The Radon–Nikodym Theorem for a σ-Finite Measure, and Uniqueness of Densities

In the setting of Measurable Functions and the Lebesgue Integral: Standing Notation, let $(X,\mathcal{F},\mu)$ be a measure space.

(i) Uniqueness of integrable densities. Let $f,g\in\mathcal{L}^{1}(\mu)$ with

$$\int_{A}f\,d\mu=\int_{A}g\,d\mu\qquad(A\in\mathcal{F})$$

(Integrals over Measurable Subsets (vi)). Then the set $\{x\in X:f(x)\neq g(x)\}$ belongs to $\mathcal{F}$ and has $\mu$-measure $0$.

(ii) Uniqueness of nonnegative densities. Let $\mu$ be $\sigma$-finite, and let $f,g:X\to\mathbb{R}$ be measurable with $f(x)\ge0$ and $g(x)\ge0$ for every $x\in X$, and with

$$\int_{A}f\,d\mu=\int_{A}g\,d\mu\ \text{ in }[0,\infty]\qquad(A\in\mathcal{F}).$$

Then the set $\{x\in X:f(x)\neq g(x)\}$ belongs to $\mathcal{F}$ and has $\mu$-measure $0$.

(iii) Existence of a density. Assume the axiom of dependent choice, and let $\mu$ be $\sigma$-finite. Let $\nu$ be a finite measure on $(X,\mathcal{F})$ that is absolutely continuous with respect to $\mu$. Then there is $f\in\mathcal{L}^{1}(\mu)$ with $f(x)\ge0$ for every $x\in X$ and

$$\nu(A)=\int_{A}f\,d\mu\qquad(A\in\mathcal{F}).$$

(iv) Finite signed measures. Assume the axiom of dependent choice, and let $\mu$ be $\sigma$-finite. Let $\nu$ be a finite signed measure on $(X,\mathcal{F})$ that is absolutely continuous with respect to $\mu$. Then there is $f\in\mathcal{L}^{1}(\mu)$ with

$$\nu(A)=\int_{A}f\,d\mu\qquad(A\in\mathcal{F}).$$