[proofplan] The strategy is to reduce all claims to the measurability of the supremum, which in turn reduces to the preimage characterisation of measurability. For the supremum of a countable family, the preimage $\{\sup_k f_k > a\}$ is the countable union $\bigcup_k \{f_k > a\}$, which lies in $\mathcal{F}$ since $\sigma$-algebras are closed under countable unions. The infimum follows by negation, the limsup and liminf by iterating suprema and infima over countable index sets, and the pointwise limit is the common value of limsup and liminf. [/proofplan] [step:Express the superlevel set of the supremum as a countable union] Define $g: X \to \overline{\mathbb{R}}$ by $g(x) := \sup_{k \ge 1} f_k(x)$. For any $a \in \mathbb{R}$, we have $g(x) > a$ if and only if $f_k(x) > a$ for at least one index $k \ge 1$. Therefore \begin{align*} \{x \in X : g(x) > a\} = \bigcup_{k=1}^{\infty} \{x \in X : f_k(x) > a\}. \end{align*} Each set $\{f_k > a\}$ belongs to $\mathcal{F}$ because $f_k$ is $\mathcal{F}$-measurable. Since $\mathcal{F}$ is a $\sigma$-algebra, it is closed under countable unions, so $\{g > a\} \in \mathcal{F}$ for every $a \in \mathbb{R}$. By the [Generator Criterion for Measurability](/theorems/525), a function into $\overline{\mathbb{R}}$ is $\mathcal{F}$-measurable if and only if $\{g > a\} \in \mathcal{F}$ for every $a \in \mathbb{R}$. (The sets $(a, \infty]$ for $a \in \mathbb{R}$ generate the Borel $\sigma$-algebra on $\overline{\mathbb{R}}$.) We have verified this condition, so $g = \sup_{k \ge 1} f_k$ is $\mathcal{F}$-measurable. [guided] The key observation is that the supremum "sees" its value exceed a threshold $a$ precisely when at least one term in the family exceeds $a$. This converts the superlevel set of the supremum into a countable union — the operation that $\sigma$-algebras are designed to handle. Define $g: X \to \overline{\mathbb{R}}$ by $g(x) := \sup_{k \ge 1} f_k(x)$. Fix any $a \in \mathbb{R}$. Then $g(x) > a$ means $\sup_{k \ge 1} f_k(x) > a$, which by the definition of the supremum holds if and only if there exists $k \ge 1$ with $f_k(x) > a$. This is exactly the condition $x \in \bigcup_{k=1}^{\infty} \{f_k > a\}$, so \begin{align*} \{x \in X : g(x) > a\} = \bigcup_{k=1}^{\infty} \{x \in X : f_k(x) > a\}. \end{align*} Each $f_k$ is $\mathcal{F}$-measurable, so $\{f_k > a\} \in \mathcal{F}$. A $\sigma$-algebra is closed under countable unions, hence $\{g > a\} \in \mathcal{F}$. By the [Generator Criterion for Measurability](/theorems/525), to verify $\mathcal{F}$-measurability of a function $g: X \to \overline{\mathbb{R}}$, it suffices to check that $\{g > a\} \in \mathcal{F}$ for every $a \in \mathbb{R}$. (The half-lines $(a, \infty]$ generate $\mathcal{B}(\overline{\mathbb{R}})$.) We have verified precisely this condition, so $g = \sup_{k \ge 1} f_k$ is $\mathcal{F}$-measurable. This establishes part (i) for the supremum. [/guided] [/step] [step:Deduce measurability of the infimum by negation] For the infimum, write $\inf_{k \ge 1} f_k = -\sup_{k \ge 1}(-f_k)$. Each $-f_k$ is $\mathcal{F}$-measurable (since $\{-f_k > a\} = \{f_k < -a\} = X \setminus \{f_k \ge -a\} \in \mathcal{F}$). By the result of the previous step, $\sup_{k \ge 1}(-f_k)$ is $\mathcal{F}$-measurable. Multiplying a measurable function by $-1$ preserves measurability (by the same preimage argument), so $\inf_{k \ge 1} f_k$ is $\mathcal{F}$-measurable. This completes part (i). [/step] [step:Express limsup and liminf as iterated suprema and infima over countable indices] For part (ii), recall the definitions: \begin{align*} \limsup_{k \to \infty} f_k(x) &= \inf_{m \ge 1}\, \sup_{k \ge m}\, f_k(x), \\ \liminf_{k \to \infty} f_k(x) &= \sup_{m \ge 1}\, \inf_{k \ge m}\, f_k(x). \end{align*} For the limsup: for each fixed $m \ge 1$, the function $g_m: X \to \overline{\mathbb{R}}$ defined by $g_m(x) := \sup_{k \ge m} f_k(x)$ is $\mathcal{F}$-measurable by the first step (applied to the countable family $(f_k)_{k \ge m}$). Then $\limsup_{k \to \infty} f_k = \inf_{m \ge 1} g_m$ is the infimum of a countable family of measurable functions, hence $\mathcal{F}$-measurable by the second step. The argument for the liminf is identical with the roles of supremum and infimum exchanged: $h_m(x) := \inf_{k \ge m} f_k(x)$ is measurable for each $m$, and $\liminf_{k \to \infty} f_k = \sup_{m \ge 1} h_m$ is measurable. [/step] [step:Conclude measurability of the pointwise limit] For part (iii), suppose $f(x) = \lim_{k \to \infty} f_k(x)$ exists in $\overline{\mathbb{R}}$ for every $x \in X$. When a limit exists, it equals both the limsup and the liminf: \begin{align*} f(x) = \limsup_{k \to \infty} f_k(x) = \liminf_{k \to \infty} f_k(x) \quad \text{for every } x \in X. \end{align*} Since the limsup is $\mathcal{F}$-measurable by part (ii), $f$ is $\mathcal{F}$-measurable. [/step]