Proof of Finite Signed Measures from Finite Measures and from Integrals

Throughout, the field and order laws of $\mathbb{R}$ are used freely (The Structure of ℝ as a Field and an Ordered Ring (i), The Structure of ℝ as a Field and an Ordered Ring (ii), Rules of Arithmetic and Order in an Ordered Field (iv)).

Step 1: finite measures. Let $\mu$ be a finite measure on $(X,\mathcal{F})$. For $A\in\mathcal{F}$ we have $\mu(A)\le\mu(X)<\infty$ (Basic Properties of a Measure (iii)), so $\mu(A)\in\mathbb{R}$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (iii)); thus $\mu$ is a map $\mathcal{F}\to\mathbb{R}$, and $\mu(\emptyset)=0$. Let $(A_{m})$ be a pairwise disjoint sequence in $\mathcal{F}$ with union $A$. Countable additivity of $\mu$ says that the sum $\sum_{m\in\mathbb{N}}\mu(A_{m})$ in $[0,\infty]$ equals the real number $\mu(A)$; as its terms are real, the real series $\sum_{m=1}^{\infty}\mu(A_{m})$ converges with sum $\mu(A)$ (Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence (vi)). This proves (ii).

Step 2: differences of finite measures. Let $\mu_{1},\mu_{2}$ be as in (i). By Step 1 their values are real, so $\mu_{1}-\mu_{2}$ is a map $\mathcal{F}\to\mathbb{R}$, and $(\mu_{1}-\mu_{2})(\emptyset)=0$. Let $(A_{m})$ be a pairwise disjoint sequence in $\mathcal{F}$ with union $A$. By Step 1 the series $\sum_{m}\mu_{i}(A_{m})$ converge with sums $\mu_{i}(A)$ ($i=1,2$), hence (Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping (i))

$$\sum_{m=1}^{\infty}\bigl(\mu_{1}(A_{m})-\mu_{2}(A_{m})\bigr)=\mu_{1}(A)-\mu_{2}(A).$$

Step 3: restriction to $\mathcal{G}$. Let $\mu$ be a measure and $\nu$ a finite signed measure on $(X,\mathcal{F})$. Since $\mathcal{G}\subseteq\mathcal{F}$, the restrictions are maps $\mu|_{\mathcal{G}}:\mathcal{G}\to[0,\infty]$ and $\nu|_{\mathcal{G}}:\mathcal{G}\to\mathbb{R}$ with the same values as $\mu$ and $\nu$ on $\mathcal{G}$ (Domains and Restrictions of Maps (ii), Domains and Restrictions of Maps (iii)). As $\emptyset,X\in\mathcal{G}$, we get $\mu|_{\mathcal{G}}(\emptyset)=0=\nu|_{\mathcal{G}}(\emptyset)$ and $\mu|_{\mathcal{G}}(X)=\mu(X)$, so $\mu|_{\mathcal{G}}$ is finite if $\mu$ is, and a probability measure if $\mu$ is one. A pairwise disjoint sequence $(A_{m})$ in $\mathcal{G}$ is one in $\mathcal{F}$, and its union lies in $\mathcal{G}$ (σ-Algebra and Measurable Space (i)); since all values involved agree, countable additivity of $\mu$ and of $\nu$ at $(A_{m})$ is countable additivity of $\mu|_{\mathcal{G}}$ and of $\nu|_{\mathcal{G}}$ at $(A_{m})$. This proves (iii).

Step 4: indefinite integrals. Let $\mu$, $f$ and $\nu_{f}$ be as in (iv); $f$ is a measurable map $X\to\mathbb{R}$ (The Borel Sets of ℝ Form a σ-Algebra, and Integrable Functions Are Measurable Real-Valued Maps (iii)). Let $A\in\mathcal{G}$. As $A\in\mathcal{F}$, $f$ is integrable on $A$ (Integrals over Measurable Subsets (vi)), so $\nu_{f}(A)$ is a real number and $\nu_{f}$ is a map $\mathcal{G}\to\mathbb{R}$. Moreover $f\mathbf{1}_{A}\in\mathcal{L}^{1}(\mu)$ and

$$\int_{A}f\,d\mu=\int_{X}f\mathbf{1}_{A}\,d\mu$$

(Integrals over Measurable Subsets (ii)); in particular $f\mathbf{1}_{A}$ is measurable (The Borel Sets of ℝ Form a σ-Algebra, and Integrable Functions Are Measurable Real-Valued Maps (iii)). As $\mu(\emptyset)=0$, $\nu_{f}(\emptyset)=0$ by Integrals over Measurable Subsets (ix). Let $(A_{m})$ be a pairwise disjoint sequence in $\mathcal{G}$ with union $A\in\mathcal{G}$, and put $B_{n}=A_{1}\cup\dots\cup A_{n}\in\mathcal{G}$. By Integrals over Measurable Subsets (vii) and induction on $n$, then by what was just shown with $B_{n}$ in place of $A$,

$$\sum_{m=1}^{n}\int_{A_{m}}f\,d\mu=\int_{B_{n}}f\,d\mu=\int_{X}f\mathbf{1}_{B_{n}}\,d\mu .$$

For each $x\in X$ the sequence $(f\mathbf{1}_{B_{n}}(x))_{n}$ is constant from some index on, with value $f\mathbf{1}_{A}(x)$: equal to $f(x)$ from the index $m$ with $x\in A_{m}$ on if $x\in A$, and $0$ if $x\notin A$. Moreover the measurable functions $f\mathbf{1}_{B_{n}}$ satisfy $|f\mathbf{1}_{B_{n}}|\le|f|$, and $|f|\in\mathcal{L}^{1}(\mu)$ by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (x). By dominated convergence (Dominated Convergence Theorem (iii)),

$$\int_{X}f\mathbf{1}_{B_{n}}\,d\mu\to\int_{X}f\mathbf{1}_{A}\,d\mu=\int_{A}f\,d\mu ,$$

so the series $\sum_{m}\nu_{f}(A_{m})$ converges with sum $\nu_{f}(A)$. Thus $\nu_{f}$ is a finite signed measure on $(X,\mathcal{G})$. Finally, $\mu|_{\mathcal{G}}$ is a measure on $(X,\mathcal{G})$ by Step 3, and if $A\in\mathcal{G}$ with $\mu|_{\mathcal{G}}(A)=\mu(A)=0$, then $\nu_{f}(A)=0$ by Integrals over Measurable Subsets (ix); so $\nu_{f}\ll\mu|_{\mathcal{G}}$.

Step 5: nonnegative integrands. Let moreover $f(x)\ge0$ for every $x\in X$. As $f$ is measurable, for $A\in\mathcal{G}$ the number $\nu_{f}(A)$ is also the integral over $A$ of $f\in\mathcal{M}^{+}$ (Integrals over Measurable Subsets (iii)), an element of $[0,\infty]$; so $\nu_{f}$ is a map $\mathcal{G}\to[0,\infty]$ with real values, and $\nu_{f}(\emptyset)=0$. For a pairwise disjoint sequence $(A_{m})$ in $\mathcal{G}$ with union $A$, the real series $\sum_{m}\nu_{f}(A_{m})$ converges with sum $\nu_{f}(A)$ by Step 4, so $\nu_{f}(A)$ is also the sum $\sum_{m\in\mathbb{N}}\nu_{f}(A_{m})$ in $[0,\infty]$ (Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence (vi)). Hence $\nu_{f}$ is a measure on $(X,\mathcal{G})$; it is finite because $\nu_{f}(X)$ is real (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (ii)), and $\nu_{f}\ll\mu|_{\mathcal{G}}$ because $\nu_{f}$ vanishes on the sets $A\in\mathcal{G}$ with $\mu|_{\mathcal{G}}(A)=0$, as shown in Step 4.