Proof of Separability of the Lebesgue Spaces of the Torus

Each result cited below is universally quantified over the data appearing in its own statement, and is applied to the data named here. The measure space is $(Q,\mathcal{B}_{Q},\lambda_{Q})$ of clause 3, with $\lambda_{Q}(Q)=1$. We record once, for nonnegative reals $a$ and $b$, that $a\le b$ implies $a^{2}\le b^{2}$ and conversely: claim 5 of Elementary Arithmetic in an Ordered Field, applied with the nonnegative multipliers $a$ and then $b$, gives $a\cdot a\le a\cdot b\le b\cdot b$ by transitivity; and if $a^{2}\le b^{2}$ but $b<a$, then $0<a$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, so $b\cdot b\le b\cdot a$ by claim 5 of Elementary Arithmetic in an Ordered Field and $b\cdot a<a\cdot a$ by claim 10 of Elementary Order Arithmetic in an Ordered Field, giving $b^{2}<a^{2}$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, a contradiction.

Claim 1. Finiteness of $N_{k}$. Put $Z_{k}=\{a\in\mathbb{Z}:0\le a\le2^{k}-1\}$, so that $N_{k}$ is exactly the set of $n$-tuples with all entries in $Z_{k}$. We show $Z_{k}$ is the image of the initial segment $[2^{k}]$ under the map $l\mapsto l-1$, natural numbers being read as real numbers through the canonical map as fixed in clause 1. If $l\in[2^{k}]$ then $l-1$ is an integer by claim 2 of Arithmetic, Order, Discreteness and Intervals of the Integers, and $1\le l\le2^{k}$ gives $0\le l-1\le2^{k}-1$, so $l-1\in Z_{k}$. Conversely let $a\in Z_{k}$. Then $a+1$ is an integer with $0<a+1$, so by claim 1 of Arithmetic, Order, Discreteness and Intervals of the Integers it is the image of a natural number $l$; from $a+1\le2^{k}$ and the strict monotonicity of the canonical map we get $l\le2^{k}$, so $l\in[2^{k}]$ and $a=l-1$. The set $[2^{k}]$ is finite by claim 1 of Basic Properties of Finite Sets, so $Z_{k}$ is finite by claim 4 there, and $N_{k}$ is finite by Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets.

The cells partition $Q$. Let $x\in Q$ and put $j_{i}=\lfloor2^{k}x_{i}\rfloor$ for $i\in[n]$, using the integer part, so that $j_{i}\le2^{k}x_{i}<j_{i}+1$ and hence $2^{-k}j_{i}\le x_{i}<2^{-k}(j_{i}+1)$, the number $2^{k}$ being positive. From $0\le x_{i}<1$ we get $0\le2^{k}x_{i}<2^{k}$; the left inequality gives $-1<j_{i}$ and hence $0\le j_{i}$, and the right gives $j_{i}<2^{k}$ and hence $j_{i}\le2^{k}-1$, both by claim 3 of Arithmetic, Order, Discreteness and Intervals of the Integers applied to integers. So $j\in N_{k}$ and $x\in Q_{k,j}$. If also $x\in Q_{k,j'}$ with $j'\in N_{k}$, then $j'_{i}\le2^{k}x_{i}<j'_{i}+1$ for every $i$, so $j'_{i}=\lfloor2^{k}x_{i}\rfloor=j_{i}$ by the uniqueness in Existence and Uniqueness of the Integer Part of a Real Number; thus the cells are pairwise disjoint and their union contains $Q$. Conversely if $j\in N_{k}$ and $x\in Q_{k,j}$ then $0\le2^{-k}j_{i}\le x_{i}<2^{-k}(j_{i}+1)\le2^{-k}2^{k}=1$ for every $i$, so $x\in Q$. Each $Q_{k,j}$ is the Borel rectangle whose $i$th factor is the interval $\{t:2^{-k}j_{i}\le t<2^{-k}(j_{i}+1)\}$, hence lies in $\mathcal{B}(\mathbb{R}^{n})$ by claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ and claim 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, and being contained in $Q$ it lies in $\mathcal{B}_{Q}$.

The diameter bound. If $x,y\in Q_{k,j}$ then for every $i$ both $x_{i}$ and $y_{i}$ lie between $2^{-k}j_{i}$ and $2^{-k}(j_{i}+1)$, so $|x_{i}-y_{i}|\le2^{-k}$, and hence $(x_{i}-y_{i})^{2}=|x_{i}-y_{i}|^{2}\le(2^{-k})^{2}$ by claim 4 of Properties of the Absolute Value in an Ordered Field and the squaring fact recorded above. Summing, $\lVert x-y\rVert^{2}\le n(2^{-k})^{2}=(\sigma_{n}2^{-k})^{2}$ by clause 3, and both $\lVert x-y\rVert$ and $\sigma_{n}2^{-k}$ are nonnegative, so $\lVert x-y\rVert\le\sigma_{n}2^{-k}$ by the converse half of that fact.

The correspondence. Since the cells $Q_{k,j}$, $j\in N_{k}$, are pairwise disjoint with union $Q$, a family $(c_{j})_{j\in N_{k}}$ in $\mathbb{Q}$ determines a well-defined map $s:Q\to\mathbb{R}$ with $s(x)=c_{j}$ for $x\in Q_{k,j}$, and this $s$ lies in $D_{k}$. Conversely each $Q_{k,j}$ is nonempty, containing the point whose $i$th coordinate is $2^{-k}j_{i}$, so a member of $D_{k}$ determines its family uniquely.

Claim 2. Let $s\in D_{k}$ with family $(c_{j})_{j\in N_{k}}$. For a real $c$ the set $\{x\in Q:c<s(x)\}$ is the union of those finitely many cells $Q_{k,j}$ with $c<c_{j}$, hence lies in $\mathcal{B}_{Q}$; so $s$ is measurable, and it takes only finitely many values, so it is a simple function. Let $C$ be the greatest element of the finite set $\{|c_{j}|:j\in N_{k}\}$. Then $|s|\le C$ on $Q$, so the integral of $|s|^{p}$ over $Q$ is at most $C^{p}\lambda_{Q}(Q)=C^{p}<\infty$, by Properties of Real Powers (v), The Integral of an Indicator Function is the Measure of the Set and claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion; hence $s\in\mathcal{L}^{p}(\mathbb{T}^{n})$ by Power-Integrable Functions and the p-Seminorm (ii).

For countability, fix $k$ and let $r$ be the number of elements of the finite set $N_{k}$, so that there is a bijection from the initial segment $[r]$ onto $N_{k}$ by the definition of the number of elements. Composing a family $(c_{j})_{j\in N_{k}}$ with this bijection identifies such families with $r$-tuples in $\mathbb{Q}$, that is, with members of $\mathbb{Q}^{r}$. The set $\mathbb{Q}$ is countable by claim 2 of The Integers and the Rational Numbers are Countable, so $\mathbb{Q}^{r}$ is countable by claim 2 of Products and Powers of Countable Sets. By claim 1 the map sending a family to the member of $D_{k}$ it determines is a surjection of $\mathbb{Q}^{r}$ onto $D_{k}$, so $D_{k}$ is countable by claim 4 of Basic Properties of Countable Sets. Hence $D=\bigcup_{k\in\mathbb{N}}D_{k}$ is countable by A Countable Union of Countable Sets is Countable, and $\{[s]:s\in D\}$, being the image of $D$ under $s\mapsto[s]$, is countable by claim 4 of Basic Properties of Countable Sets.

Claim 3. Let $f\in\mathcal{L}^{p}(\mathbb{T}^{n})$ and let $\varepsilon$ be a positive real.

By Continuous Periodic Functions are Power-Integrable and Dense on the Torus (iv) there is $u\in C_{\mathrm{per}}$ with $\lVert[f]-[u|_{Q}]\rVert_{L^{p}(\mathbb{T}^{n})}\le\varepsilon/2$. The set $\overline{Q}$ is compact by The Half-Open Unit Cell Tiles Euclidean Space (i), and $u$ is continuous on $\overline{Q}$ relative to $\overline{Q}$, the $\delta$ furnished by continuity on $\mathbb{R}^{n}$ serving verbatim in Continuous Map Between Metric Spaces. So by Heine-Cantor Theorem: Continuity on a Compact Metric Space or a Compact Subset Implies Uniform Continuity there is a positive real $\delta$ such that $|u(x)-u(y)|<\varepsilon/4$ whenever $x,y\in\overline{Q}$ and $d_{E}(x,y)<\delta$.

Choose $k\in\mathbb{N}$ with $\sigma_{n}2^{-k}<\delta$, as follows. First, $k\le2^{k}$ for every $k\in\mathbb{N}$, by induction over the natural numbers, which begin at $1$: the base case reads $1\le2$; and if $k\le2^{k}$ then, since $1\le2^{k}$, we get $k+1\le2^{k}+2^{k}=2^{k+1}$. The product $k\,2^{k}$ is positive by claim 5 of Elementary Order Arithmetic in an Ordered Field. Multiplying the inequality $k\le2^{k}$ by $(k\,2^{k})^{-1}$, which is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, and using claim 5 of Elementary Arithmetic in an Ordered Field, gives $2^{-k}\le1/k$. By The Archimedean Property of the Real Numbers there is $k\in\mathbb{N}$ with $\sigma_{n}<k\delta$, and then $\sigma_{n}2^{-k}\le\sigma_{n}/k<\delta$.

For each $j\in N_{k}$ let $y^{j}$ be the point of $\mathbb{R}^{n}$ whose $i$th coordinate is $2^{-k}j_{i}$; it lies in $Q_{k,j}\subseteq Q\subseteq\overline{Q}$. By claim 1 of The Rational Numbers are Dense in the Real Numbers there is a rational $c_{j}$ with $u(y^{j})-\varepsilon/4<c_{j}<u(y^{j})+\varepsilon/4$, hence $|c_{j}-u(y^{j})|<\varepsilon/4$ by claim 9 of Properties of the Absolute Value in an Ordered Field; the choice of one such $c_{j}$ for each of the finitely many $j$ is a finite choice. Let $s\in D_{k}$ be the member determined by the family $(c_{j})_{j\in N_{k}}$, as in claim 1.

Let $x\in Q$ and let $j\in N_{k}$ be the index with $x\in Q_{k,j}$. Both $x$ and $y^{j}$ lie in $Q_{k,j}$, so $d_{E}(x,y^{j})=\lVert x-y^{j}\rVert\le\sigma_{n}2^{-k}<\delta$ by claim 1, and both lie in $\overline{Q}$; hence $|u(x)-u(y^{j})|<\varepsilon/4$ and

$$|u(x)-s(x)|=|u(x)-c_{j}|\le|u(x)-u(y^{j})|+|u(y^{j})-c_{j}|<\frac{\varepsilon}{4}+\frac{\varepsilon}{4}=\frac{\varepsilon}{2},$$

by claim 5 of Properties of the Absolute Value in an Ordered Field. Therefore $|u|_{Q}-s|\le\varepsilon/2$ pointwise on $Q$, and, by the same three results used in claim 2, the integral of $|u|_{Q}-s|^{p}$ over $Q$ is at most $(\varepsilon/2)^{p}\lambda_{Q}(Q)=(\varepsilon/2)^{p}$, so $\lVert u|_{Q}-s\rVert_{p}\le\bigl((\varepsilon/2)^{p}\bigr)^{1/p}=\varepsilon/2$ by Properties of Real Powers (v) and Properties of Real Powers (vi). Finally, by Minkowski's Inequality (i) and The Lebesgue Space Lᵖ(μ) (iii),

$$\bigl\lVert[f]-[s]\bigr\rVert_{L^{p}(\mathbb{T}^{n})}\le\bigl\lVert[f]-[u|_{Q}]\bigr\rVert_{L^{p}(\mathbb{T}^{n})}+\bigl\lVert u|_{Q}-s\bigr\rVert_{p}\le\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .$$

Claim 4. Let $F\in L^{p}(\mathbb{T}^{n})$ and choose a representative $f$ with $F=[f]$. For each $k\in\mathbb{N}$, claim 3 applied with $\varepsilon=1/k$ provides $s_{k}\in D$ with $\lVert F-[s_{k}]\rVert_{L^{p}(\mathbb{T}^{n})}\le1/k$; the choice of one such $s_{k}$ for each $k$ uses countable choice. Given a positive real $\varepsilon$, The Archimedean Property of the Real Numbers provides $k_{0}\in\mathbb{N}$ with $1<k_{0}\varepsilon$, and then $\lVert F-[s_{k}]\rVert_{L^{p}(\mathbb{T}^{n})}<\varepsilon$ for every $k\ge k_{0}$. So $([s_{k}])_{k\in\mathbb{N}}$ converges to $F$ in the metric of $L^{p}(\mathbb{T}^{n})$, and $F$ lies in the closure of $\{[s]:s\in D\}$ by Sequential Characterization of the Closure in a Metric Space. As $F$ was arbitrary, that set is dense. Being countable by claim 2, it witnesses that $L^{p}(\mathbb{T}^{n})$ is separable.