∈ X it means that f(x, ·) ∈ L1(ν) for x, similarly for h(y). The conclusion of the theorem follows since Z Z Z f d(µ × ν) = f + d(µ × ν) − f − d(µ × ν).

 In the Tonelli and Fubini theorems the assumptions that measures are σ-finite are important, and in the Fubini theorem the assumption f ∈ L1 is important.