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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=1An)=n=1μ(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 ABA \subseteq B, then μ(A)μ(B)\mu(A) \leq \mu(B).

Proof. B=A(BA)B = A \cup (B \setminus A) is a disjoint union, so μ(B)=μ(A)+μ(BA)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=1An)n=1μ(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=Ank=1n1AkB_n = A_n \setminus \bigcup_{k=1}^{n-1} A_k for n2n \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 A1A2A_1 \subseteq A_2 \subseteq \cdots, then

μ(n=1An)=limnμ(An)\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \lim_{n \to \infty} \mu(A_n)

Proof. Write An=A1(A2A1)(A3A2)\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=1μ(An+1An)=limnμ(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 A1A2A_1 \supseteq A_2 \supseteq \cdots and μ(A1)<\mu(A_1) < \infty, then

μ(n=1An)=limnμ(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 x0Xx_0 \in X, define δx0(A)=1\delta_{x_0}(A) = 1 if x0Ax_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])=bam([a, b]) = b - a for all aba \leq b. On Rn\mathbb{R}^n, the Lebesgue measure satisfies m([a1,b1]××[an,bn])=i=1n(biai)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)=limδ0inf{i(diamUi)α:AUi,diamUi<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 NXN \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={AN:AF,NBF,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,FG)(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 AFA \in \mathcal{F}, BGB \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×Yfd(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=PNX = P \cup N where PP is positive (every measurable subset has ν0\nu \geq 0) and NN is negative (every subset has ν0\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 xyx \sim y iff xyQx - 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 3n3^{-n} remain. The measure of the removed set is:

n=02n3n+1=13n=0(23)n=13112/3=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)=11=0m(C) = 1 - 1 = 0. The Cantor set is an uncountable null set.

\blacksquare