Proof of Nonnegative Combinations of Finite Measures and Their Integrals

Throughout, the values of a finite measure are real numbers $\ge0$, on which the sum and the product of $[0,\infty]$ are those of $\mathbb{R}$ (routine facts and background of the setting). Let $\mu=a_{1}\mu_{1}+a_{2}\mu_{2}$.

Step 1: a finite measure. The values of $\mu$ are real numbers $\ge0$, $\mu(\emptyset)=0$, and $\mu(X)<\infty$. For a pairwise disjoint sequence $(A_{m})$ in $\mathcal{F}$, countable additivity of $\mu_{1}$ and $\mu_{2}$ and the additivity and homogeneity of sums of sequences in $[0,\infty]$ give

$$\mu\Bigl(\bigcup_{m\in\mathbb{N}}A_{m}\Bigr)=a_{1}\sum_{m\in\mathbb{N}}\mu_{1}(A_{m})+a_{2}\sum_{m\in\mathbb{N}}\mu_{2}(A_{m})=\sum_{m\in\mathbb{N}}\bigl(a_{1}\mu_{1}(A_{m})+a_{2}\mu_{2}(A_{m})\bigr)=\sum_{m\in\mathbb{N}}\mu(A_{m}).$$

This proves (i).

Step 2: nonnegative simple functions. For $s\in\mathcal{S}^{+}$, the simple integral is a finite sum over the image $s[X]$ whose terms are real, and

$$I_{\mu}(s)=\sum_{y\in s[X]}y\,\mu\bigl(s^{-1}(\{y\})\bigr)=a_{1}\sum_{y\in s[X]}y\,\mu_{1}\bigl(s^{-1}(\{y\})\bigr)+a_{2}\sum_{y\in s[X]}y\,\mu_{2}\bigl(s^{-1}(\{y\})\bigr)=a_{1}I_{\mu_{1}}(s)+a_{2}I_{\mu_{2}}(s).$$

Step 3: nonnegative measurable functions. Let $g\in\mathcal{M}^{+}$, and let $(s_{m})$ be its sequence of dyadic approximations: $s_{m}\in\mathcal{S}^{+}$, $s_{m}\le s_{m+1}$ pointwise, and $g(x)=\sup_{m}s_{m}(x)$ for every $x\in X$. For each of the measures $\rho=\mu,\mu_{1},\mu_{2}$, the integral of a nonnegative simple function is its simple integral, and the monotone convergence theorem gives $\int g\,d\rho=\sup_{m}I_{\rho}(s_{m})$, the sequence $(I_{\rho}(s_{m}))_{m}$ being nondecreasing. Since the supremum of the sum of two nondecreasing sequences in $[0,\infty]$ is the sum of their suprema, and suprema commute with multiplication by $a_{i}\in[0,\infty]$, Step 2 gives

$$\int g\,d\mu=\sup_{m}\bigl(a_{1}I_{\mu_{1}}(s_{m})+a_{2}I_{\mu_{2}}(s_{m})\bigr)=a_{1}\sup_{m}I_{\mu_{1}}(s_{m})+a_{2}\sup_{m}I_{\mu_{2}}(s_{m})=a_{1}\int g\,d\mu_{1}+a_{2}\int g\,d\mu_{2}.$$

Step 4: integrable functions. Let $f\in\mathcal{L}^{1}(\mu_{1})\cap\mathcal{L}^{1}(\mu_{2})$. Then $f$ is measurable, $f^{+},f^{-}\in\mathcal{M}^{+}$, and their integrals against $\mu_{1}$ and $\mu_{2}$ are real numbers. By Step 3 with $g=f^{+}$ and $g=f^{-}$,

$$\int f^{\pm}\,d\mu=a_{1}\int f^{\pm}\,d\mu_{1}+a_{2}\int f^{\pm}\,d\mu_{2}<\infty,$$

so $f\in\mathcal{L}^{1}(\mu)$ by Integrable Functions and Their Lebesgue Integral (i), and subtracting the identity for $f^{-}$ from that for $f^{+}$ gives, by Integrable Functions and Their Lebesgue Integral (iii),

$$\int f\,d\mu=a_{1}\Bigl(\int f^{+}\,d\mu_{1}-\int f^{-}\,d\mu_{1}\Bigr)+a_{2}\Bigl(\int f^{+}\,d\mu_{2}-\int f^{-}\,d\mu_{2}\Bigr)=a_{1}\int f\,d\mu_{1}+a_{2}\int f\,d\mu_{2}.$$

With Step 3 this proves (ii).