Proof of Measurability of Countable Suprema, Bounded Pointwise Limits, Monotone Functions, and Continuous Functions

Throughout we use the criterion recorded with the definition of measurability: a map $v:X\to\mathbb{R}$ is measurable if and only if $\{x\in X:v(x)>c\}\in\mathcal{F}$ for every real number $c$.

Claim 1. Fix $x\in X$. The set $\{f_n(x):n\in\mathbb{N}\}$ is nonempty and bounded above by $K$ and below by $-K$, so by the least upper bound property of the real numbers it has a real supremum $g(x)$ and a real infimum $h(x)$, and $|g(x)|\le K$, $|h(x)|\le K$.

Fix a real number $c$. If $g(x)>c$ then $c$ is not an upper bound of $\{f_n(x):n\in\mathbb{N}\}$, so $f_n(x)>c$ for some $n$; conversely, if $f_n(x)>c$ for some $n$ then $g(x)\ge f_n(x)>c$. Therefore

$$\{x\in X:g(x)>c\}=\bigcup_{n\in\mathbb{N}}\{x\in X:f_n(x)>c\},$$

a countable union of members of $\mathcal{F}$, which lies in $\mathcal{F}$ because $\mathcal{F}$ is a $\sigma$-algebra. Hence $g$ is measurable.

Next, each $-f_n$ is measurable: for a real number $c$,

$$\{x\in X:-f_n(x)>c\}=\{x\in X:f_n(x)<-c\}=\bigcup_{k\in\mathbb{N}}\Big(X\setminus\{x\in X:f_n(x)>-c-\tfrac{1}{k}\}\Big),$$

since $f_n(x)<-c$ holds if and only if $f_n(x)\le-c-\frac{1}{k}$ for some natural number $k$; each set on the right lies in $\mathcal{F}$, and $\mathcal{F}$ is closed under complements and countable unions. The maps $-f_n$ are bounded in absolute value by $K$, so by the paragraph above $\sup_n(-f_n)$ is measurable. For any nonempty set of real numbers bounded above and below, the infimum of the set equals the negative of the supremum of the set of negatives, so $h=-\sup_n(-f_n)$; applying the negation argument once more, $h$ is measurable.

Claim 2. For each natural number $n$ put $g_n(x)=\sup_{k\ge n}f_k(x)$. Claim 1, applied to the family $(f_{n+j})_{j\in\mathbb{N}}$, which is measurable and bounded in absolute value by $K$, shows that $g_n$ is real-valued and measurable with $|g_n|\le K$.

Fix $x\in X$. Since $g_n(x)\ge f_k(x)$ for every $k\ge n$, and $f_k(x)\to f(x)$, passing to the limit in $k$ gives $g_n(x)\ge f(x)$ for every $n$. Let $\varepsilon>0$ and choose $N$ with $|f_k(x)-f(x)|<\varepsilon$ for all $k\ge N$; then $f_k(x)\le f(x)+\varepsilon$ for $k\ge N$, so $g_N(x)\le f(x)+\varepsilon$. Consequently $f(x)\le\inf_n g_n(x)\le f(x)+\varepsilon$ for every $\varepsilon>0$, whence $f(x)=\inf_n g_n(x)$. By Claim 1 applied to the sequence $(g_n)$, the map $f=\inf_n g_n$ is measurable.

Claim 3. Fix a real number $c$ and set $E=\{t\in[a,b]:u(t)>c\}$. If $t\in E$ and $t\le t'\le b$, then $u(t')\ge u(t)>c$, so $t'\in E$. If $E$ is empty it is a Borel subset of $[a,b]$. Otherwise $E$ is nonempty and bounded below by $a$, so it has a real infimum $s\in[a,b]$, and the implication just proved gives

$$(s,b]\subseteq E\subseteq[s,b].$$

Indeed, if $s<t'\le b$ then $t'$ is not a lower bound of $E$, so some $t\in E$ satisfies $t<t'$, and hence $t'\in E$; and every element of $E$ is at least $s$ and at most $b$. Therefore $E$ equals $(s,b]$ or $[s,b]$, and in either case $E$ is the intersection of $[a,b]$ with an interval of the real line, hence a member of the trace Borel $\sigma$-algebra on $[a,b]$. By the criterion recalled at the start, $u$ is measurable.

Claim 4. Let $u:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$, the interval regarded as a subset of the real line with the absolute value metric, let $c$ be a real number, and set $U=\{t\in[a,b]:u(t)>c\}$. Let $t_0\in U$. By continuity at $t_0$ relative to $[a,b]$, applied with the positive number $u(t_0)-c$ in the role of $\varepsilon$, there is $\delta>0$ such that $|u(t)-u(t_0)|<u(t_0)-c$, and hence $u(t)>c$, for every $t\in[a,b]$ with $|t-t_0|<\delta$. Choosing rational numbers $p,q$ with $t_0-\delta<p<t_0<q<t_0+\delta$ gives $t_0\in(p,q)\cap[a,b]\subseteq U$. Therefore

$$U=\bigcup\Big\{(p,q)\cap[a,b]\ :\ p,q\text{ rational},\ (p,q)\cap[a,b]\subseteq U\Big\},$$

a union of a countable collection of members of the trace Borel $\sigma$-algebra, hence itself a member. By the criterion recalled at the start, $u$ is measurable. $\blacksquare$