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13 Integration

We now come to the main use of measure theory: to define a general theory of integration.

13.1 Measurable functions

From now on, by a measure space we shall mean a triple (X,L,µ), where X is a set, L is a σ-algebra on X, and µ is a σ-additive measure defined on L. We say that the members of L are measurable, or L-measurable, if necessary to avoid confusion.

Definition 1  A function f:X→ℝ is measurable if
    Ec(f)={x∈ X: f(x)<c}    that is    Ec(f)=f−1((−∞,c))
is in L (that is Ec(f) is a measurable set) for any c∈ℝ.

A complex-valued function is measurable if its real and imaginary parts are measurable.

Lemma 2  The following are equivalent:
  1. A function f is measurable;
  2. For any a<b the set f−1((a,b)) is measurable;
  3. For any open set U⊂ ℝ the set f−1(U) is measurable.
Proof. To show 1 ⇒  2 we note that
    f−1((a,b))  = Eb(f)∖ ⎛
⎜
⎜
⎝
 
∩
n
 Ea+1/n(f) ⎞
⎟
⎟
⎠		.
For 2 ⇒  3 use that any open set U⊂ ℝ is a union of countable set of intervals (a,b), cf. proof of Cor. 25.

The final implication 3 ⇒  1 directly follows from openness of (−∞,a). □

Corollary 3  Let f: X → ℝ be measurable and g: ℝ → ℝ be continuous, then the composition g(f(x)) is measurable.
Proof. The preimage of the open set (−∞,c) under a continuous g is an open set, say U. The preimage of U under f is measurable by Lem. 3. Thus, the preimage of (−∞,c) under the composition g ∘ f is measurable, thereafter g ∘ f is a measurable function. □
Theorem 4  Let f,g:X→ℝ be measurable. Then af (a∈ℝ), f+g, fg, max(f,g) and min(f,g) are all measurable. That is measurable functions form an algebra and this algebra is closed under convergence a.e.
Proof. Use Cor. 3 to show measurability of λ f, | f | and f2. The measurability of a sum f1 + f2 follows from the relation
    Ec(f1+f2)=⋃r∈ℚ (Er(f1)⋂ Ec−r(f2)).
Next use the following identities:
    f1f2=
(f1+f2)2−(f1−f2)2
4
,
    max(f1,f2)=
(f1+f2)+⎪
⎪		f1−f2 ⎪
⎪
2
.

If (fn) is a non-increasing sequence of measurable functions converging to f. Than Ec(f)=∪n Ec(fn).

Moreover any limit can be replaced by two monotonic limits:

 
lim
n→ ∞
 fn(x)=
 
lim
n→ ∞
 
 
lim
k→ ∞
 max (fn(x), fn+1(x),…,fn+k(x)). (70)

Finally if f1 is measurable and f2=f1 almost everywhere, then f2 is measurable as well. □

We can define several types of convergence for measurable functions.

Definition 5  We say that sequence (fn) of functions converges
  1. uniformly to f (notated fn⇉ f) if
          
     
    sup
    x∈ X
    		⎪
    ⎪    		fn(x)−f(x) ⎪
    ⎪    		 → 0;
  2. almost everywhere to f (notated fn→a.e.f) if
          fn(x)→ f(x)    for all  x∈ X∖ A,  µ(A)=0;
  3. in measure µ to f (notated fn→µf) if for all ε>0
    µ({x∈ X: ⎪
    ⎪    		fn(x)−f(x) ⎪
    ⎪    		>ε }) → 0. (71)

Clearly uniform convergence implies both convergences a.e and in measure.

Theorem 6  On finite measures convergence a.e. implies convergence in measure.
Proof. Define An(ε)={x∈ X: | fn(x)−f(x) |≥ ε}. Let Bn(ε)=∪k≥ n Ak(ε). Clearly Bn(ε)⊃ Bn+1(ε), let B(ε)=∩1∞Bn(ε). If x∈ B(ε) then fn(x)↛f(x). Thus µ(B(ε))=0, but µ(B(ε))=limn→ ∞µ(Bn(ε)), cf. (67). Since An(ε)⊂ Bn(ε) we see that µ(An(ε))→ 0 as required for (71) □

Note, that the construction of sets Bn(ε) is just another implementation of the “two monotonic limits” trick (70) for sets.

Exercise 7  Present examples of sequences (fn) and functions f such that:
  1. fn→ µf but not fn→ a.e.f.
  2. fn→ a.e.f but not fn⇉ f.

However we can slightly “fix” either the set or the sequence to “upgrade” the convergence as shown in the following two theorems.

Theorem 8 (Egorov)  If fn→ a.e.f on a finite measure set X then for any σ>0 there is Eσ⊂ X with µ(Eσ)<σ and fn⇉ f on X∖ Eσ.
Proof. We use An(ε) and Bn(ε) from the proof of Thm. 6. Observe that | f(x)−fk(x) |< ε uniformly for all x ∈ X∖ Bn(ε) and k>n. For every ε>0 we seen that µ(Bn(ε))→ 0, thus for each k there is N(k) such that µ(BN(k)(1/k))<σ/2k. Put Eσ=∪k BN(k)(1/k). □
Theorem 9  If fn→ µf then there is a subsequence (nk) such that fnk→ a.e.f for k→ ∞.
Proof. In the notations of two previous proofs: for every natural k take nk such that µ(Ank(1/k))< 1/2k, which is possible since µ(An(ε))→ 0. Define Cm=∪k=m∞Ank(1/k) and C=∩ Cm. Then, µ(Cm)=1/2m−1 and, thus, µ(C)=0 by (67). If x∉C then there is such N that x∉Ank(1/k) for all k>N. That means that | fnk(x)−f(x) |<1/k for all such k, i.e fnk(x)→ f(x). Thus, we have the point-wise convergence everywhere except the zero-measure set C. □

It is worth to note, that we can use the last two theorem subsequently and upgrade the convergence in measure to the uniform convergence of a subsequence on a subset.

Exercise 10  For your counter examples from Exercise 7, find
  1. a subsequence fnk of the sequence from 1 which converges to f a.e.;
  2. a subset such that sequence from 2 converges uniformly.
Exercise 11  Read about Luzin’s C-property.

13.2 Lebesgue Integral

First we define a sort of “basis” for the space of integral functions.

Definition 12  For A⊆ X, we define χA to be the indicator function of A, by
    χA(x) = 
⎧
⎨
⎩
1: x∈ A, 
0: x∉A. 

Then, if χA is measurable, then χA−1( (1/2,3/2) ) = A ∈ L; conversely, if A∈L, then X∖ A∈L, and we see that for any U⊆ℝ open, χA−1(U) is either ∅, A, X∖ A, or X, all of which are in L. So χA is measurable if and only if A∈L.

Definition 13  A measurable function f:X→ℝ is simple if it attains only a countable number of values.
Lemma 14  A function f:X→ℝ is simple if and only if
f = 
∞
∑
k=1
tk χAk  (72)
for some (tk)k=1∞⊆ℝ and Ak∈L. That is, simple functions are linear combinations of indicator functions of measurable sets.

Moreover in the above representation the sets Ak can be pair-wise disjoint and all tk≠ 0 pair-wise different. In this case the representation is unique.

Notice that it is now obvious that

Corollary 15  The collection of simple functions forms a vector space: this wasn’t clear from the original definition.
Definition 16  A simple function in the form (72) with disjoint Ak is called summable if the following series converges:
∞
∑
k=1
⎪
⎪		tk ⎪
⎪		 µ(Ak)   if f has the above unique representation   f = 
∞
∑
k=1
tk χAk . (73)

It is another combinatorial exercise to show that this definition is independent of the way we write f.

Definition 17  We define the integral of a simple function f=∑k tk χAk (72) over a measurable set A by setting
    ∫
 


A
 f  d µ = 
∞
∑
k=1
tk µ(Ak⋂ A).

Clearly the series converges for any simple summable function f. Moreover

Lemma 18  The value of integral of a simple summable function is independent from its representation by the sum of indicators (72). In particular, we can evaluate the integral taking the canonical representation over pair-wise disjoint sets having pair-wise different values.
Proof. This is another slightly tedious combinatorial exercise. You need to prove that the integral of a simple function is well-defined, in the sense that it is independent of the way we choose to write the simple function. □
Exercise 19  Let f be the function on [0,1] which take the value 1 in all rational points and 0—everywhere else. Find the value of the Lebesgue integral ∫[0,1] f,dµ with respect to the Lebesgue measure on [0,1]. Show that the Riemann upper- and lower sums for f converges to different values, so f is not Riemann-integrable.
Remark 20  The previous exercise shows that the Lebesgue integral does not have those problems of the Riemann integral related to discontinuities. Indeed, most of function which are not Riemann-integrable are integrable in the sense of Lebesgue. The only reason, why a measurable function is not integrable by Lebesgue is divergence of the series (73). Therefore, we prefer to speak that the function is summable rather than integrable. However, those terms are used interchangeably in the mathematical literature.

We will denote by S(X) the collection of all simple summable functions on X.

Proposition 21  Let f, g:X→ ℝ be in S(X) (that is simple summable), let a, b∈ ℝ and A is a measurable set. Then:
  1. ∫A af+bg  d µ = a∫A f  d µ + b∫A g  d µ, that is S(X) is a linear space;
  2. The correspondence f→ ∫A f d µ is a linear functional on S(X);
  3. The correspondence A → ∫A f d µ is a charge;
  4. If f≤ g then ∫X f  d µ ≤ ∫X g  d µ, that is integral is monotonic;
  5. The function
    d1(f,g)=∫
     


    X
    		⎪
    ⎪    		f(x)−g(x) ⎪
    ⎪    		 d µ(x) (74)
    has all properties of a metric (distance) on S(X) probably except separation, but see the next item.
  6. For f≥ 0 we have ∫X f  d µ=0 if and only if µ( { x∈ X : f(x)≠0 } ) = 0. Therefore for the function d1 (74):
          d1(f,g)=0    if and only if    f
    a.e.
    =
     
    		g.
  7. The integral is uniformly continuous with respect the above metric d1 (74):
          ⎪
    ⎪
    ⎪
    ⎪
    ⎪
    ⎪    		∫
     


    A
    		 f(x) d µ(x)−∫
     


    A
    		 g(x) d µ(x) ⎪
    ⎪
    ⎪
    ⎪
    ⎪
    ⎪    		 ≤ d1(f,g).
Proof. The proof is almost obvious, for example the Property 1 easily follows from Lem. 18.

We will outline 3 only. Let f is an indicator function of a set B, then A→ ∫A f d µ=µ(A∩ B) is a σ-additive measure (and thus—a charge). By the Cor. 35 the same is true for finite linear combinations of indicator functions and their limits in the sense of distance d1. □

We can identify functions which has the same values a.e. Then S(X) becomes a metric space with the distance d1 (74). The space may be incomplete and we may wish to look for its completion. However, if we will simply try to assign a limiting point to every Cauchy sequence in S(X), then the resulting space becomes so huge that it will be impossible to realise it as a space of functions on X.

Exercise 22  Use ideas of Ex. 1 to present a sequence of simple functions which has the Cauchy property in metric d1 (74) but does not have point-wise limits anywhere.

To reduce the number of Cauchy sequences in S(X) eligible to have a limit, we shall ask an additional condition. A convenient reduction to functions on X appears if we ask both the convergence in d1 metric and the point-wise convergence on X a.e.

Definition 23  A function f is summable by a measure µ if there is a sequence (fn)⊂S(X) such that
  1. the sequence (fn) is a Cauchy sequence in S(X);
  2. fn→a.e. f.

Clearly, if a function is summable, then any equivalent function is summable as well. Set of equivalent classes of summable functions will be denoted by L1(X).

Lemma 24  If the measure µ is finite then any bounded measurable function is summable.
Proof. Define Ekn(f)={x∈ X: k/n≤ f(x)< (k+1)/n} and fn=∑k k/n χEkn (note that the sum is finite due to boundedness of f).

Since | fn(x)−f(x) |<1/n we have uniform convergence (thus convergence a.e.) and (fn) is the Cauchy sequence: d1(fn,fm)=∫X| fn−fm | d µ≤ (1/n+1/m)µ(X). □

Remark 25  This Lemma can be extended to the space of essentially bounded functions L∞(X), that is functions which are bounded a.e. In other words, L∞(X)⊂L1(X) for finite measures.

Another simple result, which is useful on many occasions is as follows.

Lemma 26  If the measure µ is finite and fn⇉ f then d1(fn,f)→ 0.
Corollary 27  For a convergent sequence fn→a.e. f, which admits the uniform bound | fn(x) |<M for all n and x, we have d1(fn,f)→ 0.
Proof. For any ε>0, by the Egorov’s theorem 8 we can find E, such that
  1. µ(E)< ε/2M; and
  2. from the uniform convergence on X∖ E there exists N such that for any n>N we have | f(x)−fn(x) |<ε /2µ(X).
Combining this we found that for n>N, d1(fn,f)< M ε/2M + µ(X) ε /2µ(X) < ε . □
Exercise 28  Convergence in the metric d1 and a.e. do not imply each other:
  1. Give an example of fn→a.e. f such that d1(fn ,f)↛0.
  2. Give an example of the sequence (fn) and function f in L1(X) such that d1(fn ,f)→ 0 but fn does not converge to f a.e.

To build integral we need the following

Lemma 29  Let (fn) and (gn) be two Cauchy sequences in S(X) with the same limit a.e., then d1(fn,gn)→ 0.
Proof. Let φn=fn−gn, then this is a Cauchy sequence with zero limit a.e. Assume the opposite to the statement: there exist δ>0 and sequence (nk) such that ∫x| φnk | d µ>δ. Rescaling-renumbering we can obtain ∫x| φn | d µ>1.

Take quickly convergent subsequence using the Cauchy property:

    d1(φnk,φnk+1)≤ 1/2k+2.

Renumbering agian assume d1(φk,φk+1)≤ 1/2k+2.

Since φ1 is a simple, take the canonical presentation φ1=∑k tk χAk, then ∑k | tk | µ(Ak)=∫X | φ1 | d µ≥ 1. Thus, there exists N, such that ∑k=1N | tk | µ(Ak)≥ 3/4. Put A=⊔k=1N Ak and C=max1≤ k ≤ N| tk |=maxx∈ A| φ1(x) |.

By the Egorov’s Theorem 8 there is E⊂ A such that µ(E)<1/(4C) and φn⇉ 0 on B=A∖ E. Then

    ∫
 


B
 ⎪
⎪		φ1 ⎪
⎪		 d µ= ∫
 


A
 ⎪
⎪		φ1 ⎪
⎪		 d µ−∫
 


E
 ⎪
⎪		φ1 ⎪
⎪		 d µ≥
3
4
−
1
4C
· C=
1
2
.

By the triangle inequality for d1:

    ⎪
⎪
⎪
⎪
⎪
⎪		∫
 


B
 ⎪
⎪		φn ⎪
⎪		 d µ−∫
 


B
 ⎪
⎪		φn+1 ⎪
⎪		 d µ ⎪
⎪
⎪
⎪
⎪
⎪		 ≤ d1(φn,φn+1)≤ 
1
2n+2

we get

    ∫
 


B
 ⎪
⎪		φn ⎪
⎪		 d µ≥ ∫
 


B
 ⎪
⎪		φ1 ⎪
⎪		 d µ−
n−1
∑
k=1
⎪
⎪
⎪
⎪
⎪
⎪		∫
 


B
 ⎪
⎪		φn ⎪
⎪		 d µ−∫
 


B
 ⎪
⎪		φn+1 ⎪
⎪		 d µ ⎪
⎪
⎪
⎪
⎪
⎪		≥
1
2
−
n−1
∑
1
1
2k+2
>
1
4
.

But this contradicts to the fact ∫B | φn | d µ → 0, which follows from the uniform convergence φn⇉ 0 on B. □

It follows from the Lemma that we can use any Cauchy sequence of simple functions for the extension of integral.

Corollary 30  The functional IA(f)=∫A f(x) d µ(x), defined on any A∈ L on the space of simple functions S(X) can be extended by continuity to the functional on L1(X,µ).
Definition 31  For an arbitrary summable f∈L1(X), we define the Lebesgue integral
    ∫
 


A
 f  d µ =
 
lim
n→ ∞
 ∫
 


A
 fn  d µ,
where the Cauchy sequence fn of summable simple functions converges to f a.e.
Theorem 32 
  1. L1(X) is a linear space.
  2. For any measurable set A⊂ X the correspondence f↦ ∫A f  d µ is a linear functional on L1(X).
  3. For any f∈L1(X) the value ν(A)=∫A f  d µ is a charge.
  4. d1(f,g)=∫A | f−g |  d µ is a distance on L1(X).
Proof. The proof is follows from Prop. 21 and continuity of extension. □

Summing up: we build L1(X) as a completion of S(X) with respect to the distance d1 such that elements of L1(X) are associated with (equivalence classes of) measurable functions on X.

13.3 Properties of the Lebesgue Integral

The space L1 was defined from dual convergence—in d1 metric and point-wise a.e. Can we get the continuity of the integral from the convergence almost everywhere alone? No, in general. However, we will state now some results on continuity of the integral under convergence a.e. with some additional assumptions. Finally, we show that L1(X) is closed in d1 metric.

Theorem 33 (Lebesgue on dominated convergence)  Let (fn) be a sequence of µ-summable functions on X, and there is φ∈L1(X) such that | fn(x) |≤ φ(x) for all x∈ X, n∈ℕ.

If fn→a.e. f, then f∈L1(X) and for any measurable A:

    
 
lim
n→∞
 ∫
 


A
 fn  d µ    =    ∫
 


A
 f  d µ.

Proof. For any measurable A the expression ν(A)=∫A φ  d µ defines a finite measure on X due to non-negativeness of φ and Thm. 32.
Lemma 34 (Change of variable)  If g is measurable and bounded then f=φ g is µ-summable and for any µ-measurable set A we have
∫
 


A
 f  d µ= ∫
 


A
 g  d ν. (75)
Proof.[Proof of the Lemma] Let M be the set of all g such that the Lemma is true. M includes any indicator functions g=χB of a measurable B:
      ∫
 


A
 f  d µ=∫
 


A
 φχB  d µ = ∫
 


A⋂ B
 φ  d µ =ν(A⋂ B)=∫
 


A
 χB d ν=∫
 


A
 g d ν.      
Thus M contains also finite linear combinations of indicators. For any n∈ℕ and a bounded g two functions g−(x)=1/n[ng(x)] and g+(x)=g−+1/n are finite linear combinations of indicators and are in M. Since g−(x)≤ g(x)≤ g+(x) we have
      ∫
 


A
 g− d ν= ∫
 


A
 φ g− d µ≤ ∫
 


A
 φ g d µ≤ ∫
 


A
 φ g+ d µ=∫
 


A
 g+ d ν. 
By squeeze rule for n→ ∞ we have the middle term tenses to ∫Ag d ν, that is g∈ M.

Note, that formula (75) is a change of variable in the Lebesgue integral of the type: ∫f(sinx) cosx d x = ∫f(sinx)  d (sinx). □

For the proof of the theorem define:
    gn(x)=
⎧
⎨
⎩
        fn(x)/φ(x),if  φ(x)≠ 0,
        0,if  φ(x)= 0,
    g(x)=
⎧
⎨
⎩
        f(x)/φ(x),if  φ(x)≠ 0,
        0,if  φ(x)= 0.
  
Then gn is bounded by 1 and gn→a.e. g. To show the theorem it will be enough to show limn→ ∞∫A gn d ν=∫A g d ν. For the uniformly bounded functions on the finite measure set this can be derived from the Egorov’s Thm. 8, see an example of this in the proof of Lemma 29. □

Note, that in the above proof summability of φ was used to obtain the finiteness of the measure ν, which is required for Egorov’s Thm. 8.

Exercise 35  Give an example of fn→a.e. f such that ∫X fn  d µ ≠ ∫X f  d µ. For such an example, try to find a function φ such that | fn | ≤ φ for all n and check either φ is summable.
Exercise 36 (Chebyshev’s inequality)  Show that: if f is non-negative and summable, then
µ{x∈ X: f(x)>c} < 
1
c
∫
 


X
 f d µ. (76)
Theorem 37 (B. Levi’s, on monotone convergence)  Let (fn) be monotonically increasing sequence of µ-summable functions on X. Define f(x)=limn→∞ fn(x) (allowing the value +∞).
  1. If all integrals ∫X fn d µ are bounded by the same value C<∞ then f is summable and ∫X f d µ=limn→∞∫X fn d µ.
  2. If limn→∞∫X fn d µ=+∞ then function f is not summable.
Proof. Replacing fn by fn−f1 and f by f−f1 we can assume fn≥ 0 and f≥ 0. Let E be the set where f is infinite, then
    E=⋂N⋃n ENn,    where   ENn={x∈ X: fn(x)≥ N}.
By Chebyshev’s inequality (76) we have
    Nµ(ENn) <  ∫
 


ENn
fn d µ ≤ ∫
 


X
fn d µ≤ C,
then µ(ENn)≤ C/N . Thus µ(E)=limN→∞limn→∞ µ(ENn)=0.

Thus f is finite a.e.

Lemma 38  Let f be a measurable non-negative function attaining only finite values. f is summable if and only if sup∫A f d µ<∞, where the supremum is taken over all finite-measure set A such that f is bounded on A.
Proof.[Proof of the Lemma] Necessity: if f is summable then for any set A⊂ X we have ∫A f d µ≤ ∫X f d µ<∞, thus the supremum is finite.

Sufficiency: let sup∫A f d µ=M<∞, define B={x∈ X: f(x)=0} and Ak={x∈ X: 2k≤ f(x)<2k+1, k∈ℤ}, by (76) we have µ(Ak)<M/2k and X=B⊔(⊔k=0∞Ak). Define

      g(x)=
⎧
⎨
⎩
          2k,if  x∈ Ak,
          0,if  x∈ B,
      fn(x)=
⎧
⎨
⎩
          f(x),if  x∈ ⊔−nn An,
          0,otherwise.
    

Then g(x)≤ f(x) < 2g(x). Function g is a simple function, its summability follows from the estimate ∫⊔−nn Ak g d µ≤∫⊔−nn Ak f d µ≤ M which is valid for any n, taking n→ ∞ we get summability of g. Furthermore, fn →a.e. f and fn(x)≤ f(x) <2g(x), so we use the Lebesgue Thm. 33 on dominated convergence to obtain the conclusion. □

Let A be a finite measure set such that f is bounded on A, then

    ∫
 


A
 f d µ
Cor. 27
=
 
 
lim
n→∞
∫
 


A
 fn d µ≤
 
lim
n→∞
∫
 


X
 fn d µ≤ C.

This show summability of f by the previous Lemma. The rest of statement and (contrapositive to) the second part follows from the Lebesgue Thm. 33 on dominated convergence. □

Now we can extend this result dropping the monotonicity assumption.

Lemma 39 (Fatou)  If a sequence (fn) of µ-summable non-negative functions is such that: then f is µ-summable and ∫X f d µ≤ C.
Proof.Let us replace the limit fn→ f by two monotonic limits. Define:
    gkn(x)=min(fn(x),…,fn+k(x)),
    gn(x)=
 
lim
k→ ∞
 gkn(x).
Then gn is a non-decreasing sequence of functions and limn→ ∞ gn(x)=f(x) a.e. Since gn≤ fn, from monotonicity of integral we get ∫X gn d µ≤ C for all n. Then Levi’s Thm. 37 implies that f is summable and ∫X f d µ≤ C. □
Remark 40  Note that the price for dropping monotonicity from Thm. 37 to Lem. 39 is that the limit ∫X fn d µ → ∫X f d µ may not hold any more.
Exercise 41  Give an example such that under the Fatou’s lemma condition we get limn→∞∫X fn d µ ≠ ∫X f d µ.

Now we can show that L1(X) is complete:

Theorem 42  L1(X) is a Banach space.
Proof. It is clear that the distance function d1 indeed define a norm ||f||1=d1(f,0). We only need to demonstrate the completeness. We again utilise the three-step procedure from Rem. 7.

Take a Cauchy sequence (fn) and building a subsequence if necessary, assume that its quickly convergent that is d1(fn,fn+1)≤ 1/2k. Put

    φ1=f1  and     φn=fn−fn−1  for   n>1. Then  fn=
n
∑
k=1
 φk .

The sequence ψn(x)=∑1n | φk(x) | is monotonic, integrals ∫X ψn d µ are bounded by the same constant ||f1||1+1. Thus, by the B. Levi’s Thm. 37 and its proof, ψn→ ψ for a summable essentially bounded function ψ. Therefore, the series ∑φk(x) converges as well to a value f(x) of a function f. But, this means that fn →a.e. f (the first step is completed).

We also notice | fn(x) |≤| ψ(x) |. Thus by the Lebesgue Thm. 33 on dominated convergence f∈ L1(X) (the second step is completed).

Furthermore,

0≤
 
lim
n→ ∞
 ∫
 


X
⎪
⎪		fn−f ⎪
⎪		 d µ≤
 
lim
n→ ∞
 
∞
∑
k=n
⎪⎪
⎪⎪		φk⎪⎪
⎪⎪		=0.

That is, fn→ f in the norm of L1(X). (That completes the third step and the whole proof). □

The next important property of the Lebesgue integral is its absolute continuity.

Theorem 43 (Absolute continuity of Lebesgue integral)  Let f∈L1(X). Then for any ε>0 there is a δ>0 such that | ∫A f d µ |<ε if µ(A)<δ.
Proof. If f is essentially bounded by M, then it is enough to set δ=ε/M. In general let:
    An=
{x∈ X: n≤⎪
⎪		f(x) ⎪
⎪		< n+1},
    Bn=⊔0n Ak,
    Cn=X∖ Bn.
Then ∫X| f | d µ=∑0∞∫Ak| f | d µ, thus there is an N such that ∑N∞∫Ak| f | d µ=∫CN| f | d µ<ε/2. Now put δ =ε/2N+2, then for any A⊂ X with µ(A)<δ:
    ⎪
⎪
⎪
⎪
⎪
⎪		∫
 


A
 f d µ ⎪
⎪
⎪
⎪
⎪
⎪		≤∫
 


A
 ⎪
⎪		f ⎪
⎪		 d µ= ∫
 


A⋂ BN
 ⎪
⎪		f ⎪
⎪		 d µ+∫
 


A⋂ CN
 ⎪
⎪		f ⎪
⎪		 d µ < 
ε
2
+
ε
2
=ε.
□

13.4 Integration on Product Measures

It is well-known geometrical interpretation of an integral in calculus as the “area under the graph”. If we advance from “area” to a “measure” then the Lebesgue integral can be treated as theory of measures of very special shapes created by graphs of functions. This shapes belong to the product spaces of the function domain and its range. We introduced product measures in Defn. 40, now we will study them in same details using the Lebesgue integral. We start from the following

Theorem 44  Let X and Y be spaces, and let S and T be semirings on X and Y respectively and µ and ν be measures on S and T respectively. If µ and ν are σ-additive, then the product measure ν× µ from Defn. 40 is σ-additive as well.
Proof. For any C=A× B∈ S× T let us define fC(x)=χA(x)ν(B). Then
    (µ×ν)(C)=µ(A)ν(B)=∫
 


X
 fC d µ.
If the same set C has a representation C=⊔k Ck for Ck∈ S× T, then σ-additivity of ν implies fC=∑k fCk. By the Lebesgue theorem 33 on dominated convergence:
    ∫
 


X
 fC d µ=
 
∑
k
∫
 


X
 fCk d µ.
Thus
    (µ×ν)(C)=
 
∑
k
(µ×ν)(Ck).
□

The above correspondence C↦ fC can be extended to the ring R(S× T) generated by S× T by the formula:

  fC=
 
∑
k
 fCk,    for  C=⊔k Ck∈ R(S× T).

We have the uniform continuity of this correspondence:

  ⎪⎪
⎪⎪		fC1−fC2⎪⎪
⎪⎪		1≤ (µ×ν)(C1▵ C2)=d1(C1,C2)

because from the representation C1=A1⊔ B and C2=A2⊔ B, where B=C1∩ C2 one can see that fC1−fC2=fA1−fA2, fC1▵ C2=fA1+fA2 together with | fA1−fA2 |≤ fA1+fA2 for non-negative functions.

Thus the map C↦ fC can be extended to the map of σ-algebra L(X× Y) of µ×ν-measurable set to L1(X) by the formula flimn Cn=limn fCn.

Exercise 45  Describe topologies where two limits from the last formula are taken.

The following lemma provides the geometric interpretation of the function fC as the size of the slice of the set C along x=const.

Lemma 46  Let C∈L(X× Y). For almost every x∈ X the set Cx={y∈ Y: (x,y)∈ C} is ν-measurable and ν(Cx)=fC(x).
Proof. For sets from the ring R(S× T) it is true by the definition. If C(n) is a monotonic sequence of sets, then ν(limn Cx(n))=limn ν(Cx(n)) by σ-additivity of measures. Thus the property ν(Cx)=fx(C) is preserved by monotonic limits. The following result of the separate interest:
Lemma 47  Any measurable set can be received (up to a set of zero measure) from elementary sets by two monotonic limits.
Proof.[Proof of Lem. 47] Let C be a measurable set, put Cn∈R(S× T) to approximate C up to 2−n in µ×ν. Let C′=∩n=1∞∪k =1∞Cn+k, then
      (µ× ν)⎛
⎝		C∖ ⋃k=1∞Cn+k⎞
⎠		=0  and   (µ× ν)⎛
⎝		⋃k=1∞Cn+k∖ C⎞
⎠		=21−n.
Then (µ×ν)(C′▵ C)≤ 21−n for any n∈ℕ. □
Coming back to Lem. 46 we notice that (in the above notations) fC=fC′ almost everywhere. Then:
    fC(x)
a.e
=
 
fC′(x)=ν(C′x)=ν(Cx).
□

The following theorem generalizes the meaning of the integral as “area under the graph”.

Theorem 48  Let µ and ν are σ-finite measures and C be a µ×ν measurable set X× Y. We define Cx={y∈ Y: (x,y)∈ C}. Then for µ-almost every x∈ X the set Cx is ν-measurable, function fC(x)=ν(Cx) is µ-measurable and
(µ×ν)(C)=∫
 


X
 fC d µ, (77)
where both parts may have the value +∞.
Proof. If C has a finite measure, then the statement is reduced to Lem. 46 and a passage to limit in (77).

If C has an infinite measure, then there exists a sequence of Cn⊂ C, such that ∪n Cn=C and (µ×ν)(Cn)→ ∞. Then fC(x)=limn fCn (x) and

    ∫
 


X
 fCn d µ=(µ×ν)(Cn)→ +∞.

Thus fC is measurable and non-summable. □

This theorem justify the well-known technique to calculation of areas (volumes) as integrals of length (areas) of the sections.

Remark 49 
  1. The role of spaces X and Y in Theorem 48 is symmetric, thus we can swap them in the conclusion.
  2. The Theorem 48 can be extended to any finite number of measure spaces. For the case of three spaces (X,µ), (Y,ν), (Z,λ) we have:
    (µ×ν×λ )(C)=∫
     


    X× Y
    		 λ(Cxy) d (µ×ν)(x,y)= ∫
     


    Z
    		 (µ×ν)(Cz) d λ(z), (78)
    where
        Cxy={z∈ Z: (x,y,z)∈ C},
        Cz={(x,y)∈ X× Y: (x,y,z) ∈ C}.
Theorem 50 (Fubini)  Let f(x,y) be a summable function on the product of spaces (X,µ) and (Y,ν). Then:
  1. For µ-almost every x∈ X the function f(x,y) is summable on Y and fY(x)=∫Y f(x,y) d ν(y) is a µ-summable on X.
  2. For ν-almost every y∈ Y the function f(x,y) is summable on X and fX(y)=∫X f(x,y) d µ(x) is a ν-summable on Y.
  3. There are the identities:
         
            ∫
     


    X× Y
    		 f(x,y) d (µ×ν)(x,y)
    		=
    ∫
     


    X
    		⎛
    ⎜
    ⎜
    ⎜
    ⎜
    ⎝    		∫
     


    Y
    		 f(x,y) d ν(y)⎞
    ⎟
    ⎟
    ⎟
    ⎟
    ⎠    		dµ(x)
    		(79)
     =
     ∫
     


    Y
    		⎛
    ⎜
    ⎜
    ⎜
    ⎜
    ⎝    		∫
     


    X
    		 f(x,y) d µ(x)⎞
    ⎟
    ⎟
    ⎟
    ⎟
    ⎠    		dν(y).  
    		 
  4. For a non-negative functions the existence of any repeated integral in (79) implies summability of f on X× Y.
Proof. From the decomposition f=f+−f− we can reduce our consideration to non-negative functions. Let us consider the product of three spaces (X,µ), (Y,ν), (ℝ,λ), with λ=dz being the Lebesgue measure on ℝ. Define
    C={(x,y,z)∈ X× Y× ℝ: 0≤ z≤ f(x,y)}.
Using the relation (78) we get:
    Cxy={z∈ ℝ: 0≤ z≤ f(x,y)},    λ(Cxy)=f(x,y)
    Cx=
{(y,z)∈ Y× ℝ: 0≤ z≤ f(x,y)},    (ν× λ)(Cx)=∫
 


Y
 f(x,y) d ν(y).
the theorem follows from those relations. □
Exercise 51 

13.5 Absolute Continuity of Measures

Here, we consider another topic in the measure theory which benefits from the integration theory.

Definition 52  Let X be a set with σ-algebra R and σ-finite measure µ and finite charge ν on R. The charge ν is absolutely continuous with respect to µ if µ(A)=0 for A∈ R implies ν(A)=0. Two charges ν1 and ν2 are equivalent if two conditions | ν1 |(A)=0 and | ν2 |(A)=0 are equivalent.

The above definition seems to be not justifying “absolute continuity” name, but this will become clear from the following important theorem.

Theorem 53 (Radon–Nikodym)  Any charge ν which absolutely continuous with respect to a measure µ has the form
    ν(A)=∫
 


A
 f d µ,
where f is a function from L1. The function f∈L1 is uniquely defined by the charge ν.
Proof.[Sketch of the proof] First we will assume that ν is a measure. Let D be the collection of measurable functions g:X→[0,∞) such that
    ∫
 


E
 g  d µ ≤ ν(E)    (E∈L). 
Let α = supg∈D ∫X g  d µ ≤ ν(X) < ∞. So we can find a sequence (gn) in D with ∫X gn  d µ → α.

We define f0(x) = supn gn(x). We can show that f0=∞ only on a set of µ-measure zero, so if we adjust f0 on this set, we get a measurable function f:X→[0,∞). There is now a long argument to show that f is as required.

If ν is a charge, we can find f by applying the previous operation to the measures ν+ and ν− (as it is easy to verify that ν+,ν−⋘µ).

We show that f is essentially unique. If g is another function inducing ν, then

    ∫
 


E
 f−g  d µ = ν(E) − ν(E) = 0    (E∈L). 

Let E = {x∈ X : f(x)−g(x)≥ 0}, so as f−g is measurable, E∈L. Then ∫E f−g  d µ =0 and f−g≥0 on E, so by our result from integration theory, we have that f−g=0 almost everywhere on E. Similarly, if F = {x∈ X : f(x)−g(x)≤ 0}, then F∈L and f−g=0 almost everywhere on F. As E∪ F=X, we conclude that f=g almost everywhere. □

Corollary 54  Let µ be a measure on X, ν be a finite charge, which is absolutely continuous with respect to µ. For any ε>0 there exists δ>0 such that µ(A)<δ implies | ν |(A)<ε .
Proof. By the Radon–Nikodym theorem there is a function f∈L1(X,µ) such that ν(A)=∫A f d µ. Then | ν |(A)=∫A | f | d µ ad we get the statement from Theorem 43 on absolute continuity of the Lebesgue integral. □
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