Real And Complex Measures

Signed Measure

Suppose that

(X, 𝒮)

is a measurable space. A signed measure on

𝒮

is a countably additive function

ν : 𝒮 \to [- \infty, \infty]

that does not take both values

\infty

and

- \infty

, and satisfies

ν (\emptyset) = 0

Complex Measure

Suppose that

(X, 𝒮)

is a measurable space. A complex measure on

𝒮

is a countably additive function

ν : 𝒮 \to ℂ

satisfying

ν (\emptyset) = 0

Positive and Negative Set for a Signed Measure

Suppose that

ν

is a signed measure on

(X, 𝒮)

. A set

P \in 𝒮

is positive for

ν

ν (E) \geq 0

for every

E \in 𝒮

with

E \subseteq P

. A set

N \in 𝒮

is negative for

ν

ν (E) \leq 0

for every

E \in 𝒮

with

E \subseteq N

Hahn Decomposition Theorem

Suppose that

ν

is a signed measure on

(X, 𝒮)

. Then there exist disjoint sets

P, N \in 𝒮

such that

X = P \cup N

P

is positive for

ν

, and

N

is negative for

ν

Let $α = \sup ν (E)$ , where $E \in 𝒮$ and $ν (E) < \infty$ . Choose sets $E_{n}$ with $ν (E_{n}) \to α$ , and put $P = ⋃_{n} E_{n}$ . Countable additivity of the signed measure shows that every measurable subset of $P$ has nonnegative measure; otherwise replacing $P$ by its complement relative to that subset would increase the supremum. Then $N = X ∖ P$ is negative by the same maximality argument. Thus $X = P \cup N$ is the desired Hahn decomposition.

Mutually Singular Measures

Measures

μ

and

ν

(X, 𝒮)

are mutually singular, written

μ ⟂ ν

, if there exists

E \in 𝒮

such that

μ (E) = 0

and

ν (X ∖ E) = 0

Absolute Continuity of Measures

Suppose that

μ

is a positive measure and

ν

is a signed or complex measure on

(X, 𝒮)

. We say that

ν

is absolutely continuous with respect to

μ

, written

ν ≪ μ

, if

μ (E) = 0

implies

ν (E) = 0

for every

E \in 𝒮

Jordan Decomposition Theorem

Suppose that

ν

is a signed measure. Then there exist unique positive measures

ν^{+}

and

ν^{-}

such that

ν = ν^{+} - ν^{-} and ν^{+} ⟂ ν^{-} .

Let $X = P \cup N$ be a Hahn decomposition. Define $ν^{+} (E) = ν (E \cap P), ν^{-} (E) = - ν (E \cap N) .$ Positivity and countable additivity follow from the definitions of positive and negative sets. Also $ν = ν^{+} - ν^{-}$ , and $ν^{+} ⟂ ν^{-}$ because $ν^{-} (P) = 0$ and $ν^{+} (N) = 0$ . If another mutually singular decomposition existed, comparing on the positive and negative supports forces the same values on every measurable set, giving uniqueness.

Total Variation Measure

Suppose that

ν

is a signed or complex measure. The total variation measure

| ν |

is the positive measure defined by

| ν | (E) : = \sup \sum_{k = 1}^{n} | ν (E_{k}) |,

where the supremum is over all finite measurable partitions

E_{1}, \dots, E_{n}

E

Radon-Nikodym Theorem

Suppose that

μ

is a

σ

-finite positive measure and

ν

is a

σ

-finite positive measure such that

ν ≪ μ

. Then there exists a nonnegative measurable function

f

such that

ν (E) = \int_{E} f d μ

for every measurable set

E

. The function

f

is unique up to equality almost everywhere.

For finite $μ$ , consider the measure $ρ = μ + ν$ and the closed convex set of functions $h \in L^{2} (ρ)$ satisfying $\int_{E} h d ρ \leq ν (E)$ for all measurable $E$ . The Hilbert-space projection argument produces $h$ with $0 \leq h \leq 1$ and $ν (E) = \int_{E} h d ρ$ . Since $ν ≪ μ$ , the set where $h = 1$ has $ν$ - and $μ$ -measure $0$ , so $f = h / (1 - h)$ satisfies $ν (E) = \int_{E} f d μ$ . The $σ$ -finite case follows by applying the finite case on a countable finite-measure exhaustion and patching the densities. If two densities give the same integrals over all $E$ , their positive difference sets have integral $0$ , so they are equal almost everywhere.

Lebesgue Decomposition Theorem

Suppose that

μ

and

ν

are

σ

-finite positive measures on

(X, 𝒮)

. Then there exist unique positive measures

ν_{a}

and

ν_{s}

such that

ν = ν_{a} + ν_{s}, ν_{a} ≪ μ, ν_{s} ⟂ μ .

Let $ρ = μ + ν$ . By the Radon-Nikodym theorem, write $ν (E) = \int_{E} f d ρ$ and $μ (E) = \int_{E} g d ρ$ . Put $A = {g > 0}$ and $S = {g = 0}$ . Define $ν_{a} (E) = ν (E \cap A)$ and $ν_{s} (E) = ν (E \cap S)$ . Then $ν_{a} ≪ μ$ , while $ν_{s} ⟂ μ$ . If $ν = η_{a} + η_{s}$ is another such decomposition, absolute continuity and singularity force $η_{a} = ν_{a}$ and $η_{s} = ν_{s}$ on $A$ and $S$ , proving uniqueness.