Proof of Properties of the Integral with Respect to a Finite Signed Measure: Linearity, Bounds, Bounded Functions, Linearity in the Measure, Indicators and Finite Measures

Throughout, a measurable function $X\to\mathbb{R}$ with values $\ge0$ lies in $\mathcal{M}^{+}$, and when it is integrable its integrals as an element of $\mathcal{M}^{+}$ and of $\mathcal{L}^{1}$ coincide; and for a measure $\rho$ and $f\in\mathcal{L}^{1}(\rho)$, $\int f\,d\rho=\int f^{+}\,d\rho-\int f^{-}\,d\rho$ (base). By the Hahn decomposition theorem (which uses the axiom of dependent choice), every finite signed measure on $(X,\mathcal{F})$ has a Hahn set, so its Jordan decomposition exists. Let $\nu^{+},\nu^{-}$ and $\nu'^{+},\nu'^{-}$ be the Jordan decompositions of $\nu$ and $\nu'$; these are finite measures, and $|\nu|=\nu^{+}+\nu^{-}$ setwise by Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (ii).

Step 1: linearity. Let $f,g\in\mathcal{L}^{1}(\nu)=\mathcal{L}^{1}(|\nu|)$ and $a,b\in\mathbb{R}$. By linearity of the Lebesgue integral, $af+bg\in\mathcal{L}^{1}(|\nu|)=\mathcal{L}^{1}(\nu)$. By Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (viii), $f$ and $g$ lie in $\mathcal{L}^{1}(\nu^{+})$ and $\mathcal{L}^{1}(\nu^{-})$, and linearity of the integrals against $\nu^{+}$ and $\nu^{-}$ gives

$$\int(af+bg)\,d\nu=\int(af+bg)\,d\nu^{+}-\int(af+bg)\,d\nu^{-}=a\Bigl(\int f\,d\nu^{+}-\int f\,d\nu^{-}\Bigr)+b\Bigl(\int g\,d\nu^{+}-\int g\,d\nu^{-}\Bigr)=a\int f\,d\nu+b\int g\,d\nu .$$

This proves (i).

Step 2: the bound. Let $f\in\mathcal{L}^{1}(\nu)$. Then $|f|\in\mathcal{L}^{1}(|\nu|)$, and $f\in\mathcal{L}^{1}(\nu^{\pm})$ by Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (viii). The inequality $|\int f\,d\rho|\le\int|f|\,d\rho$ for $\rho=\nu^{+},\nu^{-}$ and the identity of Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (viii) give

$$\Bigl|\int f\,d\nu\Bigr|\le\Bigl|\int f\,d\nu^{+}\Bigr|+\Bigl|\int f\,d\nu^{-}\Bigr|\le\int|f|\,d\nu^{+}+\int|f|\,d\nu^{-}=\int|f|\,d|\nu| .$$

This proves (ii).

Step 3: bounded functions. Let $f$ and $C$ be as in (iii). Since $|\nu|$ is a finite measure (Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (i)), $f\in\mathcal{L}^{1}(|\nu|)=\mathcal{L}^{1}(\nu)$ by Bounded Measurable Functions on a Finite Measure Space Are Integrable (i). The function with constant value $C=C\,\mathbf{1}_{X}(x)$ lies in $\mathcal{M}^{+}$ and has integral $C\,|\nu|(X)=C\,\|\nu\|$ against $|\nu|$, and $|f|\le C$ pointwise; by monotonicity of the integral on $\mathcal{M}^{+}$ and Step 2,

$$\Bigl|\int f\,d\nu\Bigr|\le\int|f|\,d|\nu|\le C\,\|\nu\| .$$

This proves (iii).

Step 4: differences of finite measures. Let $\mu_{1},\mu_{2}$, $f$ and $C$ be as in (iv). For every finite measure $\rho$ on $(X,\mathcal{F})$, $f\in\mathcal{L}^{1}(\rho)$ by Bounded Measurable Functions on a Finite Measure Space Are Integrable (i); and $f\in\mathcal{L}^{1}(\nu)$ by Step 3. Since $\nu^{+}-\nu^{-}=\nu=\mu_{1}-\mu_{2}$ setwise, $\nu^{+}(A)+\mu_{2}(A)=\nu^{-}(A)+\mu_{1}(A)$ for every $A\in\mathcal{F}$: the finite measures $\nu^{+}+\mu_{2}$ and $\nu^{-}+\mu_{1}$ of Nonnegative Combinations of Finite Measures and Their Integrals (i) are equal. So Nonnegative Combinations of Finite Measures and Their Integrals (ii) gives

$$\int f\,d\nu^{+}+\int f\,d\mu_{2}=\int f\,d(\nu^{+}+\mu_{2})=\int f\,d(\nu^{-}+\mu_{1})=\int f\,d\nu^{-}+\int f\,d\mu_{1},$$

and hence

$$\int f\,d\nu=\int f\,d\nu^{+}-\int f\,d\nu^{-}=\int f\,d\mu_{1}-\int f\,d\mu_{2}.$$

This proves (iv).

Step 5: linearity in the measure. Let $f$ and $C$ be as in (v). By (iii), applied to $\nu$, $\nu'$ and to the finite signed measures $\nu+\nu'$ and $c\nu$, $f$ is integrable with respect to each of them. For finite measures $\rho_{1},\rho_{2}$ and real $a_{1},a_{2}\ge0$, $f$ lies in $\mathcal{L}^{1}(\rho_{1})\cap\mathcal{L}^{1}(\rho_{2})$ by Bounded Measurable Functions on a Finite Measure Space Are Integrable (i), so Nonnegative Combinations of Finite Measures and Their Integrals (ii) gives

$$\int f\,d(a_{1}\rho_{1}+a_{2}\rho_{2})=a_{1}\int f\,d\rho_{1}+a_{2}\int f\,d\rho_{2}.\tag{$*$}$$

Now $\nu+\nu'=(\nu^{+}+\nu'^{+})-(\nu^{-}+\nu'^{-})$ setwise, a difference of finite measures, so Step 4 applied to $\nu+\nu'$, then $(*)$, give

$$\int f\,d(\nu+\nu')=\int f\,d\nu^{+}+\int f\,d\nu'^{+}-\int f\,d\nu^{-}-\int f\,d\nu'^{-}=\int f\,d\nu+\int f\,d\nu' .$$

If $c\ge0$, then $c\nu=c\nu^{+}-c\nu^{-}$ setwise, a difference of finite measures (Nonnegative Combinations of Finite Measures and Their Integrals (i) with $\mu_{2}=\mu_{1}$ and $a_{2}=0$), and Step 4 and $(*)$ give $\int f\,d(c\nu)=c\int f\,d\nu^{+}-c\int f\,d\nu^{-}=c\int f\,d\nu$. If $c<0$, then $c\nu=|c|\nu^{-}-|c|\nu^{+}$ setwise, and likewise $\int f\,d(c\nu)=|c|\int f\,d\nu^{-}-|c|\int f\,d\nu^{+}=c\int f\,d\nu$. This proves (v).

Step 6: indicators. Let $A\in\mathcal{F}$. The indicator $\mathbf{1}_{A}$ is a nonnegative simple function, in particular measurable as a function into $\mathbb{R}$, with $|\mathbf{1}_{A}|\le1$; so $\mathbf{1}_{A}\in\mathcal{L}^{1}(\nu)$ by Step 3. For every finite measure $\rho$, $\int\mathbf{1}_{A}\,d\rho=\rho(A)$, so

$$\int\mathbf{1}_{A}\,d\nu=\nu^{+}(A)-\nu^{-}(A)=\nu(A).$$

This proves (vi).

Step 7: finite measures. Suppose $\nu(A)\ge0$ for every $A\in\mathcal{F}$. By Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (iv), $\nu$ is a finite measure with $|\nu|=\nu$, so the set $\mathcal{L}^{1}(|\nu|)$ of Integrable Functions and the Integral with Respect to a Finite Signed Measure (i) is the set $\mathcal{L}^{1}(\nu)$ of the base; and for $f$ in it, by the same clause, $\int f\,d\nu^{+}-\int f\,d\nu^{-}$ is the Lebesgue integral of $f$ against $\nu$. This proves (vii).