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Measures

A measure on a measurable space (X,F)(X, \mathcal{F}) is a function μ:F→[0,∞]\mu : \mathcal{F} \to [0, \infty] satisfying:

  1. μ(∅)=0\mu(\varnothing) = 0.
  2. Countable additivity: if {An}n=1∞\{A_n\}_{n=1}^{\infty} are pairwise disjoint sets in F\mathcal{F}, then

μ(⋃n=1∞An)=∑n=1∞μ(An)\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \sum_{n=1}^{\infty} \mu(A_n)

The triple (X,F,μ)(X, \mathcal{F}, \mu) is called a measure space.

Proposition 2.1 (Monotonicity). If A⊆BA \subseteq B, then μ(A)≤μ(B)\mu(A) \leq \mu(B).

Proof. B=A∪(B∖A)B = A \cup (B \setminus A) is a disjoint union, so μ(B)=μ(A)+μ(B∖A)≥μ(A)\mu(B) = \mu(A) + \mu(B \setminus A) \geq \mu(A). ■\blacksquare

Proposition 2.2 (Countable Subadditivity). For any sequence {An}⊆F\{A_n\} \subseteq \mathcal{F}:

μ(⋃n=1∞An)≤∑n=1∞μ(An)\mu\left(\bigcup_{n=1}^{\infty} A_n\right) \leq \sum_{n=1}^{\infty} \mu(A_n)

Proof. Define B1=A1B_1 = A_1 and Bn=An∖⋃k=1n−1AkB_n = A_n \setminus \bigcup_{k=1}^{n-1} A_k for n≥2n \geq 2. Then {Bn}\{B_n\} are pairwise disjoint with ⋃Bn=⋃An\bigcup B_n = \bigcup A_n. By countable additivity and monotonicity, μ(⋃An)=∑μ(Bn)≤∑μ(An)\mu(\bigcup A_n) = \sum \mu(B_n) \leq \sum \mu(A_n). ■\blacksquare

Proposition 2.3 (Continuity from Below). If A1⊆A2⊆⋯A_1 \subseteq A_2 \subseteq \cdots, then

μ(⋃n=1∞An)=lim⁡n→∞μ(An)\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \lim_{n \to \infty} \mu(A_n)

Proof. Write ⋃An=A1∪(A2∖A1)∪(A3∖A2)∪⋯\bigcup A_n = A_1 \cup (A_2 \setminus A_1) \cup (A_3 \setminus A_2) \cup \cdots, a disjoint union. Then μ(⋃An)=μ(A1)+∑n=1∞μ(An+1∖An)=lim⁡n→∞μ(An)\mu(\bigcup A_n) = \mu(A_1) + \sum_{n=1}^{\infty} \mu(A_{n+1} \setminus A_n) = \lim_{n \to \infty} \mu(A_n). ■\blacksquare

Proposition 2.4 (Continuity from Above). If A1⊇A2⊇⋯A_1 \supseteq A_2 \supseteq \cdots and μ(A1)<∞\mu(A_1) < \infty, then

μ(⋂n=1∞An)=lim⁡n→∞μ(An)\mu\left(\bigcap_{n=1}^{\infty} A_n\right) = \lim_{n \to \infty} \mu(A_n)

Example (Counting Measure). On any set XX with F=P(X)\mathcal{F} = \mathcal{P}(X), define μ(A)=∣A∣\mu(A) = |A| (the cardinality, ∞\infty for infinite sets). This is a measure.

Example (Dirac Measure). For x0∈Xx_0 \in X, define δx0(A)=1\delta_{x_0}(A) = 1 if x0∈Ax_0 \in A, and 00 otherwise. This is a measure.

Example (Lebesgue Measure). The Lebesgue measure mm on R\mathbb{R} is the completion of a measure on B(R)\mathcal{B}(\mathbb{R}) satisfying m([a,b])=b−am([a, b]) = b - a for all a≤ba \leq b. On Rn\mathbb{R}^n, the Lebesgue measure satisfies m([a1,b1]×⋯×[an,bn])=∏i=1n(bi−ai)m([a_1, b_1] \times \cdots \times [a_n, b_n]) = \prod_{i=1}^n (b_i - a_i).

Example (Hausdorff Measure). For α>0\alpha > 0, the α\alpha-dimensional Hausdorff measure Hα\mathcal{H}^\alpha on Rn\mathbb{R}^n generalises Lebesgue measure to non-integer dimensions: Hα(A)=lim⁡δ→0inf⁡{∑i(diam⁡Ui)α:A⊆⋃Ui,diam⁡Ui<δ}\mathcal{H}^\alpha(A) = \lim_{\delta \to 0} \inf\{\sum_i (\operatorname{diam} U_i)^\alpha : A \subseteq \bigcup U_i, \operatorname{diam} U_i < \delta\}. For α=n\alpha = n, Hn\mathcal{H}^n coincides with the Lebesgue measure up to a constant factor.

Definition. A set N⊆XN \subseteq X is null (or μ\mu-negligible) if μ(N)=0\mu(N) = 0. A property holds almost everywhere (a.e.) if it holds on the complement of a null set.

Proposition 2.5. A countable union of null sets is null.

Definition. A measure space (X,F,μ)(X, \mathcal{F}, \mu) is complete if every subset of a null set is measurable (and hence null). The Lebesgue measure is the completion of the Borel measure.

Theorem 2.6 (Completion). Every measure μ\mu on a σ\sigma-algebra F\mathcal{F} has a unique completion μ‾\overline{\mu} on the σ\sigma-algebra F‾={A∪N:A∈F,N⊆B∈F,μ(B)=0}\overline{\mathcal{F}} = \{A \cup N : A \in \mathcal{F}, N \subseteq B \in \mathcal{F}, \mu(B) = 0\}.

Theorem 2.7. Given σ\sigma-finite measure spaces (X,F,μ)(X, \mathcal{F}, \mu) and (Y,G,ν)(Y, \mathcal{G}, \nu), there exists a unique product measure μ×ν\mu \times \nu on (X×Y,F⊗G)(X \times Y, \mathcal{F} \otimes \mathcal{G}) satisfying (μ×ν)(A×B)=μ(A)ν(B)(\mu \times \nu)(A \times B) = \mu(A)\nu(B) for all A∈FA \in \mathcal{F}, B∈GB \in \mathcal{G}.

Theorem 2.8 (Fubini-Tonelli). If f(x,y)f(x, y) is nonnegative and measurable (Tonelli) or integrable (Fubini), then:

∫X×Yf d(μ×ν)=∫X(∫Yf(x,y) dν(y))dμ(x)=∫Y(∫Xf(x,y) dμ(x))dν(y)\int_{X \times Y} f\, d(\mu \times \nu) = \int_X \left(\int_Y f(x, y)\, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f(x, y)\, d\mu(x)\right) d\nu(y)

Definition. A signed measure ν\nu on (X,F)(X, \mathcal{F}) is a countably additive function ν:F→(−∞,∞]\nu : \mathcal{F} \to (-\infty, \infty] that can take at most one of the values ±∞\pm\infty.

Theorem 2.9 (Hahn Decomposition). For any signed measure ν\nu, there exists a Hahn decomposition X=P∪NX = P \cup N where PP is positive (every measurable subset has ν≥0\nu \geq 0) and NN is negative (every subset has ν≤0\nu \leq 0). This decomposition is unique up to null sets.

Theorem 2.10 (Jordan Decomposition). Every signed measure ν\nu can be uniquely expressed as ν=ν+−ν−\nu = \nu^+ - \nu^- where ν+\nu^+ and ν−\nu^- are positive measures called the positive and negative variations.

Problem. Show that the Vitali set is not Lebesgue measurable.

Solution

Define an equivalence relation on [0,1][0, 1] by x∼yx \sim y iff x−y∈Qx - y \in \mathbb{Q}. Choose one representative from each equivalence class to form the Vitali set VV. For rationals q∈[−1,1]∩Qq \in [-1, 1] \cap \mathbb{Q}, define Vq=V+qV_q = V + q (mod 1). These are pairwise disjoint and ⋃qVq=[0,1]\bigcup_q V_q = [0, 1].

By translation invariance of Lebesgue measure, m(Vq)=m(V)m(V_q) = m(V) for all qq. If VV were measurable, then 1=m([0,1])=∑qm(Vq)=∑qm(V)1 = m([0, 1]) = \sum_q m(V_q) = \sum_q m(V). The right side is 00 if m(V)=0m(V) = 0 or ∞\infty if m(V)>0m(V) > 0, both contradictions. Hence VV is not measurable.

■\blacksquare

2.8 Worked Example: Measure of the Cantor Set

Section titled “2.8 Worked Example: Measure of the Cantor Set”

Problem. Compute the Lebesgue measure of the Cantor set CC.

Solution

The Cantor set is constructed by removing the middle third (1/3,2/3)(1/3, 2/3) from [0,1][0, 1], then removing the middle third of each remaining interval, ad infinitum. After nn stages, 2n2^n intervals each of length 3−n3^{-n} remain. The measure of the removed set is:

∑n=0∞2n3n+1=13∑n=0∞(23)n=13⋅11−2/3=1\sum_{n=0}^\infty \frac{2^n}{3^{n+1}} = \frac{1}{3}\sum_{n=0}^\infty \left(\frac{2}{3}\right)^n = \frac{1}{3} \cdot \frac{1}{1-2/3} = 1

Therefore m(C)=1−1=0m(C) = 1 - 1 = 0. The Cantor set is an uncountable null set.

■\blacksquare