[proofplan] We encode the set $A$ as a point $x_A$ in the two-sided symbolic space $\{0,1\}^{\mathbb{Z}}$ and take the orbit closure under the shift. Along intervals on which the density of $A$ approaches $\overline d(A)$, we average Dirac masses over the orbit of $x_A$ and pass to a weak subsequential limit. The endpoint error in the empirical averages proves that the limiting measure is shift-invariant, and cylinder sets in the symbolic system exactly record finite intersections of shifts of $A$. Passing to the limit gives the correspondence inequality, while the one-coordinate cylinder gives $\mu(B)=\overline d(A)$. [/proofplan] [step:Choose density-realising intervals and build the symbolic orbit closure] Set \begin{align*} d := \overline d(A). \end{align*} By the definition of the limit superior, choose a strictly increasing sequence $(N_j)_{j=1}^{\infty}$ in $\mathbb{N}$ such that \begin{align*} \lim_{j \to \infty} \frac{|A \cap \{1,\dots,N_j\}|}{N_j} = d. \end{align*} Let $\Omega := \{0,1\}^{\mathbb{Z}}$ with the [product topology](/page/Product%20Topology) and its Borel $\sigma$-algebra $\mathcal{B}(\Omega)$. Define the shift map \begin{align*} \sigma: \Omega &\to \Omega \\ x &\mapsto \sigma x, \end{align*} where \begin{align*} (\sigma x)(m) := x(m+1) \end{align*} for every $x \in \Omega$ and every $m \in \mathbb{Z}$. The map $\sigma$ is a homeomorphism, with inverse $\sigma^{-1}$ given by $(\sigma^{-1}x)(m)=x(m-1)$. Define the indicator sequence of $A$ as the map \begin{align*} x_A: \mathbb{Z} &\to \{0,1\} \\ m &\mapsto x_A(m), \end{align*} where $x_A(m)=1$ if $m \in A$ and $x_A(m)=0$ otherwise. Since $A \subset \mathbb{N}$, this convention also gives $x_A(m)=0$ for $m \leq 0$. Let \begin{align*} X := \overline{\{\sigma^n x_A : n \in \mathbb{Z}\}} \subset \Omega, \end{align*} where the closure is taken in the product topology. Then $X$ is compact, metrizable, and $\sigma(X)=X$. Let \begin{align*} T := \sigma|_X: X \to X \end{align*} be the restricted shift homeomorphism, and let $\mathcal{B}:=\mathcal{B}(X)$ be the Borel $\sigma$-algebra of $X$. [/step] [step:Extract a weak limit of empirical orbit measures] For each $j \in \mathbb{N}$, define the Borel probability measure \begin{align*} \mu_j := \frac{1}{N_j}\sum_{n=1}^{N_j} \delta_{T^n x_A} \end{align*} on $(X,\mathcal{B})$, where $\delta_y$ denotes the Dirac probability measure at $y \in X$. Since $X$ is compact and metrizable, the space of Borel probability measures on $X$ is sequentially compact in the [weak topology](/page/Weak%20Topology) (citing a result not yet in the wiki: weak compactness of probability measures on compact metric spaces). Therefore there is a subsequence $(N_{j_k})_{k=1}^{\infty}$ and a Borel probability measure $\mu$ on $X$ such that \begin{align*} \mu_{j_k} \to \mu \end{align*} weakly, meaning that for every continuous function $\varphi: X \to \mathbb{R}$, \begin{align*} \lim_{k \to \infty} \int_X \varphi(y)\, d\mu_{j_k}(y) = \int_X \varphi(y)\, d\mu(y). \end{align*} Because $(N_{j_k})_{k=1}^{\infty}$ is a subsequence of the density-realising sequence, we still have \begin{align*} \lim_{k \to \infty} \frac{|A \cap \{1,\dots,N_{j_k}\}|}{N_{j_k}} = d. \end{align*} [guided] For each interval $\{1,\dots,N_j\}$, we put equal mass on the orbit points \begin{align*} T x_A,\; T^2x_A,\; \dots,\; T^{N_j}x_A. \end{align*} This produces the probability measure \begin{align*} \mu_j := \frac{1}{N_j}\sum_{n=1}^{N_j} \delta_{T^n x_A}. \end{align*} The compactness input is used only to guarantee a limiting probability measure. Since $X$ is compact and metrizable, the family of Borel probability measures on $X$ is weakly sequentially compact (citing a result not yet in the wiki: weak compactness of probability measures on compact metric spaces). Hence, after passing to a subsequence $(N_{j_k})_{k=1}^{\infty}$, there is a Borel probability measure $\mu$ on $X$ such that for every continuous function $\varphi: X \to \mathbb{R}$, \begin{align*} \lim_{k \to \infty} \int_X \varphi(y)\, d\mu_{j_k}(y) = \int_X \varphi(y)\, d\mu(y). \end{align*} Passing to this subsequence does not disturb the density limit, because every subsequence of a convergent sequence has the same limit: \begin{align*} \lim_{k \to \infty} \frac{|A \cap \{1,\dots,N_{j_k}\}|}{N_{j_k}} = d. \end{align*} [/guided] [/step] [step:Show that the limiting measure is shift-invariant] Let $\varphi: X \to \mathbb{R}$ be continuous. For each $j \in \mathbb{N}$, \begin{align*} \int_X \varphi(Ty)\, d\mu_j(y) &= \frac{1}{N_j}\sum_{n=1}^{N_j} \varphi(T^{n+1}x_A),\\ \int_X \varphi(y)\, d\mu_j(y) &= \frac{1}{N_j}\sum_{n=1}^{N_j} \varphi(T^n x_A). \end{align*} Subtracting the two finite sums gives \begin{align*} \left| \int_X \varphi(Ty)\, d\mu_j(y) - \int_X \varphi(y)\, d\mu_j(y) \right| = \frac{1}{N_j} \left| \varphi(T^{N_j+1}x_A)-\varphi(Tx_A) \right| \leq \frac{2\|\varphi\|_\infty}{N_j}. \end{align*} Letting $j=j_k$ and then $k \to \infty$, [weak convergence](/page/Weak%20Convergence) gives \begin{align*} \int_X \varphi(Ty)\, d\mu(y) = \int_X \varphi(y)\, d\mu(y). \end{align*} Since this holds for every continuous function $\varphi: X \to \mathbb{R}$, the measure $\mu$ is $T$-invariant, by the fact that continuous functions determine Borel probability measures on compact metric spaces (citing a result not yet in the wiki: uniqueness of Borel probability measures from integrals against continuous functions). Thus $(X,\mathcal{B},\mu,T)$ is an invertible measure-preserving system. [guided] The only possible failure of exact shift-invariance for the empirical measures comes from the two endpoints of the finite orbit segment. Let $\varphi: X \to \mathbb{R}$ be continuous. Because $\mu_j$ is the average of the Dirac masses at $T^n x_A$ for $1 \leq n \leq N_j$, integrating $\varphi \circ T$ against $\mu_j$ gives \begin{align*} \int_X \varphi(Ty)\, d\mu_j(y) = \frac{1}{N_j}\sum_{n=1}^{N_j} \varphi(T^{n+1}x_A). \end{align*} On the other hand, \begin{align*} \int_X \varphi(y)\, d\mu_j(y) = \frac{1}{N_j}\sum_{n=1}^{N_j} \varphi(T^n x_A). \end{align*} The two sums contain the same middle terms: \begin{align*} \varphi(T^2x_A),\dots,\varphi(T^{N_j}x_A). \end{align*} Only $\varphi(Tx_A)$ and $\varphi(T^{N_j+1}x_A)$ remain. Therefore \begin{align*} \left| \int_X \varphi(Ty)\, d\mu_j(y) - \int_X \varphi(y)\, d\mu_j(y) \right| = \frac{1}{N_j} \left| \varphi(T^{N_j+1}x_A)-\varphi(Tx_A) \right| \leq \frac{2\|\varphi\|_\infty}{N_j}. \end{align*} The right-hand side tends to $0$ along the subsequence $j=j_k$. Since $\varphi$ and $\varphi \circ T$ are continuous and $\mu_{j_k}$ converges weakly to $\mu$, we may pass to the limit and obtain \begin{align*} \int_X \varphi(Ty)\, d\mu(y) = \int_X \varphi(y)\, d\mu(y). \end{align*} This identity for all continuous test functions is exactly the weak formulation of $T$-invariance of $\mu$, using the fact that continuous functions determine Borel probability measures on compact metric spaces (citing a result not yet in the wiki: uniqueness of Borel probability measures from integrals against continuous functions). Since $T$ is a homeomorphism of $X$, the system is invertible as well as measure-preserving. [/guided] [/step] [step:Define the cylinder set representing membership in $A$] Define the coordinate-zero cylinder \begin{align*} B := \{x \in X : x(0)=1\}. \end{align*} The coordinate projection $x \mapsto x(0)$ is continuous on $X$, and $\{1\}$ is open and closed in $\{0,1\}$, so $B$ is a clopen subset of $X$. In particular, $B \in \mathcal{B}$ and its indicator function $\mathbb{1}_B: X \to \mathbb{R}$ is continuous. For every $n \in \mathbb{N}$, \begin{align*} T^n x_A \in B \iff (T^n x_A)(0)=1 \iff x_A(n)=1 \iff n \in A. \end{align*} Therefore \begin{align*} \mu_j(B) = \frac{1}{N_j}\sum_{n=1}^{N_j}\mathbb{1}_B(T^n x_A) = \frac{|A \cap \{1,\dots,N_j\}|}{N_j}. \end{align*} Passing to the subsequence and using weak convergence against the continuous function $\mathbb{1}_B$ gives \begin{align*} \mu(B) = \lim_{k \to \infty}\mu_{j_k}(B) = \lim_{k \to \infty}\frac{|A \cap \{1,\dots,N_{j_k}\}|}{N_{j_k}} = d. \end{align*} Thus $\mu(B)=\overline d(A)$. [/step] [step:Identify finite intersections of shifted cylinders with shifted intersections of $A$] Let $F \subset \mathbb{Z}$ be finite. Define the cylinder set \begin{align*} C_F := \bigcap_{h \in F} T^{-h}B. \end{align*} If $F=\varnothing$, then $C_F=X$. If $F \neq \varnothing$, then $C_F$ is a finite intersection of clopen sets, hence clopen, and $\mathbb{1}_{C_F}: X \to \mathbb{R}$ is continuous. For each $n \in \mathbb{N}$, \begin{align*} T^n x_A \in C_F &\iff T^n x_A \in T^{-h}B \quad \text{for every } h \in F\\ &\iff T^{n+h}x_A \in B \quad \text{for every } h \in F\\ &\iff x_A(n+h)=1 \quad \text{for every } h \in F\\ &\iff n+h \in A \quad \text{for every } h \in F\\ &\iff n \in \bigcap_{h \in F}(A-h). \end{align*} Consequently, \begin{align*} \mu_j(C_F) = \frac{1}{N_j} \left| \left(\bigcap_{h \in F}(A-h)\right)\cap \{1,\dots,N_j\} \right|. \end{align*} [guided] The cylinder $C_F$ asks that all coordinates indexed by $F$ are equal to $1$, after translating this condition into the dynamics. Explicitly, \begin{align*} C_F := \bigcap_{h \in F} T^{-h}B. \end{align*} For a point $x \in X$, membership in $T^{-h}B$ means $T^h x \in B$, which means the zeroth coordinate of $T^h x$ is $1$. Since $T$ is the left shift, this is the same as saying that the $h$-th coordinate of $x$ is $1$. Now evaluate this at the orbit point $T^n x_A$. For each $n \in \mathbb{N}$, \begin{align*} T^n x_A \in C_F &\iff T^{n+h}x_A \in B \quad \text{for every } h \in F\\ &\iff (T^{n+h}x_A)(0)=1 \quad \text{for every } h \in F\\ &\iff x_A(n+h)=1 \quad \text{for every } h \in F\\ &\iff n+h \in A \quad \text{for every } h \in F\\ &\iff n \in \bigcap_{h \in F}(A-h). \end{align*} This equivalence is the core of the correspondence principle: the measure of a cylinder in the dynamical system records the density of a finite intersection pattern in the original set $A$. Averaging the indicator of $C_F$ over the first $N_j$ orbit points gives \begin{align*} \mu_j(C_F) = \frac{1}{N_j}\sum_{n=1}^{N_j}\mathbb{1}_{C_F}(T^n x_A) = \frac{1}{N_j} \left| \left(\bigcap_{h \in F}(A-h)\right)\cap \{1,\dots,N_j\} \right|. \end{align*} [/guided] [/step] [step:Pass to the limit and obtain the correspondence inequality] Since $\mathbb{1}_{C_F}: X \to \mathbb{R}$ is continuous, weak convergence gives \begin{align*} \mu(C_F) = \lim_{k \to \infty} \mu_{j_k}(C_F). \end{align*} Using the counting identity from the previous step, \begin{align*} \mu(C_F) &= \lim_{k \to \infty} \frac{1}{N_{j_k}} \left| \left(\bigcap_{h \in F}(A-h)\right)\cap \{1,\dots,N_{j_k}\} \right|\\ &\leq \limsup_{N \to \infty} \frac{1}{N} \left| \left(\bigcap_{h \in F}(A-h)\right)\cap \{1,\dots,N\} \right|\\ &= \overline d\left(\bigcap_{h \in F}(A-h)\right). \end{align*} Because $C_F=\bigcap_{h \in F}T^{-h}B$, this is exactly \begin{align*} \mu\left(\bigcap_{h \in F}T^{-h}B\right) \leq \overline d\left(\bigcap_{h \in F}(A-h)\right). \end{align*} Together with $\mu(B)=\overline d(A)$ and the $T$-invariance of $\mu$, this proves the Furstenberg correspondence principle. [/step]