Measure Theory — Part 1
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
A Measurable Set is an element of the σ-algebra.
A Measurable Space is the ordered pair (X, Σ).
A Measure μ on a measurable space (X, Σ) is a function μ: Σ→ℝ∪{∞}, which satisfies the following properties:
MS1. ∀A ∈ Σ, μ(A) ≥ 0
MS2. μ(∅) = 0
MS3. 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.
A Measure Space is the triple (X, Σ, μ).
The measure μ has the following properties:
MS4. 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 ]
MS5. 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 ]
MS6.
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 measurable function f : X → ℝ is said to be a Simple Function if it takes only finitely many values:
MF2. ∀x ∈ X, f(x) = α[1] IA[1](x) + … + α[n] IA[n](x)
with A[i] = f-1({α[i]}) ∈ Σx and A[1]∪…∪A[n] = X and α[1] < α[2] < .. α[n]
Here, 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 range of the function.
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 justified.
For measurable function f that is not non-negative, it can be written as:
f(x) = f+(x) — f-(x)
f+(x) = max(f(x), 0)
f-(x) = min(f(x), 0)
Note that both f+ and f− are non-negative measurable functions. Since max(x,y), min(x,y) and constant functions are all continuous, thus f+ and f- are continuous and measurable too. Also note that:
|f(x)| = f+(x) + f-(x) [ Here, |f| is the absolute value of f ]
Lebesgue Integral of the general measurable function f : X → ℝ is defined as follows if min(∫_X f+(x)dμ(x) , ∫_X f-(x)dμ(x)) < ∞:
LB4: ∫_X f(x)dμ(x) = ∫_X f+(x)dμ(x) — ∫_X f-(x)dμ(x)
This condition is required since ∞ — ∞ is not defined if both ∫_X f+(x)dμ(x) = ∞ and ∫_X f-(x)dμ(x) = ∞.
A measurable function f: X → ℝ is said to be Lebesgue Integrable if
∫_X |f(x)| dμ(x) < ∞
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
Below are the properties and theorems of the Lebesgue Integral:
Monotone Convergence Theorem:
Let (fn(x)) be a sequence of non-negative measurable function defined on measurable set A. If the following conditions 1) and 2) are satisfied:
1) 0 ≤ fn(x) ≤ fn+1(x) ∀n∈ℕ ∀x∈A and
2) lim_n→∞ fn(x) = f(x) ∀x∈A
Then lim_n→∞ ∫_A fn(x) dμ(x) = ∫_A f(x) dμ(x)
Lebesgue Bounded Convergence Theorem:
Let (fn(x)) be a sequence of measurable function defined on measurable set A and μ(A) < ∞. If the following conditions 1) and 2) are satisfied:
1) There exists M ∈ ℝ s.t. |fn(x)| ≤ M ∀n∈ℕ ∀x∈A and
2) lim_n→∞ fn(x) = f(x) ∀x∈A
Then lim_n→∞ ∫_A fn(x) dμ(x) = ∫_A f(x) dμ(x)
LB5: ∫_A⋃B f(x)dμ(x) = ∫_A f(x)dμ(x) + ∫_B f(x)dμ(x) where A⋂B = ø
LB6: ∫_A c·(f+g)(x)dμ(x) = c ∫_A f(x)dμ(x) + c ∫_A g(x)dμ(x) where c is a constant and c ∈ℝ
LB7: | ∫_A f(x)dμ(x) | ≤ ∫_A |f(x)| dμ(x)
LB8: ∫_A f(x)dμ(x) = 0 iff f(x) = 0 a.e.
LB9: If f ≤ g, then ∫_A f(x)dμ(x) ≤ ∫_A g(x)dμ(x)
To be continued…
