Measure Theory

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12 Measure Theory

The presentation in this section is close to [, , ].

12.1 Basic Measure Theory

The following object will be the cornerstone of our construction.

Definition 1 Let X be a set. A σ-algebra R on X is a collection of subsets of X, written R⊆ 2^X, such that

X∈R;
if A,B∈R, then A∖ B∈R;
if (A_n) is any sequence in R, then ∪_n A_n∈R.

Note, that in the third condition we admit any countable unions. The usage of “σ” in the names of σ -algebra and σ-ring is a reference to this. If we replace the condition by

if (A_n)₁^m is any finite family in R, then ∪_n=1^m A_n∈R;

then we obtain definitions of an algebra.

For a σ-algebra R and A,B∈R, we have

A ⋂ B = X∖

⎛
⎝

X∖(A⋂ B)

⎞
⎠

= X ∖

⎛
⎝

(X∖ A)⋃(X∖ B)

⎞
⎠

∈R.

Similarly, R is closed under taking (countably) infinite intersections.

If we drop the first condition from the definition of (σ-)algebra (but keep the above conclusion from it!) we got a (σ-)ring, that is a (σ-)ring is closed under (countable) unions, (countable) intersections and subtractions of sets.

Exercise 2

Use the above comments to write in full the three missing definitions: of set algebra, set ring and set σ-ring.
Show that the empty set belongs to any non-empty ring.

Sets A_k are pairwise disjoint if A_n∩ A_m=∅ for n≠m. We denote the union of pairwise disjoint sets by ⊔, e.g. A ⊔ B ⊔ C.

It is easy to work with a vector space through its basis. For a ring of sets the following notion works as a helpful “basis”.

Definition 3 A semiring S of sets is a collection such that

it is closed under intersection;
for A, B∈ S we have A∖ B=C₁⊔ … ⊔ C_N with C_k∈ S.

Again, any non-empty semiring contain the empty set.

Example 4 The following are semirings but not rings:

The collection of intervals [a,b) on the real line;
The collection of all rectangles { a≤ x < b, c≤ y <d } on the plane.

As the intersection of a family of σ-algebras is again a σ-algebra, and the power set 2^X is a σ-algebra, it follows that given any collection D⊆ 2^X, there is a σ-algebra R such that D⊆R, such that if S is any other σ-algebra, with D⊆S, then R⊆S. We call R the σ-algebra generated by D.

Exercise 5 Let S be a semiring. Show that

The collection of all finite disjoint unions ⊔_k=1ⁿ A_k, where A_k∈ S, is a ring. We call it the ring R(S) generated by the semiring S.
Any ring containing S contains R(S) as well.
The collection of all finite (not necessarily disjoint!) unions ∪_k=1ⁿ A_k, where A_k∈ S, coincides with R(S).

We introduce the symbols +∞, −∞, and treat these as being “extended real numbers”, so −∞ < t < ∞ for t∈ℝ. We define t+∞ = ∞, t∞ = ∞ if t>0 and so forth. We do not (and cannot, in a consistent manner) define ∞ − ∞ or 0·∞.

Definition 6 A measure is a map µ:R→[0,∞] defined on a (semi-)ring (or σ-algebra) R, such that if A=⊔_n A_n for A∈R and a finite subset (A_n) of R, then µ (A) = ∑_n µ(A_n). This property is called additivity of a measure.

The additivity property of a measure is rather demanding. For example, let us consider the decomposition [0,1)=[0,1/2) ⊔ [1/2,1) = [0,1/3) ⊔ [1/3,2/3) ⊔ [2/3,1), then additivity puts measures of those five intervals into equations:

µ([0,

)) + µ( [

,1) ) = µ([0.1)) = µ([0,

)) + µ([

,1)).

Similar equations appear from any other (out of infinitely many) decomposition of [0,1), thus measures of various intervals are highly interconnected and very far from being arbitrary.

Exercise 7 Show that the following two conditions are equivalent:

µ(∅)=0.
There is a set A∈R such that µ(A)<∞.

The first condition often (but not always) is included in the definition of a measure.

In analysis we are interested in infinities and limits, thus the following extension of additivity is very important.

Definition 8 In terms of the previous definition we say that µ is countably additive (or σ-additive) if for any countable infinite family (A_n) of pairwise disjoint sets from R such that A=⊔_n A_n∈R we have µ(A) = ∑_n µ(A_n). If the sum diverges, then as it will be the sum of positive numbers, we can, without problem, define it to be +∞.

Note, that this property may be stated as a sort of continuity of an additive measure, cf. (7):

⎛
⎜
⎜
⎝

lim

n→∞

⊔

k=1

A_k

⎞
⎟
⎟
⎠

lim

n→∞

⎛
⎜
⎜
⎝

⊔

k=1

A_k

⎞
⎟
⎟
⎠

Example 9

Fix a point a∈ℝ and define a measure µ by the condition µ(A)=1 if a∈ A and µ(A)=0 otherwise.
For the ring obtained in Exercise 5 from semiring S in Example 1 define µ([a,b))=b−a on S. This is a measure, and we will show its σ-additivity.
For ring obtained in Exercise 5 from the semiring in Example 2, define µ(V)=(b−a)(d−c) for the rectangle V={ a≤ x < b, c≤ y <d } S. It will be again a σ-additive measure.
Let X=ℕ and R=2^ℕ, we define µ(A)=0 if A is a finite subset of X=ℕ and µ(A)=+∞ otherwise. Let A_n={n}, then µ(A_n)=0 and µ(⊔_n A_n)=µ(ℕ)=+∞≠ ∑_n µ(A_n)=0. Thus, this measure is not σ-additive.

We will see further examples of measures which are not σ-additive in Section 12.4.

Definition 10 A measure µ is finite if µ(A)<∞ for all A∈ R.

A measure µ is σ-finite if X is a union of countable number of sets X_k, such that for any A∈ R and any k∈ ℕ the intersection A∩ X_k is in R and µ(A∩ X_k)<∞.

Exercise 11 Modify the example 1 to obtain

a measure which is not finite, but is σ-finite. (Hint: let the measure count the number of integer points in a set).
a measure which is not σ-finite. (Hint: assign µ(A)=+∞ if a∈ A.)

Proposition 12 Let µ be a σ-additive measure on a σ-algebra R. Then:

If A,B∈R with A⊆ B, then µ(A)≤µ(B) [we call this property “monotonicity of a measure”];
If A,B∈R with A⊆ B and µ(B)<∞, then µ(B∖ A) = µ(B) − µ(A);
If (A_n) is a sequence in R, with A₁ ⊆ A₂ ⊆ A₃ ⊆⋯. Then

lim

n→∞

µ(A_n) = µ ⎛
⎝ ⋃_n A_n ⎞
⎠ .
If (A_n) is a sequence in R, with A₁ ⊇ A₂ ⊇ A₃ ⊇⋯. If µ(A_m)<∞ for some m, then

lim

n→∞

µ(A_n) = µ ⎛
⎝ ⋂_n A_n ⎞
⎠ . (67)

Proof. The two first properties are easy to see. For the third statement, define A=∪_n A_n, B₁=A₁ and B_n=A_n∖ A_n−1, n>1. Then A_n=⊔_k=1ⁿ B_n and A=⊔_k=1^∞B_n. Using the σ-additivity of measures µ(A)=∑_k=1^∞µ(B_k) and µ(A_n)=∑_k=1ⁿ µ(B_k). From the theorem in real analysis that any monotonic sequence of real numbers converges (recall that we admit +∞ as limits’ value) we have µ(A)=∑_k=1^∞µ(B_k)=lim_{n→ ∞} ∑_k=1ⁿ µ(B_k) = lim_{n→ ∞} µ(A_n). The last statement can be shown similarly. □

Exercise 13 Let a measure µ on ℕ be defined by µ(A)=0 for finite A and µ(A) = ∞ for infinite A. Check that µ is additive but not σ-additive. Therefore give an example that µ does not satisfies 3.

12.2 Extension of Measures

From now on we consider only finite measures, an extension to σ-finite measures will be done later.

Proposition 14 Any measure µ′ on a semiring S is uniquely extended to a measure µ on the generated ring R(S), see Ex. 5. If the initial measure was σ-additive, then the extension is σ-additive as well.

Proof. If an extension exists it shall satisfy µ(A)=∑_k=1ⁿ µ′(A_k), where A_k∈ S. We need to show for this definition two elements:

Consistency, i.e. independence of the value from a presentation of A∈ R(S) as A=⊔_k=1ⁿ A_k, where A_k∈ S. For two different presentation A=⊔_j=1ⁿ A_j and A=⊔_k=1^m B_k define C_jk=A_j∩ B_k, which will be pair-wise disjoint. By the additivity of µ′ we have µ′(A_j)=∑_kµ′(C_jk) and µ′(B_k)=∑_jµ′(C_jk). Then

∑

j

µ′(A_j)=

∑

j

∑

k

µ′(C_jk) =

∑

k

∑

j

µ′(C_jk)=

∑

k

µ′(B_k).
Additivity. For A=⊔_k=1ⁿ A_k, where A_k∈ R(S) we can present A_k=⊔_j=1^n(k) C_jk, C_jk∈ S. Thus A=⊔_k=1ⁿ ⊔_j=1^n(k) C_jk and:
µ(A)=
n

∑

k=1

n(k)

∑

j=1

µ′(C_jk)=
n

∑

k=1

µ(A_k).

Finally, show the σ-additivity. For a set A=⊔_k=1^∞A_k, where A and A_k∈ R(S), find presentations A=⊔_j=1ⁿ B_j, B_j∈ S and A_k=⊔_l=1^m(k) B_lk, B_lk∈ S. Define C_jlk=B_j ∩ B_lk∈ S, then B_j=⊔_k=1^∞⊔_l=1^m(k) C_jlk and A_k= ⊔_j=1ⁿ ⊔_l=1^m(k) C_jlk. Then, from σ-additivity of µ′:

µ(A)

∑

j=1

µ′(B_j)=

∑

j=1

∞

∑

k=1

m(k)

∑

l=1

µ′(C_jlk)=

∞

∑

k=1

∑

j=1

m(k)

∑

l=1

µ′(C_jlk) =

∞

∑

k=1

µ(A_k),

where we changed the summation order in series with non-negative terms. □

In a similar way we can extend a measure from a semiring to corresponding σ-ring, however it can be done even for a larger family. The procedure recall the famous story on Baron Munchausen saves himself from being drowned in a swamp by pulling on his own hair. Indeed, initially we knew measure for elements of semiring S or their finite disjoint unions from R(S). For an arbitrary set A we may assign a measure from an element of R(S) which “approximates” A. But how to measure such approximation? Well, to this end we use the measure on R(S) again (pulling on his own hair)!

Coming back to exact definitions, we introduce the following notion.

Definition 15 Let S be a semi-ring of subsets in X, and µ be a measure defined on S. An outer measure µ^* on X is a map µ^*:2^X→[0,∞] defined by:

µ^*(A)=inf

⎧
⎪
⎨
⎪
⎩

∑

µ(A_k), such that A⊆ ⋃_k A_k, A_k∈ S

⎫
⎪
⎬
⎪
⎭

Proposition 16 An outer measure has the following properties:

µ^*(∅)=0;
if A⊆ B then µ^*(A)≤µ^*(B), this is called monotonicity of the outer measure;
if (A_n) is any sequence in 2^X, then µ^*(∪_n A_n) ≤ ∑_n µ^*(A_n).

The final condition says that an outer measure is countably sub-additive. Note, that an outer measure may be not a measure in the sense of Defn. 6 due to a luck of additivity.

Example 17 The Lebesgue outer measure on ℝ is defined out of the measure from Example 2, that is, for A⊆ℝ, as

µ^*(A) = inf

⎧
⎪
⎨
⎪
⎩

∞

∑

j=1

(b_j−a_j) : A⊆ ⋃_j=1^∞[a_j,b_j)

⎫
⎪
⎬
⎪
⎭

We make this definition, as intuitively, the “length”, or measure, of the interval [a,b) is (b−a).

For example, for outer Lebesgue measure we have µ^*(A)=0 for any countable set, which follows, as clearly µ^*({x})=0 for any x∈ℝ.

Lemma 18 Let a<b. Then µ^*([a,b])=b−a.

Proof. For є>0, as [a,b] ⊆ [a,b+є), we have that µ^*([a,b])≤ (b−a)+є. As є>0, was arbitrary, µ^*([a,b]) ≤ b−a.

To show the opposite inequality we observe that [a,b)⊂[a,b] and µ^*[a,b) =b−a (because [a,b) is in the semi-ring) so µ^*[a,b]≥ b−a by 2. □

Our next aim is to construct measures from outer measures. We use the notation A▵ B=(A∪ B)∖ (A∩ B) for symmetric difference of sets.

Definition 19 Given an outer measure µ^* defined by a measure µ on a semiring S, we define A⊆ X to be Lebesgue measurable if for any ε >0 there is a finite union B of elements in S (in other words: B∈R(S) by Lem. 3), such that µ^*(A▵ B)<ε .

Figure 18: Approximating area by refined simple sets arrangements.

See Fig. 18 for an illustration of the concept of measurable sets. Obviously all elements of S and R(S) are measurable.

Exercise 20

Define a function of pairs of Lebesgue measurable sets A and B as the outer measure of the symmetric difference of A and B:
d(A,B)=µ^*(A▵ B). (68)

Show that d is a metric on the collection of equivalence classes with respect to the equivalence relation: A∼ B if d(A,B)=0. Hint: to show the triangle inequality use the inclusion:
A▵ B ⊆ (A▵ C) ⋃ (C▵ B)
Let a sequence (ε_n)→ 0 be monotonically decreasing. For a Lebesgue measurable A there exists a sequence (A_n)⊂ R(S) such that d(A,A_n)< ε_n for each n. Show that (A_n) is a Cauchy sequence for the distance d (68).

An alternative definition of a measurable set is due to Carathéodory.

Definition 21 Given an outer measure µ^*, we define E⊆ X to be Carathéodory measurable if

µ^*(A) = µ^*(A⋂ E) + µ^*(A∖ E),

for any A⊆ X.

As µ^* is sub-additive, this is equivalent to

µ^*(A) ≥ µ^*(A⋂ E) + µ^*(A∖ E) (A⊆ X),

as the other inequality is automatic.

Exercise^* 22

Show that for a Lebesgue measurable set A and any ε>0 there exist two elements B₁ and B₂ of the ring R(S) such that B₁⊂ A ⊂ B₂ and µ(B₂∖ B₁) < ε, cf. areas shadowed in darker and lighter colours on Fig. 18.
Hint: For a set B∈ R(S) such that µ^*(A▵ B)<ε/2 from Defn. 19 shall exists C∈ R(S) such that C ⊃ A▵ B and µ(C) < µ^*(A▵ B)+ε/2. Put B₁ = B ∖ C and B₂ = B ∪ C.
Let µ(X)<∞ show that A is Lebesgue measurable if and only if µ(X) = µ^*(A)+µ^*(X∖ A).
Show that measurability by Lebesgue and Carathéodory are equivalent.

Suppose now that the ring R(S) is an algebra (i.e., contains the maximal element X). Then, the outer measure of any set is finite, and the following theorem holds:

Theorem 23 (Lebesgue) Let µ^* be an outer measure on X defined by a semiring S, and let L be the collection of all Lebesgue measurable sets for µ^*. Then L is a σ-algebra, and if µ′ is the restriction of µ^* to L, then µ′ is a measure. Furthermore, µ′ is σ-additive on L if µ is σ-additive on S.

Proof.[Sketch of proof] Clearly, R(S)⊂ L. Now we show that µ^*(A)=µ(A) for a set A∈ R(S). If A⊂ ∪_k A_k for A_k ∈ S, then µ(A)≤ ∑_k µ(A_k), taking the infimum we get µ(A)≤µ^*(A). For the opposite inequality, any A∈ R(S) has a disjoint representation A=⊔_k A_k, A_k∈ S, thus µ^*(A)≤ ∑_k µ(A_k)=µ(A).

Now we will show that R(S) with the distance d (68) is an incomplete metric space, with the measure µ being uniformly continuous functions. Measurable sets make the completion of R(S) (cf. Ex. 2) with µ being continuation of µ^* to the completion by continuity, cf. Ex. 62.

Then, by the definition, Lebesgue measurable sets make the closure of R(S) with respect to this distance.

We can check that measurable sets form an algebra. To this end we need to make estimations, say, of µ^*((A₁∩ A₂)▵ (B₁∩ B₂)) in terms of µ^*(A_i▵ B_i). A demonstration for any finite number of sets is performed through mathematical inductions. The above two-sets case provide both: the base and the step of the induction.

Now, we show that L is σ-algebra. Let A_k∈ L and A=∪_k A_k. Then for any ε>0 there exists B_k∈ R(S), such that µ^*(A_k▵ B_k)<ε/2^k. Define B=∪_k B_k. Then

⎛
⎝

⋃_k A_k

⎞
⎠

▵

⎛
⎝

⋃_k B_k

⎞
⎠

⊂ ⋃_k

⎛
⎝

A_k ▵ B_k

⎞
⎠

implies µ^*(A▵ B)<ε.

We cannot stop at this point since B=∪_k B_k may be not in R(S). Thus, define B′₁=B₁ and B′_k=B_k∖ ∪_i=1^k−1 B_i, so B′_k are pair-wise disjoint. Then B=⊔_k B′_k and B′_k∈R(S). From the convergence of the series there is N such that ∑_k=N^∞µ(B′_k)<ε . Let B′=∪_k=1^N B′_k, which is in R(S). Then µ^*(B▵ B′)≤ ε and, thus, µ^*(A▵ B′)≤ 2ε.

To check that µ^* is measure on L we use the following

Lemma 24 | µ^*(A)−µ^*(B) |≤ µ^*(A▵ B), that is µ^* is uniformly continuous in the metric d(A,B) (68).

Proof.[Proof of the Lemma] Use inclusions A⊂ B∪(A▵ B) and B⊂ A∪(A▵ B). □

To show additivity take A_1,2∈L , A=A₁⊔ A₂, B_1,2∈R(S) and µ^*(A_i▵ B_i)<ε. Then µ^*(A▵(B₁∪ B₂))<2ε and | µ^*(A) − µ^*(B₁∪ B₂) |<2ε. Thus µ^*(B₁∪ B₂)=µ(B₁∪ B₂)=µ (B₁) +µ (B₂)−µ (B₁∩ B₂), but µ (B₁∩ B₂)=d(B₁∩ B₂,∅)=d(B₁∩ B₂,A₁∩ A₂)<2ε. Therefore

⎪
⎪

µ^*(B₁⋃ B₂)−µ (B₁) −µ (B₂)

⎪
⎪

<2ε.

Combining everything together we get (this is a sort of ε/3-argument):

⎪
⎪

µ^*(A)−µ^*(A₁)−µ^*(A₂)

⎪
⎪

⎪ ⎪	µ^(A)−µ^(B₁⋃ B₂) +µ^*(B₁⋃ B₂) −(µ (B₁) +µ (B₂))

+µ (B₁) +µ (B₂)−µ^*(A₁)−µ^*(A₂)

⎪
⎪

≤

⎪
⎪

µ^*(A)−µ^*(B₁⋃ B₂)

⎪
⎪

µ^*(B₁⋃ B₂)−(µ (B₁) +µ (B₂))

⎪
⎪

µ (B₁) +µ (B₂)−µ^*(A₁)−µ^*(A₂)

⎪
⎪

≤

6ε.

Thus µ^* is additive on L.

Check the countable additivity for A=⊔_k A_k. The inequality µ^*(A)≤ ∑_kµ^*(A_k) follows from countable sub-additivity. The opposite inequality is the limiting case of the finite inequality µ^*(A)≥ µ^*(⊔_k=1^N A_k)=∑_k=1^Nµ^*(A_k) following from monotonicity and additivity of µ^*. □

Corollary 25 Let E⊆ℝ be open or closed. Then E is Lebesgue measurable.

Proof. As σ-algebras are closed under taking complements, we need only show that open sets are Lebesgue measurable. For the latter we will use a common trick, using the density and the countability of the rationals.

Intervals (a,b) are Lebesgue measurable because they are countable unions of measurable half-open intervals from the semiring, e.g.:

(0,1) =

∞

⊔

k=1

⎡
⎢
⎢
⎣

k+1

⎞
⎟
⎟
⎠

Now let U⊆ℝ be open. For each x∈ U, there exists a_x<b_x with x∈(a_x,b_x)⊆ U. By making a_x slightly larger, and b_x slightly smaller, we can ensure that a_x,b_x∈ℚ. Thus U = ∪_x (a_x, b_x). Each interval is measurable, and there are at most a countable number of them (endpoints make a countable set) thus U is the countable (or finite) union of Lebesgue measurable sets, and hence U is Lebesgue measurable itself. □

We perform now an extension of finite measure to σ-finite one. Let µ be a σ-additive and σ-finite measure defined on a semiring in X=⊔_k X_k, such that the restriction of µ to every X_k is finite. Consider the Lebesgue extension µ_k of µ defined within X_k. A set A⊂ X is measurable if every intersection A∩ X_k is µ_k measurable. For a such measurable set A we define its measure by the identity:

µ(A)=

∑

µ_k(A⋂ X_k).

We call a measure µ defined on L complete if whenever E⊆ X is such that there exists F∈L with µ(F)=0 and E⊆ F, we have that E∈L. Measures constructed from outer measures by the above theorem are always complete. On the example sheet, we saw how to form a complete measure from a given measure. We call sets like E null sets: complete measures are useful, because it is helpful to be able to say that null sets are in our σ-algebra. Null sets can be quite complicated. For the Lebesgue measure, all countable subsets of ℝ are null, but then so is the Cantor set, which is uncountable.

Definition 26 If we have a property P(x) which is true except possibly x∈ A and µ(A)=0, we say P(x) is almost everywhere or a.e..

12.3 Complex-Valued Measures and Charges

We start from the following observation.

Exercise 27 Let µ₁ and µ₂ be measures on a same σ-algebra. Define µ₁+µ₂ and λµ₁, λ>0 by (µ₁+µ₂)(A)=µ₁(A)+µ₂(A) and (λµ₁)(A)=λ(µ₁(A)). Then µ₁+µ₂ and λµ₁ are measures on the same σ-algebra as well.

In view of this, it will be helpful to extend the notion of a measure to obtain a linear space.

Definition 28 Let X be a set, and R be a σ-ring. A real- (complex-) valued function ν on R is called a charge (or signed measure) if it is countably additive as follows: for any A_k∈R the identity A=⊔_k A_k implies the series ∑_k ν(A_k) is absolute convergent and has the sum ν(A).

In the following “charge” means “real charge”.

Example 29 Any linear combination of σ-additive measures on ℝ with real (complex) coefficients is real (complex) charge.

The opposite statement is also true:

Theorem 30 Any real (complex) charge ν has a representation ν=µ₁−µ₂ (ν=µ₁−µ₂+iµ₃−iµ₄), where µ_k are σ-additive measures.

To prove the theorem we need the following definition.

Definition 31 The variation of a charge on a set A is | ν |(A)=sup ∑_k| ν(A_k) | for all disjoint splitting A=⊔_k A_k.

Example 32 If ν=µ₁−µ₂, then | ν |(A)≤ µ₁(A)+µ₂(A). The inequality becomes an identity for disjunctive measures on A (that is there is a partition A=A₁⊔ A₂ such that µ₂(A₁)=µ₁(A₂)=0).

The relation of variation to charge is as follows:

Theorem 33 For any charge ν the function | ν | is a σ-additive measure.

Finally to prove the Thm. 30 we use the following

Proposition 34 For any charge ν the function | ν |−ν is a σ-additive measure as well.

From the Thm. 30 we can deduce

Corollary 35 The collection of all charges on a σ-algebra R is a linear space which is complete with respect to the distance:

d(ν₁,ν₂)=

sup

A∈R

⎪
⎪

ν₁(A)−ν₂(A)

⎪
⎪

The following result is also important:

Theorem 36 (Hahn Decomposition) Let ν be a charge. There exist A,B∈L, called a Hahn decomposition of (X,ν), with A∩ B=∅, A∪ B= X and such that for any E∈L,

ν (A⋂ E) ≥ 0, ν(B⋂ E)≤ 0.

This need not be unique.

Proof.[Sketch of proof] We only sketch this. We say that A∈L is positive if

ν(E⋂ A)≥0 (E∈L),

and similiarly define what it means for a measurable set to be negative. Suppose that ν never takes the value −∞ (the other case follows by considering the charge −ν).

Let β = infν(B₀) where we take the infimum over all negative sets B₀. If β=−∞ then for each n, we can find a negative B_n with ν(B_n)≤ −n. But then B=∪_n B_n would be negative with ν(B)≤ −n for any n, so that ν(B)=−∞ a contradiction.

So β>−∞ and so for each n we can find a negative B_n ν(B_n) < β+1/n. Then we can show that B = ∪_n B_n is negative, and argue that ν(B) ≤ β. As B is negative, actually ν(B) = β.

There then follows a very tedious argument, by contradiction, to show that A=X∖ B is a positive set. Then (A,B) is the required decomposition. □

12.4 Constructing Measures, Products

Consider the semiring S of intervals [a,b). There is a simple description of all measures on it. For a measure µ define

F_µ(t)=

⎧
⎪
⎨
⎪
⎩

µ([0,t))	if t>0,
0	if t=0,
−µ([t,0))	if t<0,

(69)

F_µ is monotonic and any monotonic function F defines a measure µ on S by the by µ([a,b))=F(b)−F(a). The correspondence is one-to-one with the additional assumption F(0)=0.

Theorem 37 The above measure µ is σ-additive on S if and only if F is continuous from the left: F(t−0)=F(t) for all t∈ℝ.

Proof. Necessity: F(t)−F(t−0)=lim_ε→
0µ([t−ε,t))=µ(lim_ε→
0[t−ε,t))=µ(∅)=0, by the continuity of a σ-additive measure, see 4.

For sufficiency assume [a,b)=⊔_k [a_k,b_k). The inequality µ([a,b))≥ ∑_k µ([a_k,b_k)) follows from additivity and monotonicity. For the opposite inequality take δ_k s.t. F(b)−F(b−δ)<ε and F(a_k)−F(a_k−δ_k)<ε/2^k (use left continuity of F). Then the interval [a,b−δ] is covered by (a_k−δ_k,b_k), due to compactness of [a,b−δ] there is a finite subcovering. Thus µ([a,b−δ ))≤∑_j=1^N µ([a_{k_j}−δ_{k_j},b_{k_j})) and µ([a,b))≤∑_j=1^N µ([a_{k_j},b_{k_j}))+2ε . □

Exercise 38

Give an example of function discontinued from the left at 1 and show that the resulting measure is additive but not σ-additive.
Check that, if a function F is continuous at point a then µ({a})=0.

Example 39

Take F(t)=t, then the corresponding measure is the Lebesgue measure on ℝ.
Take F(t) be the integer part of t, then µ counts the number of integer within the set.
Define the Cantor function as follows α(x)=1/2 on (1/3,2/3); α(x)=1/4 on (1/9,2/9); α(x)=3/4 on (7/9,8/9), and so for. This function is monotonic and can be continued to [0,1] by continuity, it is know as Cantor ladder. The resulting measure has the following properties:
- The measure of the entire interval is 1.
- Measure of every point is zero.
- The measure of the Cantor set is 1, while its Lebesgue measure is 0.

Another possibility to build measures is their product. In particular, it allows to expand various measures defined through (69) on the real line to ℝⁿ.

Definition 40 Let X and Y be spaces, and let S and T be semirings on X and Y respectively. Then S× T is the semiring consisting of { A× B : A∈ S, B∈ T } (“generalised rectangles”). Let µ and ν be measures on S and T respectively. Define the product measure µ×ν on S× T by the rule (µ× ν)(A× B)=µ(A) ν(B).

Example 41 The measure from Example 3 on the semiring of half-open rectangles is the product of two copies of pre-Lebesgue measures from Example 2 on the semiring of half-open intervals.


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Last modified: November 6, 2024.