Proof of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions

Throughout, $(X,\mathcal{A})$ is the given measurable space, and measurability of a real-valued function on $X$ may be checked on rays: a function $h:X\to\mathbb{R}$ is measurable if and only if $\{x\in X:h(x)>a\}\in\mathcal{A}$ for every real number $a$, by claim 3 of the generator lemma for the Borel $\sigma$-algebra of the real line. We use both descriptions, and we use freely that $\mathcal{A}$ contains $\emptyset$ and $X$ and is closed under complements, countable unions, and countable intersections (the last because a countable intersection is the complement of the countable union of the complements), by the axioms of a $\sigma$-algebra.

Claim 1. Let $h$ be the constant function $x\mapsto c$ and let $B$ be a member of the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$. Then $h^{-1}(B)$ equals $X$ when $c\in B$ and $\emptyset$ otherwise, and both belong to $\mathcal{A}$; so $h$ is measurable. For $h=\mathbf{1}_A$ with $A\in\mathcal{A}$: according to which of the values $1$ and $0$ belong to $B$, the preimage $h^{-1}(B)$ is $X$ (both), $A$ (only $1$), $X\setminus A$ (only $0$), or $\emptyset$ (neither); all four belong to $\mathcal{A}$.

Claims 2 and 3. Consider the maps $s,p:\mathbb{R}^2\to\mathbb{R}$ on Euclidean space $\mathbb{R}^2$ defined by $s(u,v)=u+v$ and $p(u,v)=uv$. Both are sequentially continuous in the sense of the composition lemma: let $(u_n,v_n)_{n\in\mathbb{N}}$ be a sequence in $\mathbb{R}^2$ whose Euclidean distance to a point $(u,v)$ converges to $0$. Each coordinate of a vector of $\mathbb{R}^2$ is bounded in absolute value by the Euclidean norm (claim 4 of the norm properties), and the Euclidean distance of two vectors is the norm of their difference, so $|u_n-u|$ and $|v_n-v|$ are bounded by the distance from $(u_n,v_n)$ to $(u,v)$; hence $u_n\to u$ and $v_n\to v$ by domination by a null sequence (claim 3 of Order Properties of Limits of Real Sequences). Claims 1 and 2 of the arithmetic of limits now give $s(u_n,v_n)=u_n+v_n\to u+v=s(u,v)$ and $p(u_n,v_n)=u_nv_n\to uv=p(u,v)$. The pair map $(f,g):X\to\mathbb{R}^2$ has measurable components, so the compositions $f+g=s\circ(f,g)$ and $fg=p\circ(f,g)$ are measurable by the composition lemma, applied with $d=2$ and $E=\mathbb{R}^2$. The multiple $cf$ is the product of the constant function $c$ (claim 1, measurable) with $f$, hence measurable. Finally we induct on the natural number $n$: for $n=1$, $c_1f_1$ is measurable as just shown, and $f_1$ is measurable by hypothesis; if the claims hold for $n$, then $c_1f_1+\cdots+c_{n+1}f_{n+1}=(c_1f_1+\cdots+c_nf_n)+c_{n+1}f_{n+1}$ is measurable as the sum of two measurable functions, and $f_1\cdots f_{n+1}=(f_1\cdots f_n)\,f_{n+1}$ is measurable as the product of two.

Claim 4. The map $q:\mathbb{R}\to\mathbb{R}$, $q(u)=|u|$, on $\mathbb{R}=\mathbb{R}^1$ is sequentially continuous: if the Euclidean distance from $u_n$ to $u$ converges to $0$ then $u_n\to u$, since on $\mathbb{R}^1$ the Euclidean distance agrees with the absolute-value distance $|u_n-u|$ (comparison of the two metrics on the real line), and then $|u_n|\to|u|$ by claim 4 of Order Properties of Limits of Real Sequences. Hence $|h|=q\circ h$ is measurable for every measurable real-valued $h$ on $X$, by the composition lemma with $d=1$; in particular $|f|$ and $|f-g|$ are measurable, $f-g=f+(-1)g$ being measurable by claim 2. At every point $x$,

when $f(x)\ge g(x)$ the right-hand side is $\tfrac12\big(f(x)+g(x)+f(x)-g(x)\big)=f(x)$, and when $g(x)>f(x)$ it is $g(x)$ by the symmetric computation. So $\max(f,g)=\tfrac12\big(f+g+|f-g|\big)$ is measurable by claim 2, and $\min(f,g)=-\max(-f,-g)$ (at each point, $-\max(-f(x),-g(x))$ is the smaller of $f(x)$ and $g(x)$) is measurable likewise. The description of $|f|$ in the statement agrees with this: $|f(x)|=\max(f(x),-f(x))$.

Claim 5. Fix a real number $a$. We claim

the intersection running over the natural numbers $n\ge N$; write $S$ for the right-hand side.

First let $f(x)>a$. By claim 3 of the Archimedean property there is a natural number $m$ with $\tfrac1m<\tfrac{f(x)-a}{2}$, so that $f(x)-\tfrac1m>a+\tfrac1m$. Since $(f_n(x))_{n}$ converges to $f(x)$, the definition of the limit provides an $N$ with $|f_n(x)-f(x)|<\tfrac1m$ for all $n\ge N$; for such $n$, $f_n(x)>f(x)-\tfrac1m>a+\tfrac1m$. Hence $x\in S$.

Conversely let $x\in S$, with witnesses $m$ and $N$: $f_n(x)>a+\tfrac1m$ for all $n\ge N$. The shifted sequence $(f_{N+j}(x))_{j\in\mathbb{N}}$ converges to $f(x)$ directly from the definition of the limit (any threshold index for the original sequence serves, shifted, for the shifted one). Comparing it with the constant sequence with value $a+\tfrac1m$, which converges to $a+\tfrac1m$ again directly from the definition, claim 1 of Order Properties of Limits of Real Sequences gives $a+\tfrac1m\le f(x)$; hence $f(x)>a$, so $x\in\{x\in X:f(x)>a\}$.

Each set $\{x\in X:f_n(x)>a+\tfrac1m\}$ belongs to $\mathcal{A}$, being the preimage under the measurable $f_n$ of the ray $\{u\in\mathbb{R}:u>a+\tfrac1m\}$, a member of $\mathcal{B}(\mathbb{R})$ by claim 2 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line. The set $S$ is built from these sets by countably many intersections and unions ($m$, $N$, and $n$ ranging over the natural numbers), so $S\in\mathcal{A}$. As $a$ was arbitrary, $f$ is measurable by the ray criterion recalled at the outset. $\blacksquare$