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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

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: 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 $ν_{a} (E) = \int_{E} f d μ$ whenever the integral is defined. (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]

If a signed measure is absolutely continuous with respect to $μ$ , then its Jordan parts are too. (For signed and complex measures, absolute continuity is equivalent for the measure, its Jordan or real-imaginary parts, and its total variation)

[L4]

Jordan decomposition writes $ν = ν^{+} - ν^{-}$ , and for a signed measure one has $∣ ν ∣ (X) = ν^{+} (X) + ν^{-} (X)$ . (Jordan decomposition of a signed measure into unique mutually singular positive parts, For a signed measure, total variation is nu-plus plus nu-minus, finite partitions suffice, and nu-plus and nu-minus are extremal)

Proof

technique · direct

1.1

By [L1], write $ν = ν_{a} + ν_{s}$ , where $ν_{a} ≪ μ$ , $ν_{s} ⊥ μ$ , and $ν_{a} (E) = \int_{E} f d μ$ whenever the integral is defined.

L1choose

2.1

Because $ν ≪ μ$ and $ν_{a} ≪ μ$ , the difference $ν_{s} = ν - ν_{a}$ is also absolutely continuous with respect to $μ$ . 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.

step 1.1L2algebra

3.1

Suppose $g$ is another measurable real-valued function with $ν (E) = \int_{E} g d μ$ for every measurable $E$ . Fix $n$ . 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 $X$ .

step 2.1choosealgebra

4.1

If $∣ ν ∣ (X) < + \infty$ , let $ν = ν^{+} - ν^{-}$ be the Jordan decomposition from [L4]. Then [L3] makes both $ν^{+}$ and $ν^{-}$ absolutely continuous with respect to $μ$ . Apply the positive-measure part of [L1] to obtain nonnegative measurable functions $f^{+}, f^{-}$ with $ν^{+} (E) = \int_{E} f^{+} d μ, ν^{-} (E) = \int_{E} f^{-} d μ (E \in A) .$ Then $f^{+} - f^{-}$ is another representative of $ν$ , so step 3.1 gives $f = f^{+} - f^{-}$ $μ$ -almost everywhere. Hence $∣ f ∣ \leq f^{+} + f^{-}$ almost everywhere, and $\int ∣ f ∣ d μ \leq \int f^{+} d μ + \int f^{-} d μ = ν^{+} (X) + ν^{-} (X) = ∣ ν ∣ (X) < + \infty .$ Therefore $f \in L^{1} (μ)$ by the definition of integrability.

L3L4step 3.1algebra

5.1

Steps 2.1, 3.1, and 4.1 prove existence, almost-everywhere uniqueness, and the finite-measure integrability clause.

step 2.1step 3.1step 4.1∎

Depends on

Used by

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard F. Bass, Real Analysis for Graduate Students, Theorem 13.4 (standard reference, not scraped)
John K. Hunter, Measure Theory, Theorem 6.27 (standard reference, not scraped)