[proofplan] Each algebraic operation is reduced to the preimage characterisation of measurability via the [Generator Criterion for Measurability](/theorems/525). Sums are handled by writing $\{f + g < a\}$ as a countable union over rationals; products are deduced from the identity $fg = \frac{1}{4}[(f+g)^2 - (f-g)^2]$ once we establish that squares of measurable functions are measurable; and $\max$/$\min$ are expressed as preimage intersections and unions that reduce to those of $f$ and $g$ individually. [/proofplan]

[step:Establish measurability of scalar multiples and the absolute value] **(Part (i), scalar multiples.)** Let $c \in \mathbb{R}$. If $c = 0$, then $cf$ is the constant function $0$, which is measurable since $\{0 > a\}$ is either $\varnothing$ or $X$, both in $\mathcal{F}$. If $c > 0$, then for every $a \in \mathbb{R}$, \begin{align*} \{x \in X : cf(x) > a\} = \{x \in X : f(x) > a/c\} \in \mathcal{F}, \end{align*} since $f$ is $\mathcal{F}$-measurable. If $c < 0$, then \begin{align*} \{x \in X : cf(x) > a\} = \{x \in X : f(x) < a/c\} = X \setminus \{x \in X : f(x) \ge a/c\} \in \mathcal{F}, \end{align*} where the complement lies in $\mathcal{F}$ because $\{f \ge a/c\} = \bigcap_{m=1}^{\infty}\{f > a/c - 1/m\} \in \mathcal{F}$. By the [Generator Criterion for Measurability](/theorems/525), $cf$ is $\mathcal{F}$-measurable. **(Part (iv), absolute value.)** For every $a \in \mathbb{R}$: if $a < 0$, then $\{|f| > a\} = X \in \mathcal{F}$. If $a \ge 0$, then \begin{align*} \{x \in X : |f(x)| > a\} = \{x \in X : f(x) > a\} \cup \{x \in X : f(x) < -a\} \in \mathcal{F}, \end{align*} since both sets on the right belong to $\mathcal{F}$ (the second because $\{f < -a\} = X \setminus \{f \ge -a\} \in \mathcal{F}$ by the same countable intersection argument). By the generator criterion, $|f|$ is $\mathcal{F}$-measurable. [/step]

[step:Prove the sum $f + g$ is measurable using rational approximation]**(Part (i), sums.)** Fix $a \in \mathbb{R}$. Then $f(x) + g(x) < a$ if and only if $f(x) < a - g(x)$, which holds if and only if there exists a rational number $q \in \mathbb{Q}$ with $f(x) < q < a - g(x)$. (This uses the density of $\mathbb{Q}$ in $\mathbb{R}$: the open interval $(f(x), a - g(x))$ is nonempty, hence contains a rational.) Equivalently: \begin{align*} \{x \in X : f(x) + g(x) < a\} = \bigcup_{q \in \mathbb{Q}} \bigl(\{x \in X : f(x) < q\} \cap \{x \in X : g(x) < a - q\}\bigr). \end{align*} For each $q \in \mathbb{Q}$, both $\{f < q\}$ and $\{g < a - q\}$ lie in $\mathcal{F}$ (since $f$ and $g$ are $\mathcal{F}$-measurable), so their intersection lies in $\mathcal{F}$. Since $\mathbb{Q}$ is countable, the union is a countable union of $\mathcal{F}$-sets, hence belongs to $\mathcal{F}$. By the [Generator Criterion for Measurability](/theorems/525) (using the generating family $\{(-\infty, a) : a \in \mathbb{R}\}$), $f + g$ is $\mathcal{F}$-measurable. The difference $f - g = f + (-1)g$ is measurable because $-g = (-1)g$ is measurable by the scalar multiple case, and $f + (-g)$ is measurable by the sum case.[/step]

[guided]The difficulty for sums is that the superlevel set $\{f + g > a\}$ cannot be decomposed as a simple Boolean combination of superlevel sets of $f$ and $g$ separately — the values of $f$ and $g$ are coupled at each point. The trick is to decouple them by inserting a rational "separator." Fix $a \in \mathbb{R}$. We claim \begin{align*} \{x \in X : f(x) + g(x) < a\} = \bigcup_{q \in \mathbb{Q}} \bigl(\{x \in X : f(x) < q\} \cap \{x \in X : g(x) < a - q\}\bigr). \end{align*} **Forward inclusion ($\subset$).** Suppose $f(x) + g(x) < a$. Then $f(x) < a - g(x)$, so the open interval $(f(x),\, a - g(x))$ is nonempty. By the density of $\mathbb{Q}$ in $\mathbb{R}$, there exists $q \in \mathbb{Q}$ with $f(x) < q < a - g(x)$. The first inequality gives $x \in \{f < q\}$ and the second gives $g(x) < a - q$, i.e., $x \in \{g < a - q\}$. So $x$ belongs to the union on the right. **Reverse inclusion ($\supset$).** If $x \in \{f < q\} \cap \{g < a - q\}$ for some $q \in \mathbb{Q}$, then $f(x) + g(x) < q + (a - q) = a$. Now, for each fixed $q \in \mathbb{Q}$, the set $\{f < q\} \cap \{g < a - q\}$ is the intersection of two $\mathcal{F}$-sets (using measurability of $f$ and $g$), hence lies in $\mathcal{F}$. Since $\mathbb{Q}$ is countable, the union $\bigcup_{q \in \mathbb{Q}}(\cdots)$ is a countable union of $\mathcal{F}$-sets, so $\{f + g < a\} \in \mathcal{F}$. By the [Generator Criterion for Measurability](/theorems/525), $f + g$ is $\mathcal{F}$-measurable.[/guided]

[step:Prove the product $fg$ is measurable via the polarisation identity] **(Part (ii).)** We first show that $h^2$ is $\mathcal{F}$-measurable whenever $h: X \to \mathbb{R}$ is $\mathcal{F}$-measurable. For $a < 0$, $\{h^2 > a\} = X \in \mathcal{F}$. For $a \ge 0$, \begin{align*} \{x \in X : h(x)^2 > a\} = \{x \in X : h(x) > \sqrt{a}\} \cup \{x \in X : h(x) < -\sqrt{a}\} \in \mathcal{F}, \end{align*} since both sets belong to $\mathcal{F}$. By the generator criterion, $h^2$ is measurable. Now apply the polarisation identity: \begin{align*} f(x)\, g(x) = \frac{1}{4}\bigl[(f(x) + g(x))^2 - (f(x) - g(x))^2\bigr]. \end{align*} The functions $f + g$ and $f - g$ are $\mathcal{F}$-measurable by the previous step. Squaring preserves measurability (shown above). Scalar multiplication by $\frac{1}{4}$ and subtraction preserve measurability by the first step and the sum case. Therefore $fg$ is $\mathcal{F}$-measurable. [/step]

[step:Prove the quotient $f/g$ is measurable on $\{g \neq 0\}$]**(Part (iii).)** Assume $g(x) \neq 0$ for all $x \in X$. We show $1/g$ is $\mathcal{F}$-measurable, after which $f/g = f \cdot (1/g)$ is measurable by the product case. Fix $a \in \mathbb{R}$. We compute $\{1/g > a\}$ by case analysis on the sign of $g$: \begin{align*} \{x \in X : 1/g(x) > a\} = \begin{cases} \{g > 0\} \cap \{g < 1/a\} & \text{if } a > 0, \\ \{g > 0\} \cup \{g < 1/a\} & \text{if } a < 0, \\ \{g > 0\} & \text{if } a = 0. \end{cases} \end{align*} In each case, the set is a finite Boolean combination of sets of the form $\{g > c\}$ and $\{g < c\}$, each of which lies in $\mathcal{F}$. Hence $\{1/g > a\} \in \mathcal{F}$ for every $a \in \mathbb{R}$, and $1/g$ is $\mathcal{F}$-measurable by the generator criterion.[/step]

[guided]To verify the case $a > 0$: $1/g(x) > a > 0$ forces $g(x) > 0$ (since $1/g(x)$ is positive). Given $g(x) > 0$, the inequality $1/g(x) > a$ is equivalent to $g(x) < 1/a$. So $\{1/g > a\} = \{g > 0\} \cap \{g < 1/a\}$. For $a < 0$: $1/g(x) > a$ holds whenever $g(x) > 0$ (since then $1/g(x) > 0 > a$). When $g(x) < 0$, the inequality $1/g(x) > a$ reverses to $g(x) < 1/a$ (since multiplying by the negative quantity $g(x)$ flips the inequality). So $\{1/g > a\} = \{g > 0\} \cup (\{g < 0\} \cap \{g < 1/a\})$. Since $a < 0$, we have $1/a < 0$, so $\{g < 0\} \cap \{g < 1/a\} = \{g < 1/a\}$. Hence $\{1/g > a\} = \{g > 0\} \cup \{g < 1/a\}$. For $a = 0$: $1/g(x) > 0$ if and only if $g(x) > 0$, so $\{1/g > 0\} = \{g > 0\}$. All resulting sets are finite intersections, unions, and complements of $\mathcal{F}$-sets, hence lie in $\mathcal{F}$.[/guided]

[step:Prove measurability of $\max(f,g)$, $\min(f,g)$, $f^+$, and $f^-$] **(Parts (v) and (vi).)** For $\max(f, g)$: for any $a \in \mathbb{R}$, \begin{align*} \{x \in X : \max(f(x), g(x)) > a\} = \{x \in X : f(x) > a\} \cup \{x \in X : g(x) > a\} \in \mathcal{F}, \end{align*} since $\max(f(x), g(x)) > a$ if and only if at least one of $f(x), g(x)$ exceeds $a$. By the generator criterion, $\max(f, g)$ is $\mathcal{F}$-measurable. For $\min(f, g)$: $\min(f(x), g(x)) > a$ if and only if both $f(x) > a$ and $g(x) > a$, so \begin{align*} \{x \in X : \min(f(x), g(x)) > a\} = \{x \in X : f(x) > a\} \cap \{x \in X : g(x) > a\} \in \mathcal{F}. \end{align*} For $f^+ = \max(f, 0)$: the constant function $0$ is measurable, so $\max(f, 0)$ is measurable by the case just proved. Similarly, $f^- = \max(-f, 0)$ is measurable since $-f$ is measurable by the scalar multiple case and $\max(-f, 0)$ is measurable by the same argument. [/step]