Measure Theory — Part 1

Photo by Sven Mieke on Unsplash

How to measure a set ?

Everybody knows how to measure the size of a rectangle. You measure the height and width of a rectangle and the area equals the product of height and width. A rectangle is a set of points in 2D space, so can we measure the size of a set of anything?

Mathematician has invented a tool to generalise the intuitive notions of length, area, and volume, it is called Measure Theory. OK! Let’s start to learn how to measure anything!

Measure Theory 101

So, what is the meaning of the Measure ? Actually, it is the basic concept in the Measure Theory. In layman terms, it is a function that measures the the size of a measurable set and always returns a finite non-negative real number.

Let X be a set.
A σ-algebra Σ on a set X is a nonempty collection of subsets of X which satisfies the following properties:
SA1. X ∈ Σ
SA2. ∀A ∈ Σ, X\A ∈ Σ
SA3. for all countable collection A[i] ∈ Σ, A[1]∪…∪A[∞] ∈ Σ

Note that the power set of X, 2^X, is a σ-algebra defined on X.

The element of the σ-algebra is called Measurable Set. So, the complement of a Measurable Set is again a Measurable Set. The countable union of Measurable Sets is a Measurable Set. The countable intersection of Measurable Set is also a Measurable Set.

MST1. The power set on X, 2^X, is the largest σ-algebra.
Proof
SA1.
X ∈ 2^X
SA2.
∀c ∈ 2^X, X\c ∈ 2^X
SA3.
∀ c1,c2 ∈ 2^X, c1⋃c2 ∈ 2^X
Thus, 2^X is σ-algebra. Since σ-algebra is a collection of subsets of X and 2^X is the largest one, 2^X, is the largest σ-algebra.

MST2. Countable intersection of measurable set is measurable
Proof
∀A[i] ∈ Σ
⇒ A[i]c ∈ Σ
⇒ A[1]c ∪ … ∪ A[∞]c ∈ Σ
⇒ (A[1]∩ … ∩ A[∞])c ∈ Σ
⇒ A[1]∩ … ∩ A[∞] ∈ Σ

A Measurable Space is the ordered pair (X, Σ).

Consider a function f: X → Y with measurable space (X, Σ) on X. Then, the set of pre-image of all subsets in Y which is a measurable set in Σ forms a σ-algebra.

MST3. The set {c | f-1(c) ∈ Σ, c ⊆ Y, f: X → Y } is a σ-algebra.
Proof
Let SA = {c | f-1(c) ∈ Σ, c ⊆ Y, f: X → Y }
SA1.
f-1(Y) = X ∈ Σ ⇒ Y ∈ SA
SA2.
∀c ∈ SA,
f-1(Y\c) = f-1(Y)\f-1(c) = X\f-1(c) ∈ Σ [ Since f-1(c) ∈ Σ ]
⇒ Y\c ∈ SA
SA3.
∀ c1,c2 ∈ SA
f-1(c1⋃c2) = f-1(c1) ⋃ f-1(c2) ∈ Σ [ Since f-1(c1), f-1(c2) ∈ Σ ]
⇒ c1⋃c2 ∈ SA
Thus, SA is σ-algebra

Measure and Measure Space

A Measure μ on a measurable space (X, Σ) is a function μ: Σ→ℝ∪{∞}, which satisfies the following properties:
MSU1. ∀A ∈ Σ, μ(A) ≥ 0
MSU2. μ(∅) = 0
MSU3. for all countable collection A[i] ∈ Σ, if A[i]∩A[j]=∅ where i ≠ j
μ(A[1]∪…∪A[∞]) = μ(A[1])+…+μ(A[∞])

Note that ℝ∪{∞} is the Extended Real Numbers. It is constructed from the real number system ℝ by adding the element ∞ which is greater than every real number. The product of 0 * ∞ is defined to be 0. If a > 0, a * ∞ = ∞ and a* -∞ = -∞

A Measure Space is the triple (X, Σ, μ).

The measure μ has the following properties:
MSU4. μ(A) ≤ μ(B) if A ⊆ B and A, B ∈ Σ
Proof
A ⊆ B ⇒ B=(B\A)∪A
⇒ μ(B) = μ((B\A)∪A) = μ(B\A) + μ(A) [ Since B\A ∩ A = ∅ ]
⇒ μ(B) ≥ μ(A) [ By MS1 ]

MSU5. for all countable collection A[i] ∈ Σ
μ(∪A[i]) ≤ Σ μ(A[i])
Proof
Let A’[i] = A[i]\∪A[1]..∪A[i-1], A’[1]=A[1]
⇒ ∪A’[i] = ∪A[i] and A’[i]∩A’[j]=∅ where i ≠ j
⇒ μ(∪A[i]) = μ(∪A’[i])
= Σμ(A’[i]) [ Since A’[i]∩A’[j]=∅ ]
≤ Σμ(A[i]) [ Since A’[i] ≤ A[i] and by MS4 ]

Measurable Function and Simple Function

Let (X, Σx) and (Y, Σy) be the measurable spaces.
MF1. A function f : X → Y is said to be Measurable Function if
∀B ∈ Σy, f−1(B) ∈ Σx

A function f : X → ℝ is said to be a Simple Function if it takes only finitely many values f(X) = { α[1], … , α[n] }
MF2. ∀x ∈ X, f(x) = α[1] IA[1](x) + … + α[n] IA[n](x) with
1) ∀i A[i] = f-1({α[i]}) ∈ Σx and
2) A[1] ∪ … ∪ A[n] = X and
3) α[1] < ..< α[n]

Here, the symbol I denotes the Indicator Function:
Given a subset A ⊆ X, IA: X → {0, 1}
IA(x) = 1 if x ∈ A
IA(x) = 0 if x ∉ A

Lebesgue Integral and Riemann Integral

The Lebesgue Integral is a more generalized version of the Riemann Integral. For the Riemann integral, the domain is partitioned into intervals. By constructing the bars with heights that meet the graph, the integral can be approximated by the sum of the area of these bars.

For the Lebesgue integral, the range is partitioned into intervals by choosing a finite number of target values. By constructing the bars with heights that equal to these values but below the function, the integral can be approximated by the sum of the area of these bars. Note that there may be several bars with heights that equal to these values but with different domain of the function.

Riemann Integral (Upper one), Lebesgue Integral (Lower one) From wikipedia

The Lebesgue Integral is more general than the Riemann Integral since the domain of the function can be more general space, not only Euclidean space. The Lebesgue Integral is also more powerful and let’s consider the function q: [0,1] → {0, 1} below which has no Riemann Integral:

q(x) = 1 if x is a rational number
q(x) = 0 otherwise

upper Riemann sum = inf(Σ sup(q(xi*)) ∆x) = 1*(1–0) = 1
lower Riemann sum = sup(Σ inf(q(xi*)) ∆x) = 0*(1–0) = 0
⇒ upper Riemann sum ≠ lower Riemann sum ⇒ Not Riemann Integrable

Note that since rational number is dense in R, there is always a rational number between any real numbers, so sup(q(xi*)) = 1. Irrational number is also dense in R, so inf(q(xi*)) = 0.

In the Riemann Integral, “dx” is an infinitesimal change in x:
∫ f(x)dx where x∈[0, 1]

In the Lebesgue Integral, μ is a “measure” of a “measurable set” X and f(x) is a “measurable function”:
∫_X f(x)dμ(x)

Let (𝑋, Σ, 𝜇) be a measure space and 𝜇: Σ → ℝ∪{∞} be a measure.

Lebesgue Integral of the simple function f : X → ℝ+ with the canonical representation f = Σ α[i] IA[i] is defined as follows:

LB1: ∫_X f(x)dμ(x) = Σ α[i] μ(A[i])

If B is a measurable subset of X, the Lebesgue Integral becomes:
LB2: ∫_B f(x)dμ(x) = ∫_X IB f(x) dμ(x) = Σ α[i] μ(A[i] ∩ B)

Lebesgue Integral of the non-negative measurable function f : X → ℝ+ is defined as the supremum of set of Lebesgue Integral of simple functions:

LB3: ∫_X f(x)dμ(x) = sup { ∫_X s(x)dμ(x) : 0 ≤ s ≤ f, s is simple }

Note that the measure μ is defined on extended real numbers. Thus, ∫_Xs(x)dμ(x) = Σ α[i] μ(A[i]) ≤ ∞ and the set {∫_X s(x)dμ(x)} has the upper bound ∞. Since s(x) = 0 is also a simple function, so the set {∫_X s(x)dμ(x)} is non-empty. By the least upper bound axiom, the supremum of the set exists. Therefore, the definition of the Lebesgue Integral of the non-negative function is valid.

Let’s go back to the function q(x) and consider its Lebesgue Integral.
L(q) = ∫_[0,1]q(x)dμ(x)
= 0*μ([0,1]\ℚ) + 1*μ([0,1]⋂ℚ)
= 1*0 [ measure of countable set is 0 w.r.t. Lebesgue Measure ]
= 0

Conclusion

I have introduced the concept of σ-algebra Σ on a set X and the measure defined on the measurable set, the members of the σ-algebra Σ. The measurable function is a function for which the pre-image of any measurable set in the target space is a measurable set in the source space. A simple function is a measurable function that takes only finitely many values. Finally, the Lebesgue Integral for the simple function and non-negative function are introduced.

Your feedback is highly appreciated and will help me to continue my articles.

Please give this post a clap if you like this post. Thanks!!