A measure on a measurable space (X,F) is a function μ:F→[0,∞] satisfying:
- μ(∅)=0.
- Countable additivity: if {An}n=1∞ are pairwise disjoint sets in F, then
μ(⋃n=1∞An)=∑n=1∞μ(An)
The triple (X,F,μ) is called a measure space.
Proposition 2.1 (Monotonicity). If A⊆B, then μ(A)≤μ(B).
Proof. B=A∪(B∖A) is a disjoint union, so μ(B)=μ(A)+μ(B∖A)≥μ(A). ■
Proposition 2.2 (Countable Subadditivity). For any sequence {An}⊆F:
μ(⋃n=1∞An)≤∑n=1∞μ(An)
Proof. Define B1=A1 and Bn=An∖⋃k=1n−1Ak for n≥2. Then {Bn} are pairwise disjoint with ⋃Bn=⋃An. By countable additivity and monotonicity, μ(⋃An)=∑μ(Bn)≤∑μ(An). ■
Proposition 2.3 (Continuity from Below). If A1⊆A2⊆⋯, then
μ(⋃n=1∞An)=limn→∞μ(An)
Proof. Write ⋃An=A1∪(A2∖A1)∪(A3∖A2)∪⋯, a disjoint union. Then μ(⋃An)=μ(A1)+∑n=1∞μ(An+1∖An)=limn→∞μ(An). ■
Proposition 2.4 (Continuity from Above). If A1⊇A2⊇⋯ and μ(A1)<∞, then
μ(⋂n=1∞An)=limn→∞μ(An)
Example (Counting Measure). On any set X with F=P(X), define μ(A)=∣A∣ (the cardinality, ∞ for infinite sets). This is a measure.
Example (Dirac Measure). For x0∈X, define δx0(A)=1 if x0∈A, and 0 otherwise. This is a measure.
Example (Lebesgue Measure). The Lebesgue measure m on R is the completion of a measure on B(R) satisfying m([a,b])=b−a for all a≤b. On Rn, the Lebesgue measure satisfies m([a1,b1]×⋯×[an,bn])=∏i=1n(bi−ai).
Example (Hausdorff Measure). For α>0, the α-dimensional Hausdorff measure Hα on Rn generalises Lebesgue measure to non-integer dimensions: Hα(A)=limδ→0inf{∑i(diamUi)α:A⊆⋃Ui,diamUi<δ}. For α=n, Hn coincides with the Lebesgue measure up to a constant factor.
Definition. A set N⊆X is null (or μ-negligible) if μ(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) 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 on a σ-algebra F has a unique completion μ on the σ-algebra F={A∪N:A∈F,N⊆B∈F,μ(B)=0}.
Theorem 2.7. Given σ-finite measure spaces (X,F,μ) and (Y,G,ν), there exists a unique product measure μ×ν on (X×Y,F⊗G) satisfying (μ×ν)(A×B)=μ(A)ν(B) for all A∈F, B∈G.
Theorem 2.8 (Fubini-Tonelli). If 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)
Definition. A signed measure ν on (X,F) is a countably additive function ν:F→(−∞,∞] that can take at most one of the values ±∞.
Theorem 2.9 (Hahn Decomposition). For any signed measure ν, there exists a Hahn decomposition X=P∪N where P is positive (every measurable subset has ν≥0) and N is negative (every subset has ν≤0). This decomposition is unique up to null sets.
Theorem 2.10 (Jordan Decomposition). Every signed measure ν can be uniquely expressed as ν=ν+−ν− where ν+ and ν− 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] by x∼y iff x−y∈Q. Choose one representative from each equivalence class to form the Vitali set V. For rationals q∈[−1,1]∩Q, define Vq=V+q (mod 1). These are pairwise disjoint and ⋃qVq=[0,1].
By translation invariance of Lebesgue measure, m(Vq)=m(V) for all q. If V were measurable, then 1=m([0,1])=∑qm(Vq)=∑qm(V). The right side is 0 if m(V)=0 or ∞ if m(V)>0, both contradictions. Hence V is not measurable.
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Problem. Compute the Lebesgue measure of the Cantor set C.
Solution
The Cantor set is constructed by removing the middle third (1/3,2/3) from [0,1], then removing the middle third of each remaining interval, ad infinitum. After n stages, 2n intervals each of length 3−n remain. The measure of the removed set is:
∑n=0∞3n+12n=31∑n=0∞(32)n=31⋅1−2/31=1
Therefore m(C)=1−1=0. The Cantor set is an uncountable null set.
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