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Spring 2011 - Problem 6

Lebesgue-Radon-Nikodym derivative, measure theory

Suppose μ\mu and ν\nu are finite positive (regular) Borel measures on Rn\R^n. Prove the existence and uniqueness of the Lebesgue decomposition: There are a unique pair of positive Borel measures μa\mu_a and μs\mu_s so that

μ=μa+μs,μa≪ν,andμs⊥ν.\mu = \mu_a + \mu_s, \quad \mu_a \ll \nu, \quad\text{and}\quad \mu_s \perp \nu.

That is, μa\mu_a is absolutely continuous to ν\nu, while μs\mu_s is mutually singular to ν\nu.
Solution.

Let B\mathcal{B} denote the Borel σ\sigma-algebra and set

M=sup⁡ {μ(B)∣B∈B and ν(B)=0}.M = \sup\,\set{\mu\p{B} \mid B \in \mathcal{B} \text{ and } \nu\p{B} = 0}.

Observe that for any B∈BB \in \mathcal{B}, ν(B)≤μ(X)<∞\nu\p{B} \leq \mu\p{X} < \infty, so taking supremums, we see M≤μ(X)<∞M \leq \mu\p{X} < \infty. Hence, for each n≥1n \geq 1, there exists Bn∈BB_n \in \mathcal{B} with ν(Bn)=0\nu\p{B_n} = 0 and μ(Bn)≥M−1n\mu\p{B_n} \geq M - \frac{1}{n}. If we set A=⋃n=1∞BnA = \bigcup_{n=1}^\infty B_n, then

ν(A)≤∑n=1∞ν(Bn)=0  ⟹  μ(A)≤M,\nu\p{A} \leq \sum_{n=1}^\infty \nu\p{B_n} = 0 \implies \mu\p{A} \leq M,

by definition of MM. Moreover,

μ(A)=μ(⋃n=1∞Bn)=lim⁡N→∞μ(⋃n=1NBn)≥lim⁡N→∞(M−1N)=M,\mu\p{A} = \mu\p{\bigcup_{n=1}^\infty B_n} = \lim_{N\to\infty} \mu\p{\bigcup_{n=1}^N B_n} \geq \lim_{N\to\infty} \p{M - \frac{1}{N}} = M,

so μ(A)=M\mu\p{A} = M, i.e., the supremum is attained. For B∈BB \in \mathcal{B}, define

μa(B)=μ(B∖A)andμs(B)=μ(B∩A).\mu_a\p{B} = \mu\p{B \setminus A} \quad\text{and}\quad \mu_s\p{B} = \mu\p{B \cap A}.

Since μ\mu is a finite Borel measure, it's clear that these are also finite Borel measures. We will show that they satisfy the given properties:

If ν(B)=0\nu\p{B} = 0, then ν(A∪B)=0\nu\p{A \cup B} = 0 as well so

M=μ(A)≤μ(A∪B)≤MM = \mu\p{A} \leq \mu\p{A \cup B} \leq M

by definition of MM, so these are all equalities. Hence, by measurability of BB,

M=μ(A∪B)=μ(A)+μ(E∖A)=M+μa(B)  ⟹  μa(B)=0,M = \mu\p{A \cup B} = \mu\p{A} + \mu\p{E \setminus A} = M + \mu_a\p{B} \implies \mu_a\p{B} = 0,

so μa≪ν\mu_a \ll \nu. On the other hand, Rn=A∪Ac\R^n = A \cup A^\comp, A∩Ac=∅A \cap A^\comp = \emptyset, ν(A)=0\nu\p{A} = 0, and μs(Ac)=μs(∅)=0\mu_s\p{A^\comp} = \mu_s\p{\emptyset} = 0, so μs⊥ν\mu_s \perp \nu. Finally, by measurability of AA, μ(B)=μ(B∖A)+μ(B∩A)=μa(B)+μs(B)\mu\p{B} = \mu\p{B \setminus A} + \mu\p{B \cap A} = \mu_a\p{B} + \mu_s\p{B} for any BB, which proves existence.

It remains to show uniqueness. Suppose μ=μa+μs=μa′+μs′\mu = \mu_a + \mu_s = \mu_a' + \mu_s' are two Lebesgue decompositions of μ\mu. Then μa−μa′=μs′−μs\mu_a - \mu_a' = \mu_s' - \mu_s. If ν(B)=0\nu\p{B} = 0, then by absolute continuity,

0=μa(B)−μa′(B)=μs(B)−μs′(B)  ⟹  μs(B)=μs′(B).0 = \mu_a\p{B} - \mu_a'\p{B} = \mu_s\p{B} - \mu_s'\p{B} \implies \mu_s\p{B} = \mu_s'\p{B}.

On the other hand, if ν(B)>0\nu\p{B} > 0, then let A,A′A, A' be such that μs(A)=0\mu_s\p{A} = 0 and ν(Ac)=0\nu\p{A^\comp} = 0; μs′(A′)=0\mu_s'\p{A'} = 0 and ν(A′c)=0\nu\p{A'^\comp} = 0, which exist since μs,μs′⊥ν\mu_s, \mu_s' \perp \nu. If we set A′′=A∩A′A'' = A \cap A', then μs(A′′)=μs′(A′′)=0\mu_s\p{A''} = \mu_s'\p{A''} = 0 and ν(A′′c)=0\nu\p{A''^\comp} = 0. Hence,

μs(B)=μs(B∖A′′)andμs′(B)=μs′(B∖A′′),\mu_s\p{B} = \mu_s\p{B \setminus A''} \quad\text{and}\quad \mu_s'\p{B} = \mu_s'\p{B \setminus A''},

but ν(B∖A′′)=0\nu\p{B \setminus A''} = 0, so by absolutely continuity again, μa(B∖A′′)\mu_a\p{B \setminus A''}. Thus,

0=μa(B∖A′′)−μa′(B∖A′′)=μs(B∖A′′)−μs′(B∖A′′)=μs(B)−μs′(B)  ⟹  μs(B)=μs′(B).\begin{aligned} 0 &= \mu_a\p{B \setminus A''} - \mu_a'\p{B \setminus A''} \\ &= \mu_s\p{B \setminus A''} - \mu_s'\p{B \setminus A''} \\ &= \mu_s\p{B} - \mu_s'\p{B} \\ \implies \mu_s\p{B} &= \mu_s'\p{B}. \end{aligned}

Hence, μs=μs′  ⟹  μa=μa′\mu_s = \mu_s' \implies \mu_a = \mu_a', so the decompositions are the same.