[proofplan] We write both $s$ and $t$ over a common measurable partition (their joint refinement), reducing all integrals to finite sums. Linearity, monotonicity, and additivity over domains then follow from the corresponding properties of finite sums and the measure $\mu$. For part 4, we verify countable additivity of the set function $E \mapsto \int_E s \, d\mu$ by expressing it as a finite linear combination of $\mu$ restricted to the atoms, and using the countable additivity of $\mu$. [/proofplan] [step:Write $s$ and $t$ over a common partition] Let $s = \sum_{i=1}^m a_i\,\mathbb{1}_{A_i}$ and $t = \sum_{j=1}^n b_j\,\mathbb{1}_{B_j}$ be the standard-form representations, where $\{A_i\}$ and $\{B_j\}$ are measurable partitions of $X$ and $a_i, b_j \ge 0$. Form the common refinement $C_{ij} := A_i \cap B_j$. On each $C_{ij}$, $s = a_i$ and $t = b_j$. By definition of the integral of a non-negative simple function: \begin{align*} \int_X s \, d\mu = \sum_{i=1}^m a_i\,\mu(A_i), \quad \int_X t \, d\mu = \sum_{j=1}^n b_j\,\mu(B_j). \end{align*} Since $A_i = \bigsqcup_{j=1}^n C_{ij}$ (disjoint union) and $\mu$ is finitely additive, $\mu(A_i) = \sum_{j=1}^n \mu(C_{ij})$, so \begin{align*} \int_X s \, d\mu = \sum_{i,j} a_i\,\mu(C_{ij}), \quad \int_X t \, d\mu = \sum_{j,i} b_j\,\mu(C_{ij}). \end{align*} [/step] [step:Prove linearity] For $\alpha, \beta \ge 0$, the function $\alpha s + \beta t$ is non-negative simple with $(\alpha s + \beta t)(x) = \alpha a_i + \beta b_j$ on each $C_{ij}$. By definition of the integral: \begin{align*} \int_X (\alpha s + \beta t) \, d\mu &= \sum_{i,j} (\alpha a_i + \beta b_j)\,\mu(C_{ij}) \\ &= \alpha \sum_{i,j} a_i\,\mu(C_{ij}) + \beta \sum_{j,i} b_j\,\mu(C_{ij}) \\ &= \alpha \int_X s \, d\mu + \beta \int_X t \, d\mu. \end{align*} The second equality uses the distributive law for finite sums and the fact that all terms are non-negative (so no issues with $\infty - \infty$ arise). The convention $0 \cdot \infty = 0$ is used when $a_i = 0$ and $\mu(C_{ij}) = \infty$, or when $\mu(C_{ij}) = 0$ and $a_i > 0$, consistent with the definition of the simple integral. [guided] Linearity reduces to the distributive law for finite sums. The only subtlety is the convention $0 \cdot \infty = 0$: if $a_i = 0$ and $\mu(C_{ij}) = \infty$, then $a_i \cdot \mu(C_{ij}) = 0$, which is consistent because $s = 0$ on $C_{ij}$ and the integral of the zero function over any set is $0$. The restriction $\alpha, \beta \ge 0$ ensures all terms remain non-negative, avoiding $\infty - \infty$. [/guided] [/step] [step:Prove monotonicity] Suppose $s(x) \le t(x)$ for all $x \in X$. On each $C_{ij}$, this means $a_i \le b_j$. Therefore \begin{align*} \int_X s \, d\mu = \sum_{i,j} a_i\,\mu(C_{ij}) \le \sum_{i,j} b_j\,\mu(C_{ij}) = \int_X t \, d\mu. \end{align*} The inequality $a_i\,\mu(C_{ij}) \le b_j\,\mu(C_{ij})$ holds term-by-term: if $\mu(C_{ij}) = 0$, both sides are $0$; if $\mu(C_{ij}) > 0$, it follows from $a_i \le b_j$ and $\mu(C_{ij}) \ge 0$. [/step] [step:Prove additivity over domains] Let $E, F \in \mathcal{A}$ be disjoint. For any $G \in \mathcal{A}$, $\int_G s \, d\mu = \int_X s \cdot \mathbb{1}_G \, d\mu = \sum_{i=1}^m a_i\,\mu(A_i \cap G)$. Therefore: \begin{align*} \int_{E \cup F} s \, d\mu &= \sum_{i=1}^m a_i\,\mu(A_i \cap (E \cup F)) \\ &= \sum_{i=1}^m a_i\left[\mu(A_i \cap E) + \mu(A_i \cap F)\right] \\ &= \sum_{i=1}^m a_i\,\mu(A_i \cap E) + \sum_{i=1}^m a_i\,\mu(A_i \cap F) \\ &= \int_E s \, d\mu + \int_F s \, d\mu. \end{align*} The second equality uses the finite additivity of $\mu$: since $E$ and $F$ are disjoint, $A_i \cap E$ and $A_i \cap F$ are disjoint, so $\mu(A_i \cap (E \cup F)) = \mu(A_i \cap E) + \mu(A_i \cap F)$. The third equality uses the distributive law for finite sums (again, all terms are non-negative, so rearrangement is valid). [/step] [step:Prove the set function $E \mapsto \int_E s \, d\mu$ is a measure] Define $\nu : \mathcal{A} \to [0, \infty]$ by $\nu(E) := \int_E s \, d\mu = \sum_{i=1}^m a_i\,\mu(A_i \cap E)$. We verify the two axioms of a measure. **$\nu(\varnothing) = 0$:** $\nu(\varnothing) = \sum_i a_i\,\mu(A_i \cap \varnothing) = \sum_i a_i \cdot 0 = 0$, using $\mu(\varnothing) = 0$ and the convention $0 \cdot 0 = 0$. **Countable additivity:** Let $(E_k)_{k=1}^\infty$ be a sequence of pairwise disjoint sets in $\mathcal{A}$, and set $E := \bigsqcup_{k=1}^\infty E_k$. Then \begin{align*} \nu(E) &= \sum_{i=1}^m a_i\,\mu(A_i \cap E) = \sum_{i=1}^m a_i\,\mu\!\left(\bigsqcup_{k=1}^\infty (A_i \cap E_k)\right) = \sum_{i=1}^m a_i \sum_{k=1}^\infty \mu(A_i \cap E_k). \end{align*} The second equality uses $A_i \cap E = \bigsqcup_k (A_i \cap E_k)$ (disjoint since the $E_k$ are disjoint), and the third uses the countable additivity of $\mu$. Since $a_i \ge 0$ and $\mu(A_i \cap E_k) \ge 0$ for all $i, k$, all terms are non-negative. By Tonelli's theorem for non-negative double series (interchanging a finite sum with a countable sum of non-negative terms): \begin{align*} \nu(E) = \sum_{k=1}^\infty \sum_{i=1}^m a_i\,\mu(A_i \cap E_k) = \sum_{k=1}^\infty \nu(E_k). \end{align*} [guided] The set function $\nu(E) = \sum_i a_i\,\mu(A_i \cap E)$ is a finite non-negative linear combination of the measures $E \mapsto \mu(A_i \cap E)$. Each of these is itself a measure (the restriction of $\mu$ to $A_i$, extended by $\mu_{A_i}(E) := \mu(A_i \cap E)$), since $\mu$ is countably additive and intersection with $A_i$ preserves disjointness. A non-negative linear combination of measures is again a measure: $\nu(\varnothing) = 0$ is immediate, and countable additivity follows by interchanging the finite sum over $i$ with the countable sum over $k$, which is valid because all terms are non-negative. [/guided] [/step]