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A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density

Statement

Assume the Axiom of Choice. Let $μ$ be a positive measure and let $ν$ be a signed measure on $(X, A)$ . Assume there is an increasing measurable exhaustion $(X_{n})$ with $⋃_{n} X_{n} = X$ , $μ (X_{n}) < + \infty$ , and $∣ ν ∣ (X_{n}) < + \infty$ for every $n$ . If $ν ≪ μ$ , then there exists a measurable real-valued function $f$ , unique up to $μ$ -almost-everywhere equality, such that $ν (E) = \int_{E} f d μ (E \in A) .$ If in addition $∣ ν ∣ (X) < + \infty$ , then $f \in L^{1} (μ)$ .

Facts & Assumptions

Given: AC, a positive measure $μ$ , a signed measure $ν$ with a common finite exhaustion $(X_{n})$ , and the hypothesis $ν ≪ μ$ .

[L1]

The Lebesgue decomposition theorem gives $ν = ν_{a} + ν_{s}$ with $ν_{a} ≪ μ$ , $ν_{s} ⊥ μ$ , and $ν_{a} (E) = \int_{E} f d μ$ for every measurable $E$ , with $f$ real-valued and $\int_{X_{n}} ∣ f ∣ d μ < \infty$ . (Every sigma-finite signed measure admits a Lebesgue decomposition relative to a sigma-finite positive measure)

[L2]

A signed or complex measure that is both absolutely continuous and singular with respect to $μ$ is zero. (A signed or complex measure that is both absolutely continuous and singular with respect to the same positive measure is zero)

[L3]

Real integrability means finiteness of both positive and negative integrals; the integral is linear on $L^{1}$ . (Integrable real and complex functions, and their integrals, The Lebesgue integral is linear on $L^{1} (μ)$ )

[L5]

Variation dominates $∣ ν (E) ∣$ by its partition definition. (The total variation |nu|(E) from countable measurable partitions)

[L4]

AC is assumed for the selections in [L1]. (The Axiom of Choice)

Proof

technique · direct

1.1L1choose

By [L1], write $ν = ν_{a} + ν_{s}$ , where $ν_{a} ≪ μ$ , $ν_{s} ⊥ μ$ , and $ν_{a} (E) = \int_{E} f d μ$ for every measurable $E$ , with $f$ real-valued and $\int_{X_{n}} ∣ f ∣ d μ < \infty$ .

2.1step 1.1L2algebra

Because $ν ≪ μ$ and $ν_{a} ≪ μ$ , on each $μ$ -null set $E$ the decomposition gives $0 = 0 + ν_{s} (E)$ , so $ν_{s} ≪ μ$ without subtracting infinite values. Step 1.1 already gives $ν_{s} ⊥ μ$ , so [L2] forces $ν_{s} = 0$ . Hence $ν = ν_{a}$ , and the displayed integral formula for $f$ represents $ν$ on every measurable set.

3.1step 2.1choosealgebra

Suppose $g$ is another measurable real-valued function with $ν (E) = \int_{E} g d μ$ for every measurable $E$ . Fix $n$ . The formula on $X_{n}$ gives $\int_{X_{n}} g d μ = ν (X_{n}) \in R$ , so both positive and negative integrals of $g$ there are finite. Thus $f, g \in L^{1} (X_{n})$ , and [L3] permits subtraction of their integrals over every measurable subset of $X_{n}$ . For each $m \geq 1$ , let $P_{n, m} : = X_{n} \cap {f - g \geq 1 / m}, N_{n, m} : = X_{n} \cap {g - f \geq 1 / m} .$ Applying the two representation formulas to $P_{n, m}$ gives $0 = \int_{P_{n, m}} (f - g) d μ \geq μ (P_{n, m}) / m,$ so $μ (P_{n, m}) = 0$ ; the same argument with $N_{n, m}$ gives $μ (N_{n, m}) = 0$ . Therefore $X_{n} \cap {f \neq g} = ⋃_{m \geq 1} P_{n, m} \cup ⋃_{m \geq 1} N_{n, m}$ is $μ$ -null. Since $X = ⋃_{n} X_{n}$ , the functions $f$ and $g$ agree $μ$ -almost everywhere on $X$ .

3.2L3step 2.1algebra

If $∣ ν ∣ (X) < \infty$ , use the representation on the measurable sets $P = {f \geq 0}$ and $N = {f < 0}$ . It gives $\int f^{+} d μ = ν (P)$ and $\int f^{-} d μ = - ν (N)$ . Both are finite: $∣ ν (E) ∣ \leq ∣ ν ∣ (X)$ for every $E$ , directly from the partition definition of variation. Hence $f \in L^{1} (μ)$ by [L3].

4.1step 2.1step 3.1step 3.2∎

Steps 2.1, 3.1, and 3.2 prove existence, almost-everywhere uniqueness, and the finite-measure integrability clause.

A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources