[proofplan] We represent each simple function in its standard form as a finite linear combination of indicators of measurable sets, then verify each closure property by showing the resulting function is again a finite linear combination of indicators of measurable sets. The key algebraic facts are: (i) products and differences of indicators correspond to intersections and set differences, (ii) $\max$ and $\min$ reduce to sums and absolute values via the identities $\max(a,b) = \frac{a+b+|a-b|}{2}$ and $\min(a,b) = \frac{a+b-|a-b|}{2}$, and (iii) any finite linear combination of indicators of measurable sets can be rewritten in standard form (disjoint supports). [/proofplan]

[step:Write $s$ and $t$ in compatible standard form]Since $s$ and $t$ are simple functions on $(X, \mathcal{A})$, they take finitely many distinct values. Write \begin{align*} s = \sum_{i=1}^m a_i \,\mathbb{1}_{A_i}, \quad t = \sum_{j=1}^n b_j \,\mathbb{1}_{B_j}, \end{align*} where $a_1, \ldots, a_m$ are the distinct values of $s$, $b_1, \ldots, b_n$ are the distinct values of $t$, the sets $A_i := s^{-1}(\{a_i\}) \in \mathcal{A}$ are pairwise disjoint with $\bigcup_{i=1}^m A_i = X$, and the sets $B_j := t^{-1}(\{b_j\}) \in \mathcal{A}$ are pairwise disjoint with $\bigcup_{j=1}^n B_j = X$. Define the common refinement $C_{ij} := A_i \cap B_j$ for $1 \le i \le m$, $1 \le j \le n$. The sets $\{C_{ij}\}$ are pairwise disjoint, each $C_{ij} \in \mathcal{A}$ (since $\mathcal{A}$ is closed under finite intersections), and $\bigcup_{i,j} C_{ij} = X$. On each $C_{ij}$, $s$ takes the constant value $a_i$ and $t$ takes the constant value $b_j$: \begin{align*} s = \sum_{i,j} a_i \,\mathbb{1}_{C_{ij}}, \quad t = \sum_{i,j} b_j \,\mathbb{1}_{C_{ij}}. \end{align*}[/step]

[guided]Why introduce the common refinement? The sets $A_i$ and $B_j$ are each partitions of $X$, but they need not be compatible. The refinement $\{C_{ij} = A_i \cap B_j\}$ creates a single partition on which both $s$ and $t$ are constant — this is the key step that allows us to perform pointwise algebraic operations by manipulating the coefficients.[/guided]

[step:Verify scalar multiple, sum, and product are simple]**Scalar multiple.** For $\alpha \in \mathbb{R}$, $\alpha s = \sum_{i=1}^m (\alpha a_i)\,\mathbb{1}_{A_i}$, a finite linear combination of indicators of measurable sets. Hence $\alpha s$ is simple. **Sum.** Using the common refinement: \begin{align*} s + t = \sum_{i,j} (a_i + b_j)\,\mathbb{1}_{C_{ij}}. \end{align*} This is a finite linear combination of indicators of measurable sets (the $C_{ij}$ are in $\mathcal{A}$), so $s + t$ is simple. Some values $a_i + b_j$ may coincide for different pairs $(i,j)$; to obtain the standard form, merge the corresponding sets $C_{ij}$. The resulting function still takes finitely many values, each on a measurable set. **Product.** Using the common refinement: \begin{align*} s \cdot t = \sum_{i,j} (a_i \cdot b_j)\,\mathbb{1}_{C_{ij}}. \end{align*} Again a finite linear combination of indicators of measurable sets. Hence $s \cdot t$ is simple.[/step]

[guided]Each operation reduces to replacing the coefficients while keeping the same measurable partition $\{C_{ij}\}$. The sum replaces $a_i, b_j$ by $a_i + b_j$; the product replaces them by $a_i \cdot b_j$. The result is a finite linear combination of indicators of sets in $\mathcal{A}$, which is the definition of a simple function. The representation may not be in standard form (some coefficients may coincide), but merging sets with equal coefficients produces the standard form.[/guided]

[step:Verify lattice operations and absolute value are simple]**Maximum.** On each $C_{ij}$, $\max(s, t)$ takes the constant value $\max(a_i, b_j)$: \begin{align*} \max(s, t) = \sum_{i,j} \max(a_i, b_j)\,\mathbb{1}_{C_{ij}}. \end{align*} This is a finite linear combination of indicators of measurable sets, so $\max(s,t)$ is simple. The same argument applies to $\min(s,t)$ with coefficients $\min(a_i, b_j)$. **Absolute value.** $|s| = \sum_{i=1}^m |a_i|\,\mathbb{1}_{A_i}$, a finite linear combination of indicators of measurable sets. Hence $|s|$ is simple. **Positive and negative parts.** $s^+ = \max(s, 0)$ and $s^- = \max(-s, 0)$. Since the constant function $0$ is simple (it equals $0 \cdot \mathbb{1}_X$), and we have shown $\max$ preserves simplicity and scalar multiples preserve simplicity ($-s$ is simple), both $s^+$ and $s^-$ are simple.[/step]

[guided]The lattice operations act pointwise on the coefficients: on each atom $C_{ij}$, $\max(s,t)$ selects the larger of $a_i$ and $b_j$. Since there are only finitely many atoms and each yields a real number, the result is a simple function. The positive and negative parts reduce to the already-established cases: $s^+ = \max(s, 0)$ uses $\max$ with the simple function $0$, and $s^- = \max(-s, 0)$ additionally uses the scalar multiple $-s$.[/guided]

[step:Conclude the algebraic and lattice structure] Closure under addition and scalar multiplication (parts 1--2) shows that the simple functions form a **vector space** over $\mathbb{R}$. Closure under pointwise multiplication (part 3) makes it a **real algebra**. Closure under $\max$, $\min$, and $|\cdot|$ (parts 4--6) makes it a **vector lattice**: for any simple $s, t$ and $\alpha \in \mathbb{R}$, \begin{align*} \max(\alpha s + t, \alpha s + t') = \alpha s + \max(t, t'), \end{align*} and $|s| = s^+ + s^-$, with the lattice operations compatible with the vector space structure. This completes the proof that the simple functions on $(X, \mathcal{A})$ form a real algebra and a vector lattice. [/step]